# Brilliant Integration Contest - Season 3 (Part 2)

Hi Brilliant! Due to many problems in part 1 it had become slow to load. So this is a sequel of Brilliant Integration - Season 3(Part 1)

The aims of the Integration contest are to improve skills in the computation of integrals, to learn from each other as much as possible, and of course to have fun. Anyone here may participate in this contest.

The rules are as follows:

• I will start by posting the first problem. If there is a user solves it, then they must post a new one.

• You may only post a solution of the problem below the thread of problem and post your proposed problem in a new thread. Put them separately.

• Only make substantial comment that will contribute to the discussion.

• Make sure you know how to solve your own problem before posting it in case there is no one can answer it within 48 hours, then you must post the solution and you have a right to post another problem.

• If the one who solves the last problem does not post his/her own problem after solving it within a day, then the one who has a right to post a problem is the last solver before him/her.

• The scope of questions is only computation of integrals either definite or indefinite integrals.

• You are NOT allowed to post a multiple integrals problem.

• It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

• Do not copy questions from last year's contest. If anyone found to do so he/she will be banned from taking further part in this contest

Format your post is as follows:

 1 2 3 4 5 6 7 **SOLUTION OF PROBLEM xxx (number of problem) :** **[Post your solution here]** **PROBLEM xxx (number of problem) :** **[Post your problem here]** 

1 year, 6 months ago

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Problem 26 :

Show that $\int_0^\infty \ln\left(\frac{a^s + x^s}{b^s + x^s}\right)\,dx \; =\; \pi(a-b)\mathrm{cosec}\tfrac{\pi}{s} \hspace{1cm} a,b > 0\,,\,s > 1 \;.$

This problem has been solved by Fdp Dpf.

- 1 year, 6 months ago

Solution problem 26:

Let , $\displaystyle F(a)=\int_0^{+\infty} \ln\left(\dfrac{a^s+x^s}{b^s+x^s}\right)dx$

Observe that, $F(b)=0$

$\displaystyle F^\prime(a)=sa^{s-1}\int_0^{+\infty} \dfrac{1}{a^s+x^s}dx$

Perform the change of variable,

$y=\dfrac{x}{a}$,

$\displaystyle F^\prime(a)=s\int_0^{+\infty} \dfrac{1}{1+x^s}dx$

It's well known that,

$\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{\pi}{s}\mathrm{cosec}\left(\dfrac{\pi}{s}\right)$

Therefore,

$\displaystyle F^\prime(a)=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)$

Therefore, $F(a)=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)a+k$ ,

k a real constant.

Since $F(b)=0$ then $k=-\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right)b$

Therefore,

$\boxed{\displaystyle F(a)=\pi (a-b)\mathrm{cosec}\left(\dfrac{\pi}{s}\right)}$

Proof of:

$\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{\pi}{s}\mathrm{cosec}\left(\dfrac{\pi}{s}\right)$

Perform the change of variable

$y=x^s$

$\displaystyle \int_0^{+\infty} \dfrac{1}{1+x^s}dx=\dfrac{1}{s} \int_0^{+\infty} \dfrac{x^{\tfrac{1}{s}-1}}{1+x}dx$

\displaystyle \begin{align}\int_0^{+\infty} \dfrac{x^{\tfrac{1}{s}-1}}{1+x}dx&=\beta\left(\dfrac{1}{s},1-\dfrac{1}{s}\right)\\ &=\dfrac{\Gamma\left(\dfrac{1}{s}\right)\Gamma\left(1-\dfrac{1}{s}\right)}{\Gamma(1)}\\ &= \dfrac{\pi}{\sin\left(\dfrac{\pi}{s}\right)}\\ &=\pi\mathrm{cosec}\left(\dfrac{\pi}{s}\right) \end{align}

$$\beta$$ being the Beta Euler function. Third line is the use of Euler's reflection formula.

- 1 year, 6 months ago

Problem 30:

Show that $\int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\,\frac{dx}{x} \; = \; \tfrac{1}{20}\pi^2$

This problem has been solved by Fdp Dpf.

- 1 year, 6 months ago

$\displaystyle I=\int_0^1 \dfrac{\ln(\sqrt{1+\dfrac{x^2}{4}}+\dfrac{x}{2})}{x}dx$

Perform the change of variable $$y=\dfrac{x}{2}$$,

$\displaystyle I=\int_0^{\tfrac{1}{2}} \dfrac{\ln\left(\sqrt{1+x^2}+x\right)}{x}dx$

Perform the change of variable $$y=\mathrm{arsinh}(x)$$,

\begin{align}\displaystyle I&=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \dfrac{x\cosh(x)}{\sinh(x)}dx\\ &=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \dfrac{x\left(1+\mathrm{e}^{-2x}\right)}{1-\mathrm{e}^{-2x}} dx\\ &=\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \left(x\left(1+\mathrm{e}^{-2x}\right)\sum_{n=0}^{+\infty}\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+2\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} \left(x\sum_{n=1}^{+\infty}\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+2\sum_{n=1}^{+\infty} \left(\int_0^{\mathrm{arsinh}\left(\tfrac{1}{2}\right)} x\mathrm{e}^{-2nx}\right)dx\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+\dfrac{1}{2}\sum_{n=1}^{+\infty}\left(\dfrac{1}{n^2}-\dfrac{\left(1+2\mathrm{arsinh}\left(\tfrac{1}{2}\right)n\right)\mathrm{e}^{-2\mathrm{arsinh}\left(\tfrac{1}{2}\right)n}}{n^2}\right)\\ &=\dfrac{\left(\mathrm{arsinh}\left(\tfrac{1}{2}\right)\right)^2}{2}+\dfrac{1}{2}\zeta(2)-\dfrac{1}{2}\mathrm{Li}_2\left(e^{-2\mathrm{asinh}\left(\tfrac{1}{2}\right)}\right)+\mathrm{arsinh}\left(\tfrac{1}{2}\right)\ln\left(1-e^{-2\mathrm{arsinh}\left(\tfrac{1}{2}\right)}\right) \end{align}

Since, $$\mathrm{arsinh}\left(\dfrac{1}{2}\right)=\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)$$ then,

$I=\dfrac{\pi^2}{12}-\dfrac{1}{2}\mathrm{Li}_2\left(\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)-\dfrac{1}{2}\left(\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)\right)^2$

Since , $$\mathrm{Li}_2\left(\dfrac{3}{2}-\dfrac{\sqrt{5}}{2}\right)=\dfrac{\pi^2}{15}-\left(\ln\left(\dfrac{1}{2}+\dfrac{\sqrt{5}}{2}\right)\right)^2$$ then,

$I=\dfrac{\pi^2}{12}-\dfrac{\pi^2}{30}=\boxed{\dfrac{\pi^2}{20}}$

- 1 year, 5 months ago

Nice. A bit of integration by parts makes the earlier stage a little clearer, perhaps...

Note that $\frac{d}{dx}\ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right) \; = \; \frac{\frac{\frac12x}{\frac12\sqrt{1+\frac14x^2}} + \frac12}{\sqrt{1 + \frac14x^2} + \frac12x} \; = \; \frac{1}{\sqrt{4+x^2}}$ and so, integrating by parts, \begin{align} \int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\frac{dx}{x} & = \Big[(\ln x)\ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\Big]_0^1 - \int_0^1 \frac{\ln x}{\sqrt{4 + x^2}}\,dx \\ & = -\int_0^1 \frac{\ln x}{\sqrt{4 + x^2}}\,dx \end{align} Since $$\sinh^{-1}\tfrac12 = \ln\big(\tfrac12(\sqrt{5}+1)\big)$$, the substitution $$x = 2\sinh u$$ yields \begin{align} \int_0^1 \frac{\ln x}{\sqrt{4+x^2}}\,dx & = \int_0^{\sinh^{-1}\frac12} \ln\big(2\sinh u\big)\,du \; = \; \int_0^{\sinh^{-1}\frac12}\Big[u + \ln\big(1 - e^{-2u}\big)\Big]\,du \\ & = \tfrac12 \ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) + \int_0^{\ln\big(\frac{\sqrt{5}+1}{2}\big)} \frac{\ln\big(1 - e^{-2u}\big)}{e^{-2u}}\,e^{-2u}\,du \\ & = \tfrac12\ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) - \frac12\int_1^{\frac{3-\sqrt{5}}{2}} \frac{\ln(1-v)}{v}\,dv \end{align} using the substitution $$v = e^{-2u}$$. Thus $\int_0^1 \frac{\ln x}{\sqrt{4+x^2}}\,dx \; = \; \tfrac12\ln^2\left(\tfrac{\sqrt{5}+1}{2}\right) + \tfrac12\mathrm{Li}_2\left(\tfrac{3 - \sqrt{5}}{2}\right) - \tfrac12\mathrm{Li}_2(1) \; = \; \tfrac{1}{30}\pi^2 - \tfrac{1}{12}\pi^2 \; = \; -\tfrac{1}{20}\pi^2$ and so $\int_0^1 \ln\left(\sqrt{1 + \tfrac14x^2} + \tfrac12x\right)\frac{dx}{x} \; = \; \tfrac{1}{20}\pi^2$

- 1 year, 5 months ago

You're right, it's easiest this way. By the way, $$\displaystyle \int \ln\left(\sqrt{1+\dfrac{1}{4}x^2}+\dfrac{1}{2}x\right)dx=x\ln\left(\sqrt{1+\dfrac{1}{4}x^2}+\dfrac{1}{2}x\right)-\sqrt{x^2+4}$$

- 1 year, 5 months ago

Problem 29:

Show that $\displaystyle \int\limits_{0}^{\infty} \frac{e^{-tx}}{\sqrt x (x^2+1)}dx=\pi \left[ \sqrt2 C\left( \sqrt{\frac{2t}{\pi}}\right)\sin t - \sqrt2S\left( \sqrt{\frac{2t}{\pi}} \right) \cos t - \sin(t-\pi/4) \right]$

where $$C$$ and $$S$$ are Fresnel cosine integral and Fresnel sine integral, respectively.

This problem has been solved by Mark Hennings.

- 1 year, 6 months ago

In this question, we are using the Fresnel integrals $C(x) \; = \; \int_0^x \cos\big(\tfrac12\pi t^2\big)\,dt \hspace{2cm} S(x) \; = \; \int_0^x \sin\big(\tfrac12\pi t^2\big)\,dt$ If we define $A(t) \; = \; \cos t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx - \sin t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx$ for $$t > 0$$ then \begin{align} A'(t) & = \begin{array}[t]{l}\displaystyle-\sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx - \cos t \int_0^\infty \frac{e^{-tx}x\sqrt{x}}{x^2+1}\,dx \\\displaystyle - \cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx + \sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx\end{array} \\ & = -\cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}}\,dx \; = \; -\sqrt{\tfrac{\pi}{t}}\cos t \end{align} for any $$t > 0$$, so that $\frac{d}{dt}\Big[ A(t) + \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\Big] \; = \; 0$ Since $$A(t) \,\to\, 0$$ as $$t \to \infty$$, we deduce that $A(t) + \pi\sqrt{2}C\Big(\sqrt{\frac{2t}{\pi}}\Big) \; = \; \tfrac{1}{\sqrt{2}}\pi \hspace{2cm} t > 0$ Similarly, if we define $B(t) \; = \; \sin t \int_0^\infty \frac{e^{-tx}\sqrt{x}}{x^2+1}\,dx + \cos t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx$ for $$t > 0$$, then we can show that $B'(t) \; =\; -\sin t \int_0^\infty \frac{e^{-tx}}{\sqrt{x}}\,dx \; =\; - \sqrt{\tfrac{\pi}{t}}\sin t$ and hence $B(t) + \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big) \; = \; \tfrac{1}{\sqrt{2}}\pi \hspace{2cm} t > 0$ Thus we deduce that \begin{align} \int_0^\infty \frac{e^{-tx}}{\sqrt{x}(x^2+1)}\,dx & = B(t)\cos t - A(t) \sin t \\ & = \cos t\left[ \tfrac{1}{\sqrt{2}}\pi - \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\right] - \sin t\left[ \tfrac{1}{\sqrt{2}}\pi - \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big)\right] \\ & = \pi\sqrt{2}C\Big(\sqrt{\tfrac{2t}{\pi}}\Big) - \pi\sqrt{2}S\Big(\sqrt{\tfrac{2t}{\pi}}\Big) - \pi\sin\big(t - \tfrac14\pi\big) \end{align} as required.

- 1 year, 6 months ago

Problem 27:

Show that

$\displaystyle\int_0^{\tfrac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \dfrac{\ln x}{(x-1)\sqrt{x^2-2(15+8\sqrt{3})x+1}}\, dx=\dfrac{2}{3}(2-\sqrt{3})G$

This problem has been solved by Mark Hennings.

- 1 year, 6 months ago

Put $$\alpha = \tfrac{1}{12}\pi$$. With the substitution $$x = \frac{1-\cos\theta}{1+\cos\theta}$$, we obtain after much manipulation that $I \; = \; \int_0^{\frac{\sqrt{6}-\sqrt{2}-1}{\sqrt{6}-\sqrt{2}+1}} \frac{\ln x\,dx}{(x-1)\sqrt{x^2 - 2(15+8\sqrt{3})x + 1)}} \; = \; -\tfrac12\sin\alpha \int_0^\alpha \frac{\ln\left(\frac{1-\cos\theta}{1+\cos\theta}\right) \sin\theta\,d\theta}{\cos\theta\sqrt{\cos^2\theta - \cos^2\alpha}}$ Putting $$u = \cos\theta$$ now gives $I \; = \; -\tfrac12\sin\alpha \int_{\cos\alpha}^1 \frac{\ln\left(\frac{1-u}{1+u}\right)\,du}{u\sqrt{u^2 - \cos^2\alpha}}$ Now putting $$u = \cos\alpha \sec\phi$$ yields \begin{align} I & = -\tfrac12\tan\alpha \int_0^\alpha \ln\left(\frac{\cos\phi - \cos\alpha}{\cos\phi + \cos\alpha}\right)\,d\phi \; = \; -\tfrac12\tan\alpha \int_0^\alpha \ln\left(\tan\big(\tfrac{\alpha+\phi}{2}\big) \tan\big(\tfrac{\alpha-\phi}{2}\big)\right)\,d\phi \\ & = -\tfrac12\tan\alpha \int_0^\alpha \left\{ \ln\left(\tan\big(\tfrac{\alpha+\phi}{2}\big)\right) + \ln\left(\tan\big(\tfrac{\alpha-\phi}{2}\big)\right)\right\}\,d\phi \\ & = -\tan\alpha \left\{ \int_{\frac12\alpha}^\alpha \ln(\tan \phi)\,d\phi - \int_{\frac12\alpha}^0 \ln(\tan\phi)\,d\phi\right\} \\ & = -\tan\alpha \int_0^\alpha \ln(\tan\phi)\,d\phi \; = \; -\tan\alpha \int_0^{\frac{1}{12}\pi}\ln(\tan\phi)\,d\phi \\ & = \tfrac23\tan\alpha\, G \; = \; \tfrac23(2 - \sqrt{3})G \end{align} using a standard integral representation of $$G$$ at the very last stage.

- 1 year, 6 months ago

Problem 25 :

Prove That

$\dfrac{1}{\pi} \int_{0}^{\pi} x \arctan\left( \dfrac{p \sin x}{1+p \cos x} \right) \ \mathrm{d}x = \operatorname{Li}_{2}(p) \quad \forall \ |p| <1$

Notation : $$\operatorname{Li}_{2}(z)$$ denotes the Dilogarithm Function.

This problem has been solved by Mark Hennings.

- 1 year, 6 months ago

If we define $g(p) \; = \; \int_0^\pi \ln(1 + 2p \cos x + p^2)\,dx \hspace{1cm} |p| < 1$ then $$g(0) = 0$$ and, using the complex subsitution $$z = e^{ix}$$, \begin{align} g'(p) & = \int_0^\pi \frac{2\cos x + 2p}{1 + 2p\cos x + p^2}\,dx \; = \; \frac12\int_{-\pi}^\pi \frac{2\cos x + 2p}{1 + 2p\cos x + p^2}\,dx \\ & = \frac{1}{2}\int_{|z|=1} \frac{z + z^{-1} + 2p}{1 + p(z + z^{-1}) + p^2} \frac{dz}{iz} \; = \; \frac{1}{2i}\int_{|z|=1} \frac{z^2 + 2pz + 1}{z(pz+1)(z+p)}\,dz \\ & = \pi\left(\mathrm{Res}_{z=0} + \mathrm{Res}_{z=-p}\right)\frac{z^2 + 2pz + 1}{z(pz+1)(z+p)} \\ & = \pi\left(\frac{1}{p} - \frac{1-p^2}{p(1-p^2)}\right) \; = \; 0 \end{align} so that $$g(p) = 0$$ for all $$|p| < 1$$. If we now define $f(p) \; = \; \frac{1}{\pi}\int_0^\pi x \tan^{-1}\left(\frac{p \sin x}{1 + p \cos x}\right)\,dx \hspace{1cm} |p| < 1$ then $$f(0) = 0$$ and \begin{align} f'(p) & = \frac{1}{\pi}\int_0^\pi \frac{x}{1 + \left(\frac{p \sin x}{1 + p\cos x}\right)^2} \times \frac{\sin x}{(1 + p\cos x)^2}\,dx \\ & = \frac{1}{\pi}\int_0^\pi \frac{x \sin x}{1 + 2p\cos x + p^2}\,dx \\ & = \frac{1}{\pi}\Big[-\frac{x}{2p}\ln(1 + 2p\cos x + p^2)\Big]_0^\pi + \frac{1}{2\pi p}\int_0^\pi \ln(1 + 2p\cos x + p^2)\,dx \\ & = -\frac{\ln(1-p)}{p} + g(p) \; = \; -\frac{\ln(1-p)}{p} \end{align} and hence it follows that $$f(p) \,=\, \mathrm{Li}_2(p)$$.

- 1 year, 6 months ago

Aliter,

Let $f(p) = \dfrac{1}{\pi}\int_{0}^{\pi} x \arctan\left( \dfrac{p\sin x}{1+ p\cos x} \right) \ \mathrm{d}x$

Firstly, we have $$f(0) = 0$$

Note that,

$$\displaystyle \sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx) = \dfrac{1}{p}\Im \left( \sum_{k=0}^{\infty} (-p e^{ix})^{k} \right)$$

$$\displaystyle = \dfrac{1}{p} \Im \left( \dfrac{1}{1+p e^{ix}} \right)$$

$$\displaystyle = \dfrac{\sin x}{p^2 +2p \cos x +1}$$

where $$\Im(z)$$ denotes the imaginary part of $$z$$.

So we have,

$$\displaystyle f'(p) = \dfrac{1}{\pi} \int_{0}^{\pi} \dfrac{x \sin x}{p^2 +2p \cos x +1} \ \mathrm{d}x$$

$$\displaystyle = \dfrac{1}{\pi} \int_{0}^{\pi} x \left(\sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx)\right) \ \mathrm{d}x$$

$$\displaystyle = \dfrac{1}{\pi} \sum_{k=1}^{\infty} (-p)^{k-1} \int_{0}^{\pi}x \sin (kx) \ \mathrm{d}x$$

$$\displaystyle = \dfrac{1}{\pi} \sum_{k=1}^{\infty} (-p)^{k-1} \left( \dfrac{\sin(k \pi) - k \pi \cos(k \pi)}{k^2} \right)$$

Since $$\sin (k \pi) = 0$$ and $$\cos (k \pi) = (-1)^k \ \forall \ k \in \mathbb{Z}$$, we get,

$$\displaystyle f'(p) = \sum_{k=1}^{\infty} \dfrac{p^{k-1}}{k} = -\dfrac{\ln(1-p)}{p}$$

$\displaystyle \therefore f(p) = \operatorname{Li}_{2}(p) \ \square$

- 1 year, 6 months ago

Problem 36:

Show that,

$\int_0^{+\infty}\dfrac{x\ln(1+\sqrt{2}x+x^2)}{1+x^4}dx=\dfrac{\pi\ln 2}{2}$

This problem has been solved by Mark Hennings.

- 1 year, 5 months ago

Alternative solution to problem 36

Let,

$I=\int_0^{+\infty} \dfrac{x(\ln(1+\sqrt{2}x+x^2)}{1+x^4}dx$

$J=\int_0^{+\infty} \dfrac{x(\ln(1-\sqrt{2}x+x^2)}{1+x^4}dx$

$I+J=\int_0^{+\infty} \dfrac{x\ln(1+x^4)}{1+x^4}dx$

Perform the change of variable $$y=x^2$$,

$I+J=\dfrac{1}{2}\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx$

Perform the change of variable $$y=\arctan x$$,

$I+J=-\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx$

Perform the change of variable $$y=\dfrac{\pi}{2}-x$$,

$\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx=\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx$

Therefore,

\begin{align} 2(I+J)&=-\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx-\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx\\ &=-\int_0^{\tfrac{\pi}{2}} \ln(\cos(x)\sin(x) )dx\\ &=-\int_0^{\tfrac{\pi}{2}} \ln\left(\dfrac{\sin(2x)}{2} \right)dx\\ &=-\int_0^{\tfrac{\pi}{2}} \ln\left(\sin(2x) \right)dx+\dfrac{\pi\ln 2}{2}\\ \end{align}

In the latter integral perform the change of variable $$y=2x$$,

\begin{align} 2(I+J)&=-\dfrac{1}{2}\int_0^{\pi} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ &=-\dfrac{1}{2}\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx-\dfrac{1}{2}\int_{\tfrac{\pi}{2}}^{\pi} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ \end{align}

In the latter integral perform the change of variable $$y=\pi-x$$, therefore,

\begin{align} 2(I+J)&=-\dfrac{1}{2}\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx-\dfrac{1}{2}\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ &=-\int_0^{\tfrac{\pi}{2}} \ln(\sin x)dx+\dfrac{\pi\ln 2}{2}\\ &=-\int_0^{\tfrac{\pi}{2}} \ln(\cos x)dx+\dfrac{\pi\ln 2}{2}\\ &=I+J+\dfrac{\pi\ln 2}{2}\\ \end{align}

Therefore,

$\boxed{I+J=\dfrac{\pi\ln 2}{2}}$

Define on $$\Big[0;\sqrt{2}\Big]$$,

$\displaystyle F(a)= \int_0^{+\infty} \dfrac{x(\ln(1+ax+x^2)-\ln(1-ax+x^2))}{1+x^4}dx$

Observe that, $$F(\sqrt{2})=I-J$$ and $$F(0)=0$$,

$F'(a)= \int_0^{+\infty} \dfrac{2x(1+x^2)}{(x^2-ax+1)(x^2+ax+1)(x^4+1)}dx$

$F'(a)=\left[\dfrac{2 \left( \mathrm{arctan}\left( \dfrac{2x-a}{\sqrt{4-{{a}^{2}}}}\right) +\mathrm{arctan}\left( \dfrac{2x+a}{\sqrt{4-{{a}^{2}}}}\right) \right) }{\sqrt{4-{{a}^{2}}}\cdot \left( {{a}^{2}}-2\right) }-\dfrac{\sqrt{2} \left( \mathrm{arctan}\left( \sqrt{2}x-1\right) +\mathrm{arctan}\left( \sqrt{2}x+1\right) \right) }{{{a}^{2}}-2}\right]_0^{+\infty}$

Therefore,

$F'(a)=\dfrac{2\pi}{(a^2-2)\sqrt{4-a^2}}-\dfrac{\sqrt{2}\pi}{a^2-2}$

$F(\sqrt{2})=\int_0^{\sqrt{2}} F'(a)da=\dfrac{\pi}{2}\left[\mathrm{ln}\left( \dfrac{\left( \sqrt{2}+a\right)\left( \sqrt{4-{{a}^{2}}}-a\right) }{\left( \sqrt{2}-a\right)\left( \sqrt{4-{{a}^{2}}}+a\right) }\right)\right]_0^\sqrt{2}=\dfrac{\pi\ln 2}{2}$

Therefore,

$I+J=I-J=\dfrac{\pi\ln 2}{2}$

and,

$\boxed{I=\dfrac{\pi\ln 2}{2}}$ $J=0$

- 1 year, 5 months ago

You could speed up your calculation of $$I+J$$ by doing a Beta function trick. It is equal to $-\tfrac14 \frac{d}{da} B(\tfrac12,a-\tfrac12)\; \Big|_{a=1}$

- 1 year, 5 months ago

I came up with a similar solution just now, and was about to post, but you beat me to it by 4 hours!

- 1 year, 5 months ago

This one is a classic one ;)

- 1 year, 5 months ago

The substitutions $$t = 1-s$$ and $$s = \sin\theta$$ give $\int_0^1 \left(\frac{1}{\sqrt{t(2-t)}}-1\right)\,\frac{dt}{1-t} \; = \; \int_0^1 \left(\frac{1}{\sqrt{1-s^2}} - 1\right)\,\frac{ds}{s} \; = \; \int_0^{\frac12\pi}\tan\tfrac12\theta\,d\theta \; = \; \ln2$ The substitution $$x^2 = \tfrac{t}{2-t}$$ and one of the standard integral representations of the Catalan constant $$G$$ gives $\int_0^1 \frac{1}{\sqrt{t(2-t)}} \left(\tfrac14\pi - \tan^{-1}\sqrt{\tfrac{t}{2-t}}\right)\,\frac{dt}{1-t} \; = \; 2\int_0^1 \left(\tfrac14\pi - \tan^{-1}x\right)\,\frac{dx}{1-x^2} \; = \; G$ These results together give us that \begin{align} \int_0^\infty \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1}\,dx & = \int_0^\infty \frac{dx}{x^2 + \sqrt{2}x + 1}\int_0^1 \frac{x^2 + \sqrt{2}x}{tx^2 + t\sqrt{2}x + 1}\,dt \\ & = \int_0^1 \int_0^\infty \frac{x^2 + \sqrt{2}x}{(tx^2 + t\sqrt{2}x + 1)(x^2 + \sqrt{2}x + 1)}\,dx\,dt \\ & = \int_0^1 \frac{dt}{1-t}\int_0^\infty\left(\frac{1}{tx^2 + t\sqrt{2}x + 1} - \frac{1}{x^2 + \sqrt{2}x + 1}\right)\,dx \\ & = \int_0^1 \frac{dt}{1-t} \left\{ \sqrt{\tfrac{2}{t(2-t)}}\left(\tfrac12\pi - \tan^{-1}\sqrt{\tfrac{t}{2-t}}\right) - \frac{\pi}{2\sqrt{2}}\right\} \\ & = \int_0^1 \frac{dt}{1-t}\left\{ \sqrt{\tfrac{2}{t(2-t)}}\left(\tfrac14\pi - \tan^{-1}\sqrt{\tfrac{t}{2-t}}\right) + \tfrac{\pi}{2\sqrt{2}}\left(\frac{1}{\sqrt{t(2-t)}} - 1\right)\,\right\} \end{align} and so $\int_0^\infty \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1}\,dx \; = \; = \sqrt{2}G + \tfrac{\pi}{2\sqrt{2}}\ln2$ Standard substitutions show that $f(a) \; = \; \int_0^\infty \frac{x^2+1}{(x^4 + 1)^a}\,dx \; = \; \tfrac14\big[B(\tfrac34,a-\tfrac34)+B(\tfrac14,a-\tfrac14)\big]$ for $$a > \tfrac34$$, and hence $f'(a) \; = \; \frac{\Gamma(\frac34)\Gamma(a-\frac34)\psi(a-\frac34) + \Gamma(\tfrac14)\Gamma(a-\frac14)\psi(a-\frac14)}{4\Gamma(a)} - \frac{\Gamma(\frac34)\Gamma(a-\frac34) + \Gamma(\frac14)\Gamma(a-\frac14)}{4\Gamma(a)}\psi(a)$ and hence $\int_0^\infty \frac{(x^2+1)\ln(x^4+1)}{x^4+1}\,dx \; = \; -f'(1) \; = \; -\tfrac{\pi}{2\sqrt{2}}\big[\psi(\tfrac14) + \psi(\tfrac34) - 2\psi(1)\big] \; = \; \tfrac{3\pi}{\sqrt{2}}\ln2$ Next, the substitution $$x = \tan\tfrac12\theta$$ and another standard integral representation of the Catalan constant gives \begin{align} \int_0^\infty &\frac{1+x^2}{1+x^4}\ln\left(\frac{x^2 + \sqrt{2}x + 1}{x^2 - \sqrt{2}x + 1}\right)\,dx \\ & = \int_0^\pi \ln\left(\frac{1 + \frac{1}{\sqrt{2}}\sin\theta}{1 - \frac{1}{\sqrt{2}}\sin\theta}\right) \frac{d\theta}{1 + \cos^2\theta} \\ & = 2\int_0^{\frac12\pi} \ln\left(\frac{1 + \frac{1}{\sqrt{2}}\sin\theta}{1 - \frac{1}{\sqrt{2}}\sin\theta}\right) \frac{d\theta}{1 + \cos^2\theta} \; = \; 2\sqrt{2}G \end{align} Thus \begin{align} \int_0^\infty &\left(\frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1} + \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 - \sqrt{2}x + 1}\right)\,dx \\ & = 2\int_0^\infty \frac{x^2+1}{x^4+1} \ln(x^2+ \sqrt{2}x+1)\,dx \\ & = \int_0^\infty \frac{x^2+1}{x^4+1}\left[\ln\left(\frac{x^2 +\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right) + \ln(x^4+1)\right]\,dx \\ & = 2\sqrt{2}G + \tfrac{3\pi}{\sqrt{2}}\ln2 \end{align} and hence $\int_0^\infty \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 - \sqrt{2}x + 1}\,dx \; = \; \big( 2\sqrt{2}G + \tfrac{3\pi}{\sqrt{2}}\ln2 \big) - \big( \sqrt{2}G + \tfrac{\pi}{2\sqrt{2}}\ln2\big) \; = \; \sqrt{2}G + \tfrac{5\pi}{2\sqrt{2}}\ln2$ Finally (!) we deduce that \begin{align} \int_0^\infty \frac{x}{x^4+1}\ln(x^2 + \sqrt{2}x + 1)\,dx & = \frac{1}{2\sqrt{2}}\int_0^\infty\left(\frac{\ln(x^2 + \sqrt{2}x+1)}{x^2 - \sqrt{2}x + 1} - \frac{\ln(x^2 + \sqrt{2}x + 1)}{x^2 + \sqrt{2}x + 1}\right) \\ & = \tfrac12\pi \ln2 \end{align}

- 1 year, 5 months ago

Problem 31:

Show that

$\displaystyle \int_0^1 \dfrac{x-1}{(x+1)\ln x}dx=\ln\left(\dfrac{\pi}{2}\right)$

This problem has been solved by Ishan Singh.

- 1 year, 5 months ago

Let

$f(p) = \int_{0}^{1} \dfrac{x^p - 1}{(x+1) \ln x} \ \mathrm{d}x \quad ; \quad p > 0$

$$\displaystyle \implies f'(p) = \int_{0}^{1} \dfrac{x^p}{x+1} \mathrm{d}x$$

$$\displaystyle = \int_{0}^{1} \dfrac{x^p(x-1)}{x^2 -1} \mathrm{d}x$$

Substitute $$x^2 \mapsto x$$ and simplify to get,

$$\displaystyle f'(p) = \dfrac{1}{2} \int_{0}^{1} \left( \dfrac{x^{\frac{p}{2}} - 1}{x - 1} - \dfrac{x^{\frac{p-1}{2}} - 1}{x - 1} \right) \mathrm{d}x$$

$$\displaystyle = \dfrac{1}{2} \left( \psi \left( \dfrac{p}{2} + 1 \right) - \psi \left( \dfrac{p-1}{2} + 1 \right) \right)$$

Note that $$f(0) = 0$$

$$\displaystyle \implies f(1) = \int_{0}^{1} f'(p) \ \mathrm{d}p$$

$$\displaystyle = \dfrac{1}{2} \int_{0}^{1} \left( \psi \left( \dfrac{p}{2} + 1 \right) - \psi \left( \dfrac{p-1}{2} + 1 \right) \right) \mathrm{d}p$$

$$\displaystyle = \log \left(\dfrac{\pi}{2}\right)$$

where the last equality follows since $$\psi (z) = \dfrac{\mathrm{d}}{\mathrm{d}z} \log(\Gamma(z))$$

- 1 year, 5 months ago

Problem 28:

Show that,

$\int_0^1 \frac{\arctan\left(\frac{x(1-x)}{2-x}\right)}{x}\, dx=\frac{G}{3}$

$$G$$ being the Catalan constant.

This problem has been solved by Mark Hennings.

- 1 year, 6 months ago

Your derivation of this result on MSE is pretty compact, so I shall expand it. Suppose that $$a > 1$$, and let $$b = \frac{1+a^2}{a-1}$$. The substitution $$y \,=\, \frac{ax}{x+b}$$ gives $\int_0^1 \tan^{-1}\left(\frac{ax}{x+b}\right)\,\frac{dx}{x} \; = \; \int_0^{\frac{a-1}{a+1}} \tan^{-1}y\, \frac{a-y}{by} \, \frac{ab}{(a-y)^2}\,dy \; = \; \int_0^{\frac{a-1}{a+1}} \frac{a}{y(a-y)} \tan^{-1}y\,dy$ while the substitution $$z = \frac{x}{b - ax}$$ gives $\int_0^1 \tan^{-1}\left(\frac{x}{ax - b}\right)\,\frac{dx}{x} \; = \; -\int_0^{\frac{a-1}{a+1}} \tan^{-1}z \, \frac{az+a}{bz} \, \frac{b}{(az+1)^2}\,dz \; =\; -\int_0^{\frac{a-1}{a+1}} \frac{1}{z(az+1)} \tan^{-1}z\,dz$ and hence \begin{align} \int_0^1 \left\{\tan^{-1}\left(\frac{ax}{ax+b}\right) + \tan^{-1}\left(\frac{x}{ax-b}\right)\right\}\,\frac{dx}{x} & = \int_0^{\frac{a-1}{a+1}} \left(\frac{a}{y(a-y)} - \frac{1}{y(ay+1)}\right) \tan^{-1}y\,dy \\ & = \int_0^{\frac{a-1}{a+1}} \frac{a^2+1}{(a-y)(ay+1)}\,\tan^{-1}y\,dy \\ & = \int_)^{\frac{a-1}{a+1}} \left(\frac{1}{a-y} + \frac{a}{1 + ay}\right)\tan^{-1}y\,dy \\ & = \left[ \ln\left(\frac{1 + ay}{a-y}\right) \tan^{-1}y\right]_0^{\frac{a-1}{a+1}} + \int_0^{\frac{a-1}{a+1}} \frac{\ln\left(\frac{a-y}{1+ay}\right)}{1+y^2}\,dy \\ & = \int_0^{\frac{a-1}{a+1}} \frac{\ln\left(\frac{a-y}{1+ay}\right)}{1+y^2}\,dy \; = \; \int_1^a \frac{\ln x}{1+x^2}\,dx \end{align} Now \begin{align} \tan\left[\tan^{-1}\left(\frac{ax}{x+b}\right) + \tan^{-1}\left(\frac{x}{ax-b}\right)\right] & = \frac{\frac{ax}{x+b} + \frac{x}{ax-b}}{1 - \frac{ax}{x+b}\frac{x}{ax-b}} \; = \; \frac{ax(ax-b) + x(x+b)}{(x+b)(ax-b) - ax^2} \\ & = \frac{x(1-x)}{\frac{1+a^2}{(a-1)^2} - x} \end{align} and hence it follows that $\int_0^1 \tan^{-1}\left(\frac{x(1-x)}{\frac{1+a^2}{(a-1)^2} - x}\right)\,\frac{dx}{x} \; = \; \int_1^a \frac{\ln x}{1 + x^2}\,dx$ Putting $$a = 2 + \sqrt{3}$$ yields \begin{align} \int_0^1 \tan^{-1}\left(\frac{x(1-x)}{2-x}\right)\,\frac{dx}{x} & = \int_1^{2+\sqrt{3}}\frac{\ln x}{1+x^2}\,dx \; = \; \int_{\frac14\pi}^{\frac{5}{12}\pi} \ln(\tan\theta)\,d\theta \\ & = -\int_{\frac{1}{12}\pi}^{\frac14\pi} \ln(\tan\phi)\,d\phi \; = \; -\big((-G) - (-\tfrac23G)\big) \; = \; \tfrac13G \end{align} using the final substitution $$\phi = \tfrac12\pi - \theta$$.

I won't be posting the next problem until after Christmas!

- 1 year, 6 months ago

Nice. I was secretly expecting a different proof. Someone, in 2011, showed me this integral. I have tried to find out the very source for it. Since then, i enjoy to learn and sharing knowledge about computing integrals. Happy chrismas !

- 1 year, 6 months ago

PROBLEM 39:

Evaluate $\int_0^\infty\left(e^{-x^a} - \frac{1}{1+x^b}\right)\,\frac{dx}{x}$ for any $$a,b > 0$$.

- 1 year, 5 months ago

Solution to problem 39.

$J=\int_0^{+\infty} \left( \text{e}^{-x^a}-\dfrac{1}{1+x^b}\right)\dfrac{1}{x}dx$

Perform the change of variable $$y=x^a$$,

$J=\dfrac{1}{a}\int_0^{+\infty} \left( \text{e}^{-x}-\dfrac{1}{1+x^{\tfrac{a}{b}}}\right)\dfrac{1}{x}dx$

Perform integration by parts,

\begin{align}aJ&=\left[ \left( \text{e}^{-x}-\dfrac{1}{1+x^{\tfrac{a}{b}}}\right)\ln x\right]_0^{+\infty}+\int_0^{+\infty}\left( \text{e}^{-x}-\dfrac{a}{b}\dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\right)\ln x dx\\ &=\int_0^{+\infty} \text{e}^{-x}\ln x dx-\dfrac{a}{b} \int_0^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx \end{align}

\begin{align} \int_0^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx&=\int_0^{1} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx+\int_1^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx \end{align}

In the latter integral perform the change of variable $$y=\dfrac{1}{x}$$,

\begin{align} \int_0^{+\infty} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx&=\int_0^{1} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx-\int_0^{1} \dfrac{x^{\tfrac{a}{b}-1}}{\left(1+x^{\tfrac{a}{b}}\right)^2}\ln x dx\\ &=0 \end{align}

$\int_0^{+\infty} \text{e}^{-x}\ln x dx=\Gamma^\prime (1)=-\gamma$

Therefore,

$\boxed{J=-\dfrac{1}{a}\gamma}$

- 1 year, 5 months ago

Problem 37:

Evaluate $\int_0^{\frac12\pi}\, \ln(\cos x + \sin x)\,dx$

This problem has been solved by Fdp Dpf.

- 1 year, 5 months ago

Solution to problem 37.

Since,

\begin{align}\sin\left(x+\dfrac{\pi}{4}\right)&=\cos\left(\dfrac{\pi}{4}\right)\sin x+\sin \left(\dfrac{\pi}{4}\right)\cos x\\ &=\dfrac{\sqrt{2}}{2}(\sin x+\cos x)\end{align}

then,

\begin{align} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\int_0^{\tfrac{\pi}{2}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx\\ &=\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx+\int_{\tfrac{\pi}{4}}^{\tfrac{\pi}{2}} \ln\left(\sqrt{2}\sin\left(x+\dfrac{\pi}{4}\right)\right)dx\\ &=\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(\left(\dfrac{\pi}{4}-x\right)+\dfrac{\pi}{4}\right)\right)dx+\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\sin\left(\left(x+\dfrac{\pi}{4}\right)+\dfrac{\pi}{4}\right)\right)dx\\ &=2\int_0^{\tfrac{\pi}{4}} \ln\left(\sqrt{2}\cos x\right)dx\\ &=\dfrac{\pi\ln 2}{4}+2\int_0^{\tfrac{\pi}{4}} \ln\left(\cos x\right)dx\\ &=\dfrac{\pi\ln 2}{4}-\int_0^{\tfrac{\pi}{4}} \ln\left(\dfrac{1}{(\cos x)^2}\right)dx\\ &=\dfrac{\pi\ln 2}{4}-\int_0^{\tfrac{\pi}{4}} \ln\left(1+(\tan x)^2\right)dx\\ \end{align}

In the latter integral perform the change of variable $$y=\tan x$$, therefore, \begin{align} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\dfrac{\pi\ln 2}{4}-\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx \end{align}

$\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx=\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx$

In the latter integral perform the change of variable $$y=\dfrac{1}{x}$$, therefore,

\begin{align}\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx&=\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+\int_0^1 \dfrac{\ln\left(1+\dfrac{1}{x^2}\right)}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx-2\int_0^1 \dfrac{\ln x}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx+2\text{G}\\ \end{align}

$$\text{G}$$ being the Catalan constant. But, $\int_0^{+\infty} \dfrac{\ln(1+x^2)}{1+x^2}dx=\pi\ln 2$

(see my answer to problem 36)

Therefore, $\int_0^1 \dfrac{\ln(1+x^2)}{1+x^2}dx=\dfrac{\pi\ln 2}{2}-\text{G}$

Therefore,

\begin{align} \int_0^{\tfrac{\pi}{2}} \ln(\cos x+\sin x)dx&=\dfrac{\pi\ln 2}{4}-\left(\dfrac{\pi\ln 2}{2}-\text{G}\right)\\ &=\boxed{\text{G}-\dfrac{\pi\ln 2}{4}} \end{align}

- 1 year, 5 months ago

Aliter,

The integral $$\displaystyle \int_{0}^{\frac{\pi}{4}} \log(\cos x) \ \mathrm{d}x$$ can also be done using the identity $\log(\cos x) = -\log 2 - \sum_{k=1}^{\infty} \dfrac{\cos (2kx)}{k}$.

- 1 year, 5 months ago

I know but i don't like to use this identity when i can avoid to use it. I like solutions using elementary tools.

- 1 year, 5 months ago

I like solutions using elementary methods too. I consider the above identity elementary, it can be proved by writing $$\cos (2kx)$$ as $$\dfrac{e^{2ikx} + e^{-2ikx}}{2}$$ and then using the Taylor Series of logarithm. Btw, Happy New Year.

- 1 year, 5 months ago

Problem 35:

Show that

$\int_0^1 \frac{\ln x}{x^2 + x + 1}\,dx \; = \; \tfrac29\left[\tfrac23\pi^2 - \psi'(\tfrac13)\right]$

This problem has been solved by Fdp Dpf.

- 1 year, 5 months ago

\begin{align}\int_0^1 \dfrac{\ln x}{1+x+x^2}dx&=\int_0^1 \dfrac{1-x}{1-x^3}\ln x dx\\ &=\int_0^1 \dfrac{1}{1-x^3}\ln x dx-\int_0^1 \dfrac{x}{1-x^3}\ln x dx\\ &=\sum_{n=0}^{+\infty}\left(\int_0^1 x^{3n}\ln x\right)-\sum_{n=0}^{+\infty}\left(\int_0^1 x^{3n+1}\ln x\right)\\ &=-\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}+\sum_{n=0}^{+\infty} \dfrac{1}{(3n+2)^2}\\ &=-\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}+\zeta(2)-\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}-\sum_{n=1}^{+\infty} \dfrac{1}{(3n)^2}\\ &=-2\sum_{n=0}^{+\infty} \dfrac{1}{(3n+1)^2}+\dfrac{8}{9}\zeta(2)\\ &=\dfrac{4}{27}\pi^2 -\dfrac{2}{9}\sum_{n=0}^{+\infty} \dfrac{1}{\left(n+\tfrac{1}{3}\right)^2}\\ &=\boxed{\dfrac{4}{27}\pi^2 -\dfrac{2}{9}\psi^\prime \left(\dfrac{1}{3}\right)} \end{align} Since $$\displaystyle \psi^\prime \left(a\right)=\sum_{n=0}^{+\infty} \dfrac{1}{(n+a)^2}$$

- 1 year, 5 months ago

Problem 32 : Prove That

$\int_{0}^{\infty} \left( \dfrac{1-x^2}{1+x^2} \right) \cdot \ln (x) \cdot \arctan \left( \dfrac{x}{x^2 + 1} \right) \ \dfrac{\mathrm{d}x}{x} = - \dfrac{\pi^3}{20}$

This problem has been solved by Mark Hennings.

- 1 year, 5 months ago

Suppose that $$u > 0$$. Integrating by parts, we see note that \begin{align} \int_0^\infty \frac{t \,\mathrm{sech}\, t \tanh t}{1 + \frac14u^2\,\mathrm{sech}^2t}\,dt & = \Big[-\frac{2t}{u}\tan^{-1}(\tfrac12u \,\mathrm{sech}\, t)\Big]_0^\infty + \frac{2}{u}\int_0^\infty \tan^{-1}(\tfrac12u \,\mathrm{sech}\, t)\,dt \\ & = \frac{2}{u}\int_0^\infty \tan^{-1}(\tfrac12u \,\mathrm{sech}\, t)\,dt \end{align} Now the substitution $$v = \tfrac12u\mathrm{sech}\, t$$ yields \begin{align} \int_0^\infty \frac{t \,\mathrm{sech}\, t \tanh t}{1 + \frac14u^2\,\mathrm{sech}^2t}\,dt & = \frac{2}{u}\int_{\frac12u}^0 \frac{\tan^{-1}v\, (-u)}{v\sqrt{u^2-4v^2}}\,dv \\ & = 2\int_0^{\frac12u} \frac{\tan^{-1}v\,dv}{v\sqrt{u^2-4v^2}} \; = \; \frac{2}{u}\int_0^1 \frac{\tan^{-1}(\frac12 u w)}{w\sqrt{1-w^2}}\,dw \\ & = \frac{\pi}{u}\ln\Big(\sqrt{1 + \tfrac14u^2} +\tfrac12u\Big) \end{align} Note that the identity $\int_0^1 \frac{\tan^{-1} pw}{w\sqrt{1-w^2}}\,dw \; = \; \tfrac12\pi\ln\big(\sqrt{1 + p^2} + p\big)$ is a standard integral from Gradshteyn & Ryzhik at the last stage. Thus, using the substitution $$x = e^t$$, \begin{align} \int_0^\infty \frac{1-x^2}{1+x^2} \ln x \tan^{-1}\left(\frac{x}{x^2+1}\right)\,\frac{dx}{x} & = - \int_{\mathbb{R}}t \tanh t \tan^{-1}\big(\tfrac12\,\mathrm{sech}\,t\big)\,dt \\ & = -2\int_0^\infty t \tanh t \tan^{-1}\big(\tfrac12\,\mathrm{sech}\,t\big)\,dt \\ & = -2\int_0^\infty t \tanh t \left(\int_0^1 \frac{\frac12\,\mathrm{sech}\,t\,du}{1 + \frac14u^2\,\mathrm{sech}^2t}\right)\,dt \\ & = -\int_0^1\left(\int_0^\infty \frac{t\,\mathrm{sech}\,t \tanh t}{1 + \frac14u^2\,\mathrm{sech}^2t}\,dt\right)\,du \\ & = -\pi\int_0^1 \ln\Big(\sqrt{1 + \tfrac14u^2} + \tfrac12u\Big)\,\frac{du}{u} \; = \; -\tfrac{1}{20}\pi^3 \end{align} using the result of Problem 30.

- 1 year, 5 months ago

Nice. By the way, Taylor expansion of $$\dfrac{1}{x}\arctan\left(\dfrac{x}{1+x^2}\right)$$ is $$\displaystyle \sum_{n=0}^{+\infty} \dfrac{(-1)^n L_{2n+1}x^{2n}}{2n+1}$$

and,

Taylor expansion of $$\dfrac{\arctan\left(\dfrac{x}{1+x^2}\right)}{x(1+x^2)}$$ is $$\displaystyle \sum_{n=0}^{+\infty} (-1)^n\left(\sum_{p=0}^n \dfrac{L_{2p+1}}{2p+1}\right)x^{2n}$$

$$L_k$$ is the $$k$$-th Lucas number.

- 1 year, 5 months ago

(+1) Nice! My method was a bit different (used tangent half angle in the beginning), but I also reduce it to P30. The integral you mentioned can be proved using differentiation under the integral and forming a differential equation.

- 1 year, 5 months ago

Problem 43 : Prove That $\int_{0}^{\infty} \dfrac{\mathrm{d}x}{1 + (x+\tan x)^2} = \dfrac{\pi}{2}$

- 1 year, 5 months ago

Note that,

$\tan x = - \lim_{n \to \infty} \left( \dfrac{1}{x- \frac{\pi}{2}} + \dfrac{1}{x+ \frac{\pi}{2}} + \ldots + \dfrac{1}{x- (2n-1)\frac{\pi}{2}} + \dfrac{1}{x + (2n-1)\frac{\pi}{2}} \right)$

Also, by Glasser's Master Theorem (which can be proved using elementary methods by using the graphical properties of the function), we have,

$\int_{-\infty}^{+\infty}f\left(x-\frac{a_1}{x-\lambda_1}-\cdots-\frac{a_n}{x-\lambda_n}\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \quad ; \quad a_{i} >0 \ , \ \lambda_{i} \in \mathbb{R}$

Therefore,

$\int_{-\infty}^{+\infty}f\left(x+ a\tan x\right)\mathrm{d}x=\int_{-\infty}^{+\infty} f(x)\: \mathrm{d}x \quad ; \quad a >0$

Putting $$f(x) = \dfrac{1}{x^2+b^2}$$, we have,

$\int_{-\infty}^{\infty} \dfrac{\mathrm{d}x}{(x+a \tan x)^2+b^2} = \int_{-\infty}^{\infty} \dfrac{\mathrm{d}x}{x^2+b^2}$

$\therefore \int_{0}^{\infty} \dfrac{\mathrm{d}x}{(x+a \tan x)^2+b^2} = \dfrac{\pi}{2b} \quad \square$

- 1 year, 5 months ago

I think there are some big convergence issues to be settled The convergence of the series for $$\tan x$$ is anything but uniform, and there is no function that I can see that will enable the DCT to guarantee to obtain $\int_{-\infty}^\infty f(x+a\tan x)\,dx$ as the limit of $\int_{-\infty}^\infty f\left(x - \tfrac{a_1}{x-\lambda_1} - \cdots - \tfrac{a_n}{x - \lambda_n}\right)\,dx$

- 1 year, 5 months ago

Here is my take, using the ideas I had been playing with, which convert the integral into a shape for which the GMT can be used, and a limit taken, to get the answer.

Write \begin{align} I \; = \; \int_0^\infty \frac{1}{1 + (x + \tan x)^2}\,dx & = \tfrac12\int_{\mathbb{R}} \frac{dx}{1 + (x + \tan x)^2} \\ & = \tfrac12 \sum_{n \in \mathbb{Z}} \int_{-\frac12\pi}^{\frac12\pi} \frac{1}{1 + (n\pi + x + \tan x)^2}\,dx \\ & = \tfrac{1}{2\pi^2}\int_{-\frac12\pi}^{\frac12\pi} \sum_{n \in \mathbb{Z}} \frac{dx}{\big(n + \tfrac{x + \tan x}{\pi}\big)^2 + \tfrac{1}{\pi^2}} \\ & = \frac{1}{2\pi^2} \int_{-\frac12\pi}^{\frac12\pi} \frac{\pi^2 \sinh 2}{\cosh 2 - \cos(2x + 2\tan x)}\,dx \\ & = \tfrac12\sinh2 \int_{-\frac12\pi}^{\frac12\pi} \frac{dx}{\cosh 2 - \cos(2x + 2\tan x)} \end{align} Since we are integrating over a finite interval, and since the integrand is bounded on the interval, we can use the GMT, taking the limit in the manner you suggest, so that $I \; = \; \tfrac12\sinh2 \int_{-\frac12\pi}^{\frac12\pi} \frac{dx}{\cosh2 - \cos 2x} \; = \; \tfrac14\sinh2 \int_{-\pi}^\pi \frac{dx}{\cosh 2 - \cos x}$ Making the complex substitution $$z = e^{ix}$$, this becomes \begin{align} I & = \tfrac14\sinh2 \int_{|z|=1} \frac{1}{\cosh2 - \frac12(z + z^{-1})} \frac{dz}{iz} \; = \; -\frac{\sinh 2}{2i}\int_{|z|=1}\frac{dz}{(z-e^2)(z-e^{-2})} \\ & = -\pi \sinh 2\, \mathrm{Res}_{z = e^{-2}} \frac{1}{(z - e^2)(z - e^{-2})} \; = \; \tfrac12\pi \end{align} as required.

- 1 year, 5 months ago

PROBLEM 42 :

Evaluate $\int_0^1 \frac{(x^p - 1)(x^q - 1)}{(x-1) \ln x}\,dx \hspace{2cm} p,q > 0$

- 1 year, 5 months ago

Let,

$$\displaystyle \text{I} = \int_{0}^{1} \dfrac{(x^p - 1)(x^q-1)}{(x-1) \ln x} \ \mathrm{d}x$$

$$\displaystyle = - \int_{0}^{1} \dfrac{(x^p - 1)(x^q-1)}{(1-x) \ln x} \ \mathrm{d}x$$

$$\displaystyle = - \sum_{r=0}^{\infty} \int_{0}^{1} \left[ \left(\dfrac{x^{p+q+r} - x^{p+r}}{\ln x}\right) - \left(\dfrac{x^{q+r} - x^{r}}{\ln x}\right) \right] \ \mathrm{d}x$$

$$\displaystyle = -\lim_{n \to \infty} \sum_{r=0}^{n} \left[\log \left(\dfrac{p+q+r+1}{p+r+1}\right) - \log\left( \dfrac{q+r+1}{r+1} \right)\right]$$

$$\displaystyle = -\lim_{n \to \infty} \sum_{r=1}^{n} \left[\log \left(\dfrac{p+q+r}{p+r}\right) - \log\left( \dfrac{q+r}{r} \right)\right]$$

where the standard result $$\displaystyle \int_{0}^{1}\dfrac{x^a-x^b}{\ln x} \ \mathrm{d}x = \ln\left( \dfrac{a+1}{b+1} \right)$$ has been used in the above lines.

$$\displaystyle \implies \text{I} = -\lim_{n \to \infty} \log \left( \dfrac{n! \times (p+q+1)(p+q+2)\ldots(p+q+n)}{[(p+1)(p+2)\ldots(p+n)] \times [(q+1)(q+2)\ldots(q+n)]} \right)$$

$$\displaystyle = -\lim_{n \to \infty} \log \left( \dfrac{\Gamma(p+q+n+1) \Gamma(p+1) \Gamma(n+1) \Gamma(q+1) }{\Gamma(p+q+1) \Gamma(p+n+1) \Gamma(q+n+1)} \right)$$

Since $$\displaystyle \lim_{n\to \infty} \dfrac{n^x \Gamma(n+1)}{\Gamma(n+x+1)} = 1$$, we have,

$$\displaystyle \text{I} = \log\left( \dfrac{\Gamma(p+q+1)}{\Gamma(p+1) \Gamma(q+1)} \right) \quad \square$$

- 1 year, 5 months ago

One of the standard integral representations of $$\ln \Gamma(z)$$ gets you there much more easily...

- 1 year, 5 months ago

Comment deleted Jan 12, 2017

Try applying the substitution $$x = e^{-t}$$ to the standard formula $\ln\Gamma(p) \; = \; \int_0^\infty \left\{ \frac{e^{-pt} -e^{-t}}{1 - e^{-t}} + (p-1)e^{-t}\right\}\frac{dt}{t}$ When you consider $\ln\left(\frac{\Gamma(p+q+1)}{\Gamma(p+1)\Gamma(q+1)}\right)$ the second terms all cancel out...

- 1 year, 5 months ago

I think we can generalize it to $\int_{0}^{1} \dfrac{\prod_{k=1}^n (x^{a_{k}} - 1)}{(x-1) \ln x} \ \mathrm{d}x$ where $$a_{k} > 0$$

- 1 year, 5 months ago

Problem 40

Evaluate $\int_0^\infty\frac{x-1}{\sqrt{2^x-1}\ \ln\left(2^x-1\right)}dx$

- 1 year, 5 months ago

Substitute $$(2^x-1) = t^2$$ to get,

$$\text{I} = \displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t$$

Substitute $$t \mapsto \dfrac{1}{t}$$

$$\implies \text{I} = -\displaystyle \dfrac{1}{\ln^2 2} \int_{0}^{\infty} \left( \dfrac{\ln (t^2+1) - \ln 2}{(t^2 + 1) \ln t} \right) \mathrm{d}t + \dfrac{2}{\ln^2 2} \int_{0}^{\infty} \dfrac{\mathrm{d}t}{t^2+1}$$

$$\implies \text{I} = -\text{I} + \dfrac{\pi}{\ln^2 2}$$

$$\implies \text{I} = \dfrac{\pi}{2 \ln^2 2}$$

- 1 year, 5 months ago

Problem 38:

Evaluate,

$\displaystyle \int_0^{\tfrac{\pi}{4}} \dfrac{x\ln\left(\tfrac{\cos x+\sin x}{\cos x-\sin x}\right)}{\cos x(\cos x+\sin x)}\, dx$

- 1 year, 5 months ago

We note that \begin{align} I & = \int_0^{\frac14\pi}\frac{x}{\cos x(\cos x + \sin x)} \ln\left(\frac{\cos x + \sin x}{\cos x - \sin x}\right)\,dx \; = \; \int_0^{\frac14\pi} \frac{x}{1 + \tan x}\ln\left(\frac{1 + \tan x}{1 - \tan x}\right)\,\sec^2x\,dx \\ & = \int_0^1 \frac{\tan^{-1}u}{1+u}\ln\left(\frac{1+u}{1-u}\right)\,du \; = \; -\int_0^1 \tan^{-1}\left(\frac{1-y}{1+y}\right)\,\frac{\ln y}{1+y}\,dy \end{align} after the substitutions $$u = \tan x$$ and $$y = \frac{1-u}{1+u}$$. Since $\tan\big(\tfrac14\pi - \tan^{-1}y\big) \; = \; \frac{1-y}{1+y}$ we deduce that $I \; = \; -\int_0^1 \left(\tfrac14\pi - \tan^{-1}y\right)\frac{\ln y}{1+y}\,dy \; = \; -\tfrac14\pi\int_0^1 \frac{\ln y}{1+y}\,dy + \int_0^1 \frac{\tan^{-1}y \ln y}{1+y}\,dy$ Now it is elementary that $\int_0^1 \frac{\ln y}{1+y}\,dy \; = \; -\tfrac{1}{12}\pi^2$ and we see from this MSE post --- hello, it's one of yours again --- that $\int_0^1 \frac{\tan^{-1}y \ln y}{1+y}\,dy \; = \; \tfrac12 G \ln 2 - \tfrac{1}{64}\pi^3$ so it follows that $I \; = \; \tfrac{1}{192}\pi^3 + \tfrac12G \ln 2$

- 1 year, 5 months ago

Congrats ! It's like you read my mind, you know all my tricks ;) Happy new year to all !

- 1 year, 5 months ago

Problem 34:

Prove that $\large\int_{-\infty}^\infty \dfrac1{1+x^2+x^4+\cdots+x^{2k} } \, dx = \dfrac{2\pi}{k+1} \cdot \dfrac{ \cos\left( \frac{\pi}{2k+2} \right) }{ \sin \left( \frac{3\pi}{2k+2} \right) } \; .$

This problem has been solved by Mark Hennings.

- 1 year, 5 months ago

If we define $$\zeta = e^{\frac{\pi i}{k+1}}$$, then $f_k(z) \; = \; \sum_{j=0}^k z^j \; = \; \frac{z^{k+1}-1}{z-1} \; = \; \prod_{j=1}^k \big(z - \zeta^{2j}\big)$ and hence, using partial fractions, we must be able to write $\frac{1}{f_k(z)} \; = \; \sum_{j=1}^k \frac{A_j}{z - \zeta^{2j}}$ where $A_j \; = \; \lim_{z \to \zeta^{2j}} \frac{z - \zeta^{2j}}{f_k(z)} \; = \; \big[f_k'\big(\zeta^{2j}\big)\big]^{-1} \; = \; \frac{\zeta^{2j}(\zeta^{2j}-1)}{k+1}$ Thus $\frac{1}{f_k(z^2)} \; = \; \sum_{j=1}^k \frac{A_j}{z^2 - \zeta^{2j}} \; =\; \sum_{j=1}^k \frac{A_j}{2\zeta^j}\left(\frac{1}{z - \zeta^j} - \frac{1}{z + \zeta^j}\right)$ and hence the function $\frac{1}{f_k(z^2)}$ has simple poles at $$\pm \zeta^j$$ for $$1 \le j \le k$$. Thus \begin{align} \int_{-\infty}^\infty \frac{1}{f_k(x^2)}\,dx & = 2\pi i\sum_{j=1}^k \mathrm{Res}_{z = \zeta^j} \frac{1}{f_k(z^2)} \; = \; 2\pi i \sum_{j=1}^k \frac{A_j}{2\zeta^j} \\ & = \frac{\pi i}{k+1}\sum_{j=1}^k \zeta^j(\zeta^{2j} - 1) \; = \; \frac{\pi i}{k+1}\left[ \frac{\zeta^3(\zeta^{3k} - 1)}{\zeta^3 - 1} - \frac{\zeta(\zeta^k - 1)}{\zeta-1}\right] \\ & = \frac{\pi i}{k+1} \left[\frac{\zeta^3(-\zeta^{-3} - 1)}{\zeta^3-1} - \frac{\zeta(\zeta^{-1}-1)}{\zeta-1}\right] \; = \; \frac{\pi i}{k+1}\left[\frac{\zeta+1}{\zeta-1} - \frac{\zeta^3+1}{\zeta^3-1}\right] \\ & = \frac{2 \pi i}{k+1} \times \frac{\zeta(\zeta^2-1)}{(\zeta-1)(\zeta^3-1)} \; = \; \frac{2\pi i}{k+1} \times \frac{\zeta(\zeta+1)}{\zeta^3-1} \\ & = \frac{2\pi i}{k+1} \times \frac{e^{\frac{3\pi i}{2(k+1)}} 2\cos\big(\frac{\pi}{2(k+1)}\big)}{e^{\frac{3\pi i}{2(k+1)}} 2i \sin\big(\frac{3\pi}{2(k+1)}\big)} \\ & = \frac{2\pi}{k+1}\,\frac{\cos\big(\frac{\pi}{2(k+1)}\big)}{\sin\big(\frac{3\pi}{2(k+1)}\big)} \end{align} as required.

- 1 year, 5 months ago

Problem 33:

Evaluate $\large \int_0^{\frac 14\pi} \frac{\sin^px}{\cos^{p+2}x}\,dx \hspace{2cm} p > 0 \;.$

This problem has been solved by Sumanth R Hegde.

- 1 year, 5 months ago

Put $$tanx = t$$

The integral reduces to $$\int\limits_{0}^{1} t^p dt$$

This gives $$\frac{1}{p+1}$$

Anyone else may post the next question

- 1 year, 5 months ago

Problem 47:

Evaluate,

$\int_0^1 \dfrac{\arctan x}{\sqrt{1-x^2}}dx$

- 1 year, 5 months ago

The sequence of substitutions $$u = x^2$$, $$v = 1-u$$, $$w = \sqrt{v}$$, $$y = \sinh p$$ and $$q = e^p$$ give \begin{align} I \; = \; \int_0^1 \frac{\tan^{-1}x}{\sqrt{1-x^2}}\,dx & = \int_0^1 \frac{dx}{\sqrt{1-x^2}}\int_0^1 \frac{x\,dy}{1 + x^2y^2} \\ & = \int_0^1\,dy \int_0^1 \frac{x\,dx}{\sqrt{1-x^2}(1 + x^2y^2)} \; = \; \tfrac12\int_0^1\,dy \int_0^1 \frac{du}{\sqrt{1-u}(1+y^2u)} \\ & = \tfrac12\int_0^1\,dy \int_0^1 \frac{dv}{\sqrt{v}(1 + y^2 - y^2v)} \; = \; \int_0^1\,dy \int_0^1 \frac{dw}{1 + y^2 - y^2w^2} \\ & = \tfrac12\int_0^1 \ln\left(\frac{\sqrt{1+y^2} + y}{\sqrt{1+y^2} - y}\right) \frac{dy}{y\sqrt{1+y^2}} \\ & = \int_0^{\sinh^{-1}1} \frac{p}{\sinh p}\,dp \; = \; 2\int_0^{\sinh^{-1}1}\frac{pe^p\,dp}{e^{2p}-1} \\ & = 2\int_1^{1+\sqrt{2}}\frac{\ln q}{q^2-1}\,dq \; = \; -\Big[\ln q \ln(q+1) + \mathrm{Li}_2(-q) + \mathrm{Li}_2(1-q)\Big]_1^{1+\sqrt{2}} \\ & = -\ln(1+\sqrt{2})\ln(2+\sqrt{2}) - \mathrm{Li}_2(-1-\sqrt{2}) - \mathrm{Li}_2(-\sqrt{2}) - \tfrac1{12}\pi^2 \end{align} We now use some dilogarithm identities \begin{align} \mathrm{Li}_2(\sqrt{2}) + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) & = \tfrac13\pi^2 - \tfrac12\ln^2\sqrt{2} - i\pi\ln\sqrt{2} \; = \; \tfrac13\pi^2 - \tfrac18\ln^22 - \tfrac12\pi i \ln 2 \\ \mathrm{Li}_2(-\sqrt{2}) & = \tfrac12\mathrm{Li}_2(2) - \mathrm{Li}_2(\sqrt{2}) \\ &= \tfrac12\left(\tfrac14\pi^2 - \pi i \ln2\right) - \tfrac13\pi^2 + \tfrac18\ln^22 + \tfrac12\pi i \ln 2 + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) \\ & = -\tfrac{5}{24}\pi^2 + \tfrac18\ln^22 + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) \\ \mathrm{Li}_2(-1-\sqrt{2}) + \mathrm{Li}_2\big(\tfrac{1}{\sqrt{2}}\big) & = -\tfrac12\ln^2(2+\sqrt{2}) \end{align} to deduce that $\mathrm{Li}_2(-1-\sqrt{2}) + \mathrm{Li}_2(-\sqrt{2}) \; = \; -\tfrac{5}{24}\pi^2 + \tfrac18\ln^22 - \tfrac12\ln^2(2+ \sqrt{2})$ and hence that \begin{align} I & = -\ln(1 + \sqrt{2})\ln(2 + \sqrt{2}) + \tfrac{5}{24}\pi^2 - \tfrac18\ln^22 + \tfrac12\ln^2(2+\sqrt{2}) - \tfrac{1}{12}\pi^2 \\ & = \tfrac18\pi^2 - \tfrac12\ln^2(1+ \sqrt{2}) \end{align}

- 1 year, 5 months ago

$\displaystyle J=\int_0^1 \dfrac{\arctan x}{\sqrt{1-x^2}}dx$

Perform the change of variable $$y=\sqrt{\dfrac{1-x}{1+x}}$$,

\begin{align}\displaystyle J&=2\int_0^1 \dfrac{\arctan\left(\tfrac{1-x^2}{1+x^2}\right)}{1+x^2}dx\\ &=2\int_0^1 \dfrac{\arctan(1)}{1+x^2}dx-2\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx\\ &=\dfrac{\pi^2}{8}-2\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx\\ \end{align}

\begin{align} \displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx&=\Big[\arctan x\arctan(x^2)\Big]_0^1-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ &=\dfrac{\pi^2}{16}-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \end{align}

Since,

$$\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+t^2x^2}dx$$

then,

\begin{align} \displaystyle K&=\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\ \displaystyle &=\int_0^1\int_0^1 \dfrac{2x^2}{(1+t^2x^2)(1+x^4)}dtdx\\ \displaystyle &=\int_0^1\int_0^1 \left(\dfrac{2t^2}{(1+t^4)(1+x^4)}+\dfrac{2x^2}{(1+x^4)(1+t^4)}-\dfrac{2t^2}{(1+t^4)(1+t^2x^2}\right)dtdx\\ &=\displaystyle 4\left(\int_0^1 \dfrac{t^2}{1+t^4}dt\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)-K \end{align}

Therefore,

$\displaystyle K=2\left(\int_0^1 \dfrac{x^2}{1+x^4}dx\right)\left(\int_0^1 \dfrac{1}{1+x^4}dx\right)$

Since,

\begin{align}\displaystyle \int_0^1 \dfrac{x^2}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2-\sqrt{2}x+1}{x^2+\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi+\ln\left(3-2\sqrt{2}\right)\Big) \end{align}

and,

\begin{align}\displaystyle \int_0^1 \dfrac{1}{1+x^4}dx&=\left[\dfrac{1}{4\sqrt{2}}\ln\left(\dfrac{x^2+\sqrt{2}x+1}{x^2-\sqrt{2}x+1}\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x+1\right)+\dfrac{1}{2\sqrt{2}}\arctan\left(\sqrt{2}x-1\right)\right]_0^1\\ &=\dfrac{1}{4\sqrt{2}}\Big(\pi-\ln\left(3-2\sqrt{2}\right)\Big) \end{align}

Therefore,

$\boxed{K=\displaystyle \dfrac{\pi^2}{16}-\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$

Therefore,

$\boxed{\displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx=\dfrac{1}{16}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$

Therefore,

$\boxed{\displaystyle J=\dfrac{\pi^2}{8}-\dfrac{1}{8}\Big(\ln\left(3-2\sqrt{2}\right)\Big)^2}$

- 1 year, 5 months ago

You do like pulling substitutions out of nowhere! Where did $$y = \sqrt{\tfrac{1-x}{1+x}}$$ come from?

This is a nice alternative derivation. Of course, since $$3 - 2\sqrt{2} = (1 + \sqrt{2})^{-2}$$, our results are the same!

- 1 year, 5 months ago

That sub is the tangent half angle substitution in algebraic form.

- 1 year, 5 months ago

I have found out this solution after you write down yours. I was searching but i was stuck. When i have seen your solution i was pretty sure i had missed something about the use of double integration and voilà !.

The inspiration for this problem comes from: http://math.stackexchange.com/questions/2092967/help-to-prove-that-int-0-pi-over-4-arctan-cot2x-mathrm-dx-2-pi2 (the solution i have planned to post initially is the one i have mentioned in a comment)

Your solution is nice as well.

- 1 year, 5 months ago

To respond your question. When i have an integral $$\int_0^1$$ i always try the change of variable $$y=\dfrac{1-x}{1+x}$$. Sometimes it's magic. If you do that, in the present integral, it's obvious you have to continue with the change of variable $$y=\sqrt{x}$$. My tool to do such computations is Maxima.

- 1 year, 5 months ago

* PROBLEM 46: *

Show that $\int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx \; = \; \frac{\pi(\sqrt{3}-1)}{2\sqrt{2}\sqrt[4]{3}}$

- 1 year, 5 months ago

Perform the change of variable $$y=\dfrac{1-x}{1+x}$$,

\begin{align} \int_0^1 \dfrac{\sqrt{1-x^2}}{x^4+x^2+1}dx&=\int_{0}^{1}\frac{\sqrt{x}\cdot \left( 4+4\cdot x\right) }{3\cdot {{x}^{4}}+10\cdot {{x}^{2}}+3}dx\\ &=\left[\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{2\cdot \sqrt{y}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right) +\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}+2\cdot \sqrt{y}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right)+\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{2\cdot \sqrt{3}\cdot \sqrt{y}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right) +\left( {{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{3}{4}}}-{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{5}{4}}}\right) \cdot \mathrm{atan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}+2\cdot \sqrt{3}\cdot \sqrt{y}}{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}}\right) +\left( \sqrt{2}\cdot {{3}^{\frac{5}{4}}}+\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( y-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+\sqrt{3}\right) +\left( -\sqrt{2}\cdot {{3}^{\frac{5}{4}}}-\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( y+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+\sqrt{3}\right) +\left( -\sqrt{2}\cdot {{3}^{\frac{5}{4}}}-\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( \sqrt{3}\cdot y-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+1\right) +\left( \sqrt{2}\cdot {{3}^{\frac{5}{4}}}+\sqrt{2}\cdot {{3}^{\frac{3}{4}}}\right) \cdot \mathrm{log}\left( \sqrt{3}\cdot y+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot \sqrt{y}+1\right)\times -\dfrac{1}{24}\right]_0^1\\ &=\frac{\left( \sqrt{2}-\sqrt{2}\cdot \sqrt{3}\right) \cdot \mathrm{atan}\left( \frac{{{3}^{\frac{1}{4}}}-\sqrt{2}}{{{3}^{\frac{1}{4}}}}\right) +\left( \sqrt{2}\cdot \sqrt{3}-\sqrt{2}\right) \cdot \mathrm{atan}\left( \frac{\sqrt{2}+{{3}^{\frac{1}{4}}}}{{{3}^{\frac{1}{4}}}}\right) +\left( \sqrt{2}\cdot \sqrt{3}-\sqrt{2}\right) \cdot \mathrm{atan}\left( \sqrt{2}\cdot {{3}^{\frac{1}{4}}}-1\right) +\left( \sqrt{2}\cdot \sqrt{3}-\sqrt{2}\right) \cdot \mathrm{atan}\left( \sqrt{2}\cdot {{3}^{\frac{1}{4}}}+1\right) }{4\cdot {{3}^{\frac{1}{4}}}}\\ &=\dfrac{\left(\sqrt{6}-\sqrt{2}\right)\left(\arctan\left(\sqrt{3+2\sqrt{3}}\right)+\pi-\arctan\left(\sqrt{3+2\sqrt{3}}\right)\right)}{4\times 3^{\tfrac{1}{4}}}\\ &=\boxed{\dfrac{\pi \left(\sqrt{3}-1\right)}{2\sqrt{2}\sqrt[4]{3}}} \end{align}

- 1 year, 5 months ago

My approach used contour integration...

The function $$\sqrt{z - 1}$$ can be defined analytically on the cut plane $$\mathbb{C} \backslash \{(-\infty,1]$$, while the function $$\sqrt{z+1}$$ can be defined analytically on the cut plane $$\mathbb{C} \backslash (-\infty,-1]$$. Putting these two functions together, it is possible to define the function $$\sqrt{z^2-1}$$ analytically on the domain $$\mathbb{C} \backslash [-1,1]$$.

Consider the "dogbone" contour $$D_\varepsilon = -\gamma_1 + \gamma_2 + \gamma_3 + \gamma_4$$, where

• $$\gamma_1$$ is the straight line segment $$z = x$$ for $$-1+\varepsilon < x < 1-\varepsilon$$, just above the cut,
• $$\gamma_2$$ is the (almost) circular arc $$z = -1 + \varepsilon e^{i\theta}$$ for $$0 < \theta < 2\pi$$,
• $$\gamma_3$$ is the straight line segment $$z = x$$ for $$-1+\varepsilon < x < 1 - \varepsilon$$, just below the cut,
• $$\gamma_4$$ is the (almost) circular arc $$z = 1 + \varepsilon e^{i\theta}$$ for $$-\pi < \theta < \pi$$.

Then we deduce that $\lim_{\varepsilon \to 0+} \int_{D_\varepsilon} \frac{\sqrt{z^2 - 1}}{z^4+z^2+1}\,dz \; = \; -2i\int_{-1}^1 \frac{\sqrt{1-x^2}}{x^4+x^2+1}\,dx \; = \; -4i\int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx$ If we let $$\Gamma_R$$ be the circular contour $$z = Re^{i\theta}$$ for $$0 \le \theta < 2\pi$$, then we have $\left(\int_{\Gamma_R} - \int_{D_\varepsilon}\right) \frac{\sqrt{z^2-1}}{z^4 + z^2 + 1}\,dz \; = \; 2\pi i \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} \mathrm{Res}_{z=u} \frac{\sqrt{z^2-1}}{z^4+z^2+1}$ for sufficiently small $$\varepsilon>0$$ and any $$R > 1$$. Letting $$R \to \infty$$ and $$\varepsilon \to 0+$$, we deduce that \begin{align} \int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx & = \tfrac12\pi \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} \mathrm{Res}_{z=u} \frac{\sqrt{z^2-1}}{z^4+z^2+1} \; = \; \tfrac12\pi \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} \mathrm{Res}_{z=u} \frac{(z^2-1)\sqrt{z^2-1}}{z^6-1} \\ & = \tfrac{1}{12}\pi \sum_{u \in \{\pm\omega\,,\,\pm\omega^2\}} u(u^2-1)\sqrt{u^2-1} \end{align} If $$u = e^{i\theta}$$ for $$|\theta| < \pi$$, then \begin{align} u + 1 & = 2\cos\tfrac12\theta e^{\frac12i\theta} \\ u - 1 & = 2i\sin\tfrac12\theta e^{\frac12i\theta} \\ u(u^2-1) & = 2i\sin\theta e^{2i\theta} \\ \sqrt{u+1} & = \sqrt{2\cos\tfrac12\theta} e^{\frac14i\theta} \\ \sqrt{u-1} & = \sqrt{2|\sin\tfrac12\theta|}e^{\frac14i(\theta + \pi\mathrm{sgn}(\theta))} \\ \sqrt{u^2-1} & = \sqrt{2|\sin\theta|} e^{\frac14i(2\theta + \pi\mathrm{sgn}(\theta))} \\ u(u^2-1)\sqrt{u^2-1} & = 2i\sin\theta\sqrt{2|\sin\theta|} e^{\frac14i(10\theta + \pi\mathrm{sgn}(\theta))} \end{align} and hence $\int_0^1 \frac{\sqrt{1-x^2}}{x^4 + x^2 + 1}\,dx \; = \; \tfrac{1}{12}\pi \times 4 \times 3^{\frac34} \sin\tfrac{1}{12}\pi \; = \; \frac{\pi}{3^{\frac14}}\sin\tfrac{1}{12}\pi \; = \; \frac{\pi(\sqrt{3}-1)}{2^{\frac32} 3^{\frac14}}$

- 1 year, 5 months ago

Problem 45.

Evaluate $\int_0^\infty \frac{x^5}{\sinh(\pi x)(1+x^6)}dx$

- 1 year, 5 months ago

Using the Fourier transform we have $I \; = \; \int_0^\infty \frac{x^5}{\sinh \pi x(1 + x^6)}\,dx \; =\; \tfrac12\int_{-\infty}^\infty \frac{x^5}{\sinh \pi x(1 + x^6)}\,dx \; = \; \tfrac12\langle f\,,\,g\rangle \; = \; \tfrac12\langle \mathcal{F}f\,,\,\mathcal{F}g \rangle$ where $f(x) \; = \; \frac{x}{\sinh \pi x} \hspace{2cm} g(x) \; = \; \frac{x^4}{1 + x^6}$ A standard result about Fourier transforms tells us that $(\mathcal{F}f)(u) \; = \; \frac{1}{2\sqrt{2\pi}}\,\mathrm{sech}^2\tfrac12u$ and the Fourier transform of $$g$$ can be readily calculated using contour integration: for $$u > 0$$ we simply need to determine the residues of $$\frac{z^4e^{izu}}{1+z^6}$$ at the three roots of $$z^6 + 1 = 0$$ with positive imaginary part. Omitting the details, $(\mathcal{F}g)(u) \; = \; \tfrac13\sqrt{\tfrac{\pi}{2}}\Big[e^{-|u|} + e^{-\frac12|u|}\cos\big(\tfrac{\sqrt{3}}{2}|u|\big) - \sqrt{3}e^{-\frac12|u|}\sin\big(\tfrac{\sqrt{3}}{2}|u|\big)\Big]$ Thus we deduce that $I \; = \; \tfrac12\langle \mathcal{F}f\,,\,\mathcal{F}g\rangle \; = \; \tfrac{1}{12}\int_0^\infty \mathrm{sech}^2\tfrac12u\,\Big[e^{-u} + e^{-\frac12u}\cos\big(\tfrac{\sqrt{3}}{2}u\big) - \sqrt{3}e^{-\frac12u}\sin\big(\tfrac{\sqrt{3}}{2}u\big)\Big]\,du$ Now $\int_1^\infty \frac{v^a}{(1+v)^2}\,du \; = \; \tfrac12\Big(1 - aH_{-\frac12-\frac{a}{2}} + a H_{-\frac12a}\Big) \hspace{2cm} \mathfrak{Re}\,a < 1$ and so $\int_0^\infty e^{\omega u}\,\mathrm{sech}^2\tfrac12u\,du \; = \; 4\int_0^\infty \frac{e^{\omega u}}{(e^{\frac12u} + e^{-\frac12u})^2}\,du \; = \; 4\int_1^\infty \frac{v^\omega}{(1 + v)^2}\,dv \; = \; 2\Big(1 - \omega H_{\frac12\omega^2} + \omega H_{-\frac12\omega}\Big)$ where $$\omega$$ is the primitive cube root of unity. Taking real and imaginary parts appropriately, and doing a lot of simplification with polygammas, $\int_0^\infty e^{-\frac12u}\Big[\cos\big(\tfrac{\sqrt{3}}{2}u\big) - \sqrt{3}\sin\big(\tfrac{\sqrt{3}}{2}u\big)\Big]\,\mathrm{sech}^2\tfrac12u\,du \; = \; -2 + 4\pi\,\mathrm{sech}\big(\tfrac{\sqrt{3}\pi}{2}\big)$ On the other hand, $\int_0^\infty e^{-u}\,\mathrm{sech}^2\tfrac12u\,du \; = \;4\int_1^\infty \frac{dv}{v(v+1)^2} \; = \; -2 + 4\ln 2$ Putting this all together, we obtain that $I \; = \; \tfrac13\Big[-1 + \ln2 + \pi\,\mathrm{sech}\big(\tfrac{\sqrt{3}\pi}{2}\big)\Big]$ Someone else can post the next one, please.

- 1 year, 5 months ago

Here is a partial progress towards an alternate solution $f(a) = \int_0^\infty \frac{\sinh (ax)}{\sinh(\pi x)(1+x^6)} \ \mathrm{d}x$

then,

$f^{(6)} (a) + f(a) = \dfrac{1}{2} \tan\left(\dfrac{a}{2}\right)$

where $$f^{(n)} (a)$$ denotes the $$n^{\text{th}}$$ derivative of $$f$$ w.r.t. $$a$$, with $$f(0) = 0$$ and the required integral being $$f^{(5)} (0)$$.

I still have to solve that D.E.

- 1 year, 5 months ago

Comment deleted Jan 12, 2017

Using $\frac{\pi \text{csch}(\pi z)}{2}=\frac{1}{2z}+\sum_{n=1}^\infty \frac{(-1)^n z}{z^2+n^2}$

- 1 year, 5 months ago

Nice!

- 1 year, 5 months ago

Problem 41 :

Prove That

$\int_{-\infty}^{\infty} \dbinom{n}{x} \ \mathrm{d}x = 2^n \quad \Re(n) > -1$

Notation : $$\dbinom{n}{x}$$ denotes the Generalized Binomial Coefficient.

- 1 year, 5 months ago

Presumably, we mean ${n \choose x} \; = \; \frac{n!}{\Gamma(x+1)\Gamma(n+1-x)} \hspace{2cm} x \in \mathbb{R}$ Now $\Gamma(x+1)\Gamma(n+1-x) \; = \; x\Gamma(x) \prod_{j=1}^n (j-x)\Gamma(1-x) \; = \; (-1)^n \frac{\pi}{\sin\pi x}\times \prod_{j=0}^n (x-j)$ and so we can use partial fractions to write ${n \choose x} \; = \; \frac{(-1)^n n!}{\pi} \sin \pi x \left(\prod_{j=0}^n (x-j)\right)^{-1} \; = \; \frac{(-1)^n n! \sin \pi x}{\pi} \sum_{j=0}^n \frac{A_j}{x-j}$ where $A_j \; = \; \frac{(-1)^{n-j}}{j! (n-j)!} \hspace{1cm} 0 \le j \le n$ Since $\int_{\mathbb{R}} \frac{\sin \pi x}{x - j}\,dx \; = \; (-1)^j \int_{\mathbb{R}} \frac{\sin \pi x}{x}\,dx \; = \; (-1)^j \pi$ we deduce that \begin{align} \int_{\mathbb{R}} {n \choose x}\,dx & = \frac{(-1)^n n!}{\pi} \sum_{j=0}^n A_j \int_{\mathbb{R}}\frac{\sin \pi x}{x-j}\,dx \; = \; (-1)^n n! \sum_{j=0}^n (-1)^j A_j \\ & = \sum_{j=0}^n \frac{n!}{j!(n-j)!} \; = \; \sum_{j=0}^n {n \choose j} \; = \; 2^n \end{align} as required.

If we insist on being picky, we could integrate from $$-X$$ to $$X$$, and let $$X \to \infty$$, to avoid dealing with the improper integral $\int_{\mathbb{R}} \frac{\sin x}{x}\,dx$ directly. We only have to combine a finite number of integrals at the last stage, so there is no problem taking the limit.

- 1 year, 5 months ago

I think your solution can be extended to deal with non integral n if we instead expand into partial fractions using infinite product of Gamma Function. My approach was using the integral representation $\large \int_{0}^{\pi} (\sin (x))^{m-1} e^{inx} \ \mathrm{d}x = \dfrac{\pi}{2^{m-1}} \ \dfrac{e^{i n/2}}{m \operatorname{B} \left( \dfrac{1}{2} (m+n+1) , \dfrac{1}{2} (m-n+1) \right)}$

- 1 year, 5 months ago

The integral is also true for $$\Re(n) > -1$$.

- 1 year, 5 months ago

Indeed. The formulae $\int_{-\infty}^\infty \frac{dx}{\Gamma(\alpha+x)\Gamma(\beta-x)} \; = \; \frac{2^{\alpha+\beta-2}}{\Gamma(\alpha+\beta-1)}$ for $$\mathfrak{Re}(\alpha+\beta) > 1$$ is $$6.414.2$$ in Gradshteyn & Rhyzik.

Isn't it a bit late to be altering the question, though?

- 1 year, 5 months ago

Apologies, I forgot to mention that initially. But your answer is the accepted one nevertheless.

- 1 year, 5 months ago

If $$\binom{n+1}{x+1}=\binom{n}{x}+\binom{n}{x+1}$$ then i think it's straightforward to prove with a recurrence proof.

- 1 year, 5 months ago

Certainly, if we presume that the integral exists, then the recurrence relation can help us evaluate it. The RR does not prove convergence of the integral, however.

- 1 year, 5 months ago

Ohh!! I missed this season. I even didn't know that this happened. :(

- 1 year, 2 months ago

Maybe this time you can host it :)

- 1 year, 2 months ago

When is starting the season 4? :)

- 1 year, 1 month ago

I think in December 2017.

- 1 year, 1 month ago

Problem 49:

Evaluate,

$\int_{0}^{\frac{\pi }{6}}\frac{x \mathrm{sin}\left( 2 x\right) }{4 {{\mathrm{sin}^{4}x }}-2 {{\mathrm{sin^{2}}x }}+1}dx$

- 1 year, 5 months ago

\begin{align}J&=\int_0^{\tfrac{\pi}{6}} \dfrac{x\sin(2x)}{1-2\sin^2 x+4\sin^4 x}dx\\ &= \int_0^{\tfrac{\pi}{6}} \dfrac{2x\sin x\cos x}{(1-\sin^2 x)^2+3\sin^4 x}dx\\ &=\int_0^{\tfrac{\pi}{6}} \dfrac{2x\sin x\cos x}{\cos^4+3\sin^4 x}dx\\ &=\int_0^{\tfrac{\pi}{6}} \dfrac{2x\tan x}{\cos^2 x(1+3\tan^4 x)}dx\\ \end{align}

Perform the change of variable $$y=\sqrt{3}\tan x$$,

$J=\int_0^1 \dfrac{2x\arctan\left(\tfrac{x}{\sqrt{3}}\right)}{3+x^4}dx$

Since,

$\arctan u=\int_0^1 \dfrac{u}{1+t^2u^2}dt$

\begin{align}J&=2\sqrt{3}\int_0^1 \int_0^1 \dfrac{x^2}{(3+x^4)(3+t^2x^2)}dtdx\\ &=2\sqrt{3}\int_0^1 \int_0^1 \left(\frac{{{t}^{2}}+{{x}^{2}}}{\left( {{t}^{4}}+3\right)\left( {{x}^{4}}+3\right) }-\frac{{{t}^{2}}}{\left( {{t}^{4}}+3\right)\left( {{t}^{2}}{{x}^{2}}+3\right) }\right)dtdx\\ &=4\sqrt{3}\int_0^1 \int_0^1 \dfrac{t^2}{(3+t^4)(3+x^4)}dtdx-J\\ \end{align}

therefore,

\begin{align} J&=2\sqrt{3}\int_0^1 \int_0^1 \dfrac{t^2}{(3+t^4)(3+x^4)}dtdx\\ &=2\sqrt{3}\left(\int_0^1 \dfrac{x^2}{3+x^4}dx \right)\left(\int_0^1 \dfrac{1}{3+x^4}dx\right)\\ \end{align}

Since,

\begin{align} \int_0^1 \dfrac{1}{3+x^4}dx&=\left[\frac{\mathrm{arctan}\left( \frac{\sqrt{2}\cdot{{3}^{\frac{1}{4}}}x}{\sqrt{3}-{{x}^{2}}}\right) }{{{2}^{\frac{3}{2}}}{{3}^{\frac{1}{4}}}}-\frac{\mathrm{log}\left( \frac{\sqrt{3}-\sqrt{2}\cdot{{3}^{\frac{1}{4}}} x+{{x}^{2}}}{{{x}^{2}}+\sqrt{2}\cdot{{3}^{\frac{1}{4}}}x+\sqrt{3}}\right) }{{{2}^{\frac{5}{2}}}{{3}^{\frac{1}{4}}}}\right]_0^1\\ &=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{3}{4}}}\arctan\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)-\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{3}{4}}}\log\left(\dfrac{1+\sqrt{3}-\sqrt{2}\cdot 3^{\tfrac{1}{4}}}{1+\sqrt{3}+\sqrt{2}\cdot 3^{\tfrac{1}{4}}}\right) \end{align}

$$1+\sqrt{3}+\sqrt{3}\cdot 3^{\tfrac{1}{4}}$$ is a root of the polynomial $$X^4-4X^3-16X+16$$,

Therefore,

$\int_0^1 \dfrac{1}{3+x^4}dx=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{3}{4}}}\arctan\left(\dfrac{\sqrt{2}\cdot 3^{\tfrac{1}{4}}}{\sqrt{3}-1}\right)-\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{3}{4}}}\log\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right)$

and,

\begin{align} \int_0^1 \dfrac{x^2}{3+x^4}dx&=\left[\frac{\mathrm{arctan}\left( \frac{\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x}{\sqrt{3}-{{x}^{2}}}\right) }{{{2}^{\frac{3}{2}}}\cdot {{3}^{\frac{1}{4}}}}+\frac{\mathrm{log}\left( \frac{\sqrt{3}-\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x+{{x}^{2}}}{{{x}^{2}}+\sqrt{2}\cdot {{3}^{\frac{1}{4}}}\cdot x+\sqrt{3}}\right) }{{{2}^{\frac{5}{2}}}\cdot {{3}^{\frac{1}{4}}}}\right]_0^1\\ &=\dfrac{1}{2^{\tfrac{3}{2}}3^{\tfrac{1}{4}}}\arctan\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)+\dfrac{1}{2^{\tfrac{5}{2}}3^{\tfrac{1}{4}}}\log\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right) \end{align}

Therefore,

$\boxed{J=\dfrac{1}{4\sqrt{3}}\arctan^2\left(\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{1}{4}}+\dfrac{\sqrt{2}}{2}\cdot 3^{\tfrac{3}{4}}\right)-\dfrac{1}{16\sqrt{3}}\log^2\left(1+\sqrt{3}-\dfrac{\sqrt{2}}{2}3^{\tfrac{1}{4}}-\dfrac{\sqrt{2}}{2}3^{\tfrac{3}{4}}\right) }$

PS:

$F(x,t)=\dfrac{x}{(a+x^2)(a+tx)}=\dfrac{t}{a+t^2}\cdot \dfrac{1}{a+x^2} +\dfrac{1}{a+t^2}\cdot \dfrac{x}{a+x^2}-\dfrac{t}{(a+t^2)(a+tx)}$

is a nice function.

- 1 year, 5 months ago

- 1 year, 5 months ago

I am interested. My solution used/generalised a lot of the tricks that FDP uses in his solutions, so it will be interesting if he has a different approach.

- 1 year, 5 months ago

Aliter,

Let.

$\text{I} = \int_{0}^{\frac{\pi}{6}} \dfrac{x \sin(2x)}{4\sin^4 x - 2\sin^2 x + 1} \ \mathrm{d}x$

$= \dfrac{1}{4(\alpha - \beta)} \int_{0}^{\frac{\pi}{6}} x \sin (2x) \left(\dfrac{1}{\cos(2x) - 1 + \beta} - \dfrac{1}{\cos(2x) - 1 + \alpha}\right)$

where $$\alpha$$ and $$\beta$$ are the roots of the equation $$t^2 - \dfrac{1}{2} t + \dfrac{1}{4} = 0$$

Now, from Problem 25, we have,

$\dfrac{\sin x}{p^2 +2p \cos x +1} = \sum_{k=1}^{\infty} (-p)^{k-1} \sin (kx)$

Using this and interchanging, sum and integral ( and after a lot of straight forward simplification/calculations involving dilogarithm and logarithm) we get,

$\text{I} = \dfrac{1}{4\sqrt{3}}\left[\tan^{-1}\left(\frac{\sqrt{2}}{3^{\frac14} - 3^{-\frac14}}\right)\right]^2 - \frac{1}{16\sqrt{3}}\ln^2\left(\frac{3^{\frac14} + \sqrt{2} + 3^{-\frac14}}{3^{\frac14} - \sqrt{2} + 3^{-\frac14}}\right)$

- 1 year, 5 months ago

Just pipped you! You might want to clarify what $$\alpha$$ and $$\beta$$ are, though.

Why don't you post the last integral!

- 1 year, 5 months ago

Thanks for letting me do the honors. I'll post it tomorrow, I'll need some time to create a nice one :)

- 1 year, 5 months ago

For any $$a > 0$$ define \begin{align} A(a) & = \int_0^1 \frac{dx}{a^4 + x^4} \; = \; \frac{1}{a^3}\int_0^{a^{-1}} \frac{dz}{1 + z^4} \\ & = \frac{1}{2a^3\sqrt{2}}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big] + \frac{1}{4a^3\sqrt{2}} \ln\left(\frac{a + \sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \\ B(a) & = \int_0^1 \frac{x^2\,dx}{a^4 + x^4} \; = \; \frac{1}{a}\int_0^{a^{-1}} \frac{z^2\,dz}{1 + z^4} \\ & = \frac{1}{2a\sqrt{2}}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big] - \frac{1}{4a\sqrt{2}} \ln\left(\frac{a + \sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \\ a^4A(a)B(a) & = \tfrac{1}{8}\left[ \tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big) \right]^2 - \frac{1}{32} \ln^2\left(\frac{a+\sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \end{align} Thus we deduce that (putting $$z = ax$$ and playing symmetry games) \begin{align} I(a) & = \int_0^{a^{-1}} \frac{x \tan^{-1}\big(\frac{x}{a}\big)}{1 + x^4}\,dx \; = \; \int_0^{a^{-1}} \frac{x\,dx}{1 + x^4} \int_0^1 \frac{\frac{x}{a}\,dy}{1 + a^{-2}x^2y^2} \\ & = a^4 \int_0^1 \int_0^1 \frac{z^2}{(a^4 + z^4)(a^4 + y^2 z^2)}\,dy\,dz \\ & = \tfrac12a^4 \int_0^1 \int_0^1 \left\{ \frac{z^2}{(a^4 + z^4)(a^4 + y^2z^2)} + \frac{y^2}{(a^4 + y^4)(a^4 + y^2z^2)}\right\}\,dy\,dz \\ & = \tfrac12a^4 \int_0^1 \int_0^1 \frac{y^2 + z^2}{(a^4 + y^4)(a^4 + z^4)}\,dy\,dz \; = \; a^4 A(a)B(a) \\ & = \tfrac{1}{8}\Big[\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big)\Big]^2 - \frac{1}{32} \ln^2\left(\frac{a+\sqrt{2} + a^{-1}}{a - \sqrt{2} + a^{-1}}\right) \end{align} Moreover \begin{align} J(a) & = \int_0^{a^{-1}} \frac{\tan^{-1}x^2}{x^2 + a^2}\,dx \; = \; \Big[\frac{1}{a} \tan^{-1}x^2\,\tan^{-1}\big(\tfrac{x}{a}\big)\Big]_0^{a^{-1}} - \frac{2}{a} \int_0^{a^{-1}} \frac{x \tan^{-1}\big(\frac{x}{a}\big)}{1 + x^4}\,dx \\ & = \frac{1}{a}\big(\tan^{-1}a^{-2}\big)^2 - \frac{2}{a}I(a) \end{align} Thus, putting $$u = \cos2x$$ and $$v= \frac{2u-1}{\sqrt{3}}$$, we see that \begin{align} K & = \int_0^{\frac{1}{6}\pi} \frac{x \sin 2x}{4\sin^4x - 2\sin^2x + 1}\,dx \; = \; \int_0^{\frac{1}{6}\pi} \frac{x \sin 2x}{\cos^22x - \cos2x + 1}\,dx \\ & = \tfrac14\int_{\frac12}^1 \frac{\cos^{-1}u\,du}{u^2 - u + 1} \; = \; \tfrac14\Big[\tfrac{2}{\sqrt{3}}\tan^{-1}\left(\tfrac{2u-1}{\sqrt{3}}\right) \cos^{-1}u\Big]_{\frac12}^1 + \tfrac14\int_{\frac12}^1 \tfrac{2}{\sqrt{3}} \tan^{-1}\big(\tfrac{2u-1}{\sqrt{3}}\big) \frac{du}{\sqrt{1-u^2}} \\ & = \tfrac{1}{2\sqrt{3}} \int_{\frac12}^1 \tan^{-1}\left(\tfrac{2u-1}{\sqrt{3}}\right) \frac{du}{\sqrt{1-u^2}} \; = \; \frac{1}{2 \times 3^{\frac14}} \int_0^{\frac{1}{\sqrt{3}}} \frac{\tan^{-1}v\,dv}{\sqrt{(1 - \sqrt{3}v)(\sqrt{3}+v)}} \end{align} The key substitution $w \; = \; \sqrt{\frac{1 - \sqrt{3}v}{\sqrt{3}+v}}$ yields \begin{align} K & = \frac{1}{3^{\frac14}} \int_0^{3^{-\frac14}} \tan^{-1}\left(\frac{1 - \sqrt{3}w^2}{\sqrt{3} + w^2}\right) \frac{dw}{\sqrt{3} + w^2} \; = \; \frac{1}{3^{\frac14}} \int_0^{3^{-\frac14}} \Big[\tfrac16\pi - \tan^{-1}w^2\Big] \frac{dw}{\sqrt{3} + w^2} \\ & = \frac{\pi}{6 \times 3^{\frac14}} \Big[3^{-\frac14} \tan^{-1}\big(\tfrac{w}{3^{\frac14}}\big)\Big]_0^{3^{-\frac14}} - 3^{-\frac14}J(3^{\frac14}) \\ & = \frac{\pi^2}{36\sqrt{3}} - 3^{-\frac14}\Big[3^{-\frac14}\left(\tan^{-1}3^{-\frac12}\right)^2 - 2 \times 3^{-\frac14} I\big(3^{\frac14}\big)\Big] \; = \; \tfrac{2}{\sqrt{3}} I\big(3^{\frac14}\big) \\ & = \frac{1}{4\sqrt{3}}\left[\tan^{-1}\left(\frac{\sqrt{2}}{3^{\frac14} - 3^{-\frac14}}\right)\right]^2 - \frac{1}{16\sqrt{3}}\ln^2\left(\frac{3^{\frac14} + \sqrt{2} + 3^{-\frac14}}{3^{\frac14} - \sqrt{2} + 3^{-\frac14}}\right) \end{align} noting that $\tan^{-1}\big(1 + \sqrt{2}a^{-1}\big) - \tan^{-1}\big(1 - \sqrt{2}a^{-1}\big) \; =\; \tan^{-1}\left(\frac{\sqrt{2}}{a - a^{-1}}\right) \hspace{1cm} a > 1$

- 1 year, 5 months ago

Actually my solution is the same than yours. The only difference is the way you manage to transform the trigonometric integral.

- 1 year, 5 months ago

Yes, but your method of transforming the original integral into what is, apart from a variable scaling, $$\tfrac{2}{\sqrt{3}} I(3^{1/4})$$ (in my notation), is elegant.

- 1 year, 5 months ago

if $$F(a,t,x)=\frac{x}{\left( t\cdot x+a\right) \cdot \left( {{x}^{2}}+a\right) }$$, compute $$\int_0^1 F(1,t,x)dt$$ isn't nice?

Problem 47 relies on the evaluation of $$\int_0^1 \int_0^1 F(1,t^2,x^2) dtdx$$ and problem 49 relies on $$\int_0^1 \int_0^1 F(3,t^2,x^2) dtdx$$

Who will post problem 50, the last one?

- 1 year, 5 months ago

Let me clarify. Both of our methods evaluate $$\tfrac{2}{\sqrt{3}}I(3^{\frac14})$$. What I was saying was that your technique for converting the original integral to $$\tfrac{2}{\sqrt{3}}I(3^{\frac14})$$ was more elegant than mine - in Mathematics, the word "elegant" is complimentary!

- 1 year, 5 months ago

I'll post by evening, in 5 hours.

- 1 year, 5 months ago

My solution is close to yours. I will post it later (01 AM here) . This integral is based on the solution i have provided for problem 47. I have played around with it and i have build this one.

- 1 year, 5 months ago

* PROBLEM 48: *

Prove that $\int_0^\infty \frac{x^{\mu-\frac12}}{(x+r)^\mu (x+s)^\mu}\,dx \; = \; \sqrt{\pi}(\sqrt{r}+\sqrt{s})^{1-2\mu}\frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}$ for all $$r,s > 0$$ and $$\mu > \tfrac12$$.

- 1 year, 5 months ago

Someone else can post the next one...

For any $$a > 0$$ and $$0 < b < c$$ and $$|x| < 1$$ we have \begin{align} I_{a,b,c}(x) & = \int_0^1 t^{b-1}(1-t)^{c-b-1}(1 - xt)^{-a}\,dt \\ & = \int_0^1 t^{b-1}(1-t)^{c-b-1} \left(\sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n t^n\right)\,dt \\ & = \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \int_0^1 t^{n+b-1}(1-t)^{c-b-1}\,dt \; = \; \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n B(n+b,c-b) \\ & = \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \frac{\Gamma(n+b) \Gamma(c-b)}{\Gamma(n+c)} \; =\; \sum_{n=0}^\infty \frac{(a)^{(n)}}{n!} x^n \frac{(b)^{(n)} \Gamma(b)\Gamma(c-b)}{(c)^{(n)} \Gamma(c)} \\ & = B(b,c-b)\sum_{n=0}^\infty \frac{(a)^{(n)} (b)^{(n)}}{(c)^{(n)}} \frac{x^n}{n!} \; = \; B(b,c-b) \big({}_2F_1\big)(a,b;c;x) \end{align} With the substitution $$y = \tfrac{t}{1-t}$$ we see that \begin{align} \int_0^\infty y^{b-1} (1+y)^{a-c} (1 + \alpha y)^{-a}\,dy & = \int_0^1 \left(\frac{t}{1-t}\right)^{b-1}\left(\frac{1}{1-t}\right)^{a-c} \left(\frac{1 - (1-\alpha)t}{1-t}\right)^{-a}\, \frac{dt}{(1-t)^2} \\ & = \int_0^1 t^{b-1}(1-t)^{c-b-1}(1 - (1-\alpha)t)^{-a}\,dt \; = \; I_{a,b,c}(1-\alpha) \\ & = B(b,c-b) \big({}_2F_1\big)(a,b;c;1-\alpha) \end{align} for any $$a>0$$, $$0 < b < c$$ and $$0 < \alpha \le 1$$.

Since the integral is symmetric in $$r$$ and $$s$$, we may assume that $$0 < r \le s$$. Then, for any $$0 < r \le s$$ and $$\mu > \tfrac12$$ we have \begin{align} J_{\mu,r,s} & = \int_0^\infty \frac{x^{\mu-\frac12}}{(x+r)^\mu(x+s)^\mu}\,dx \; =\; \frac{\sqrt{r}}{s^\mu} \int_0^\infty t^{\mu-\frac12}(1+t)^{-\mu}\big(1 + \tfrac{r}{s}t\big)^{-\mu}\,dt \\ & = \frac{\sqrt{r}}{s^\mu} B\big(\mu+\tfrac12,\mu-\tfrac12\big) \big({}_2F_1\big)\big(\mu,\mu+\tfrac12;2\mu;1-\tfrac{r}{s}\big) \end{align} Since $\big({}_2F_1\big)(a,b;c;z) \; = \; \begin{array}{l} \displaystyle(1-z)^{-a} \frac{\Gamma(c)\Gamma(b-a)}{\Gamma(b)\Gamma(c-a)}\big({}_2F_1\big)\big(a,c-b;a-b+1;(1-z)^{-1}\big) \\ \displaystyle{}+ (1-z)^{-b} \frac{\Gamma(c)\Gamma(a-b)}{\Gamma(a)\Gamma(c-b)}\big({}_2F_1\big)\big(b,c-a;b-a+1;(1-z)^{-1}\big)\end{array}$ we deduce that \begin{align} J_{\mu,r,s} & = \frac{\sqrt{r}}{s^\mu}B\big(\mu+\tfrac12,\mu-\tfrac12\big)\left[ \begin{array}{l} \displaystyle\big(\tfrac{r}{s}\big)^{-\mu} \frac{\Gamma(2\mu)\Gamma(\tfrac12)}{\Gamma(\mu+\frac12)\Gamma(\mu)}\big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) \\ \displaystyle{} + \big(\tfrac{r}{s}\big)^{-\mu-\frac12} \frac{\Gamma(2\mu)\Gamma(-\tfrac12)}{\Gamma(\mu)\Gamma(\mu-\frac12)}\big({}_2F_1\big)\big(\mu+\tfrac12,\mu;\tfrac32;\tfrac{s}{r}\big) \end{array} \right] \\ & = \frac{2^{2\mu-1}}{r^\mu}B\big(\mu+\tfrac12,\mu-\tfrac12\big)\left[ \sqrt{r}\big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) - \sqrt{s}(2\mu-1)\big({}_2F_1\big)\big(\mu+\tfrac12,\mu;\tfrac32;\tfrac{s}{r}\big) \right]\\ & = \frac{\sqrt{\pi}}{r^\mu} \frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}\left[ \sqrt{r}\big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) - \sqrt{s}(2\mu-1)\big({}_2F_1\big)\big(\mu+\tfrac12,\mu;\tfrac32;\tfrac{s}{r}\big) \right] \end{align} Now standard hypergeometric identities give \begin{align} \big({}_2F_1\big)\big(\mu,\mu-\tfrac12;\tfrac12;\tfrac{s}{r}\big) & = \tfrac12\left[ \left(1 - \sqrt{\tfrac{s}{r}}\right)^{1-2\mu} + \left(1 + \sqrt{\tfrac{s}{r}}\right)^{1-2\mu}\right] \\ \big({}_2F_1\big)\big(\mu,\mu+\tfrac12;\tfrac32;\tfrac{s}{r}\big) & = \frac{1}{2(2\mu-1)}\sqrt{\tfrac{r}{s}}\left[\left(1 - \sqrt{\tfrac{s}{r}}\right)^{1-2\mu} - \left(1 + \sqrt{\tfrac{s}{r}}\right)^{1-2\mu}\right] \end{align} and so $J_{\mu,r,s} \; = \; \frac{\sqrt{\pi}}{r^{\mu-\frac12}} \frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}\big(1 + \sqrt{\tfrac{s}{r}}\big)^{1-2\mu} \; = \; \sqrt{\pi}\frac{\Gamma(\mu-\frac12)}{\Gamma(\mu)}(\sqrt{r}+\sqrt{s})^{1-2\mu}$ as required.

- 1 year, 5 months ago

Nice. I'm working on an alternative solution. I hope to get something soon.

- 1 year, 5 months ago

Please don't post the answer now, post the solution at the end of the week end. Thanks.

- 1 year, 5 months ago

I was aiming to allow some extra time to compensate for the typo, but will post a solution tomorrow...

- 1 year, 5 months ago

For those having a shot at this one, I need to point out a small, but vital, correction that has been made to the limits of the integral!

- 1 year, 5 months ago

It makes things easier :)

- 1 year, 5 months ago

True :)

- 1 year, 5 months ago

* PROBLEM 44 *:

Evaluate $\int_0^1 \ln\left(\frac{1 + ax}{1 - ax}\right) \frac{dx}{x\sqrt{1-x^2}} \hspace{1cm} -1 < a < 1 \;.$

- 1 year, 5 months ago

For $$-1<ax<1$$, $\log\left(\dfrac{1+ax}{1-ax}\right)=2\sum_{n=0}^{\infty} \dfrac{a^{2n+1}x^{2n+1}}{(2n+1)}$

\begin{align} \int_0^1 \dfrac{\log\left(\tfrac{1-ax}{1+ax}\right)}{x\sqrt{1-x^2}}dx&=2\int_0^1 \left(\sum_{n=0}^{\infty} \dfrac{a^{2n+1}x^{2n+1}}{(2n+1)}\cdot\dfrac{1}{{x\sqrt{1-x^2}}}\right)dx\\ &=2\sum_{n=0}^{\infty}\left(\int_0^1 \dfrac{a^{2n+1}x^{2n+1}}{(2n+1)}\cdot\dfrac{1}{{x\sqrt{1-x^2}}} dx\right)\\ &=2\sum_{n=0}^{\infty}\left(\dfrac{a^{2n+1}}{2n+1} \int_0^1 \dfrac{x^{2n}}{\sqrt{1-x^2}}dx\right)\\ &=\sum_{n=0}^{\infty}\dfrac{1}{2n+1}\cdot\text{B}\left(\tfrac{1}{2},n+\tfrac{1}{2}\right)a^{2n+1}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{2n+1}\dfrac{\Gamma\left(\tfrac{1}{2}\right)\Gamma\left(n+\tfrac{1}{2}\right)}{\Gamma(n+1)}a^{2n+1}\\ &=\sum_{n=0}^{\infty} \dfrac{1}{2n+1}\dfrac{\tfrac{(2n)!}{2^{2n}n!}\pi}{n!}a^{2n+1}\\ &=\pi \sum_{n=0}^{\infty} \dfrac{\binom{2n}{n}}{2^{2n}(2n+1)}a^{2n+1}\\ &=\boxed{\pi\arcsin(a)} \end{align}

- 1 year, 5 months ago

For variety, if we define $F(a) \; = \; \int_0^1 \ln\left(\frac{1 + ax}{1-ax}\right) \frac{dx}{x\sqrt{1-x^2}} \hspace{2cm} -1 < a < 1$ then \begin{align} F'(a) & = \int_0^1 \left(\frac{1}{1+ax} + \frac{1}{1-ax}\right) \frac{dx}{\sqrt{1-x^2}} \; =\; \int_0^{\frac12\pi} \left(\frac{1}{1 + a\cos \theta} + \frac{1}{1-a\cos\theta}\right)\,d\theta \\ & = \int_0^\pi \frac{d\theta}{1 + a\cos\theta} \; = \; \tfrac12\int_0^{2\pi} \frac{d\theta}{1 + a\cos\theta} \\ & = \tfrac12\int_{|z|=1} \frac{2}{2 + a(z+z^{-1})}\,\frac{dz}{iz} \; = \; -i\int_{|z|=1}\frac{dz}{az^2 + 2z + a} \\ & = -ia \int_{|z|=1}\frac{dz}{\big(az + 1 + \sqrt{1-a^2}\big)\big(az + 1 - \sqrt{1-a^2}\big)} \\ & = 2\pi a\,\mathrm{Res}_{z = a^{-1}(\sqrt{1-a^2}-1)} \frac{1}{\big(az + 1 + \sqrt{1-a^2}\big)\big(az + 1 - \sqrt{1-a^2}\big)} \; = \; \frac{\pi}{\sqrt{1-a^2}} \end{align} and hence $F(a) \; = \; \int_0^a \frac{\pi}{\sqrt{1-b^2}}\,db \; = \; \pi\sin^{-1}a$

- 1 year, 5 months ago

Problem 50 :

Evaluate $\int_{0}^{\infty} \dfrac{\sin(2nx) \cos(2\arctan(ax))}{(1+a^2x^2) \sin x} \ \mathrm{d}x$

where $$a>0$$ , $$n \in \mathbb{Z^+}$$

- 1 year, 5 months ago

Suppose that $$a > 0$$ and $$n \in \mathbb{N}$$. Note that $I \; = \; \int_0^\infty \frac{\sin(2nx)\cos\big(2\tan^{-1}(ax)\big)}{(1 + a^2x^2)\sin x}\,dx \; = \; \int_0^\infty \frac{1-a^2x^2}{(1 + a^2x^2)^2}\frac{\sin 2nx}{\sin x}\,dx \; = \; \lim_{\delta \to 0+}I_\delta$ where $I_\delta \; = \; \int_0^\infty \frac{1 - a^2(x+\delta)^2}{(1 + a^2(x+\delta)^2)^2}\frac{\sin 2nx}{\sin x}\,dx$ where we can use the Dominated Convergence Theorem to justify the limit, since $\left| \frac{1 - a^2(x+\delta)^2}{(1 + a^2(x+\delta)^2)^2}\frac{\sin 2nx}{\sin x}\right| \; \le \; \left| \frac{1}{1 + a^2(x+\delta)^2}\frac{\sin 2nx}{\sin x}\right| \; \le \; \frac{N_n}{1 + a^2x^2}$ for all $$x \ge 0$$ and $$\delta > 0$$, where $$\big|\tfrac{\sin 2nx}{\sin x}\big| \le N_n$$ for all $$x \ge 0$$. It is trivial that $\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-xt}\,dt \; = \; a^2 \frac{a^2x^2 - 1}{(1 + a^2x^2)^2} \hspace{2cm} a,x > 0$ and so \begin{align} I_\delta & = -a^{-2}\int_0^\infty\left(\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-(x+\delta)t}\,dt\right) \frac{\sin 2nx}{\sin x}\,dx \\ & = -a^{-2}\int_0^\infty t \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left(\int_0^\infty e^{-xt} \frac{\sin 2nx}{\sin x}\,dx\right)\,dt \end{align} Since $\int_0^\infty e^{-xt} \frac{\sin(2m+2)x}{\sin x}\,dx - \int_0^\infty e^{-xt} \frac{\sin 2mx}{\sin x}\,dx \; = \; 2\int_0^\infty e^{-xt} \cos(2m+1)x\,dx \; = \; \frac{2t}{t^2 + (2m+1)^2}$ a simple induction shows that $\int_0^\infty e^{-xt} \frac{\sin 2nx}{\sin x}\,dx \; = \; 2t\sum_{k=0}^{n-1} \frac{1}{t^2 + (2k+1)^2} \hspace{2cm} t > 0$ and hence \begin{align} I_\delta & = -2a^{-2} \int_0^\infty \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left(\sum_{k=0}^{n-1} \frac{t^2}{t^2 + (2k+1)^2}\right)\,dt \\ & = -2a^{-2} \int_0^\infty \cos\big(\tfrac{t}{a}\big) e^{-\delta t}\left( n - \sum_{k=0}^{n-1} \frac{(2k+1)^2}{t^2 + (2k+1)^2}\right)\,dt \\ & = -2a^{-2}\left[ \frac{n\delta}{\delta^2 + a^{-2}} - \sum_{k=0}^{n-1} (2k+1)^2 \int_0^\infty \frac{\cos\big(\frac{t}{a}\big) e^{-\delta t}}{t^2 + (2k+1)^2}\,dt \right] \end{align} and hence, letting $$\delta \to 0+$$, \begin{align} I & = 2a^{-2} \sum_{k=0}^{n-1} (2k+1)^2 \int_0^\infty \frac{\cos\big(\frac{t}{a}\big)}{t^2 + (2k+1)^2}\,dt \\ & = 2a^{-2} \sum_{k=0}^{n-1} (2k+1)^2 \times \frac{\pi}{2(2k+1)}e^{-\frac{2k+1}{a}} \\ & = \frac{\pi}{a^2}\sum_{k=0}^{n-1} (2k+1)e^{-\frac{2k+1}{a}} \end{align} using a standard Fourier transform to finish off. This sum can be calculated, of course, but the end result is not particularly pretty, do I shall leave things here.

- 1 year, 5 months ago

Nice! I used the expansion $\dfrac{\sin(2nx)}{2\sin x} = \sum_{k=1}^{n} \cos((2k-1)x)$

The result can also be generalized to $\int_{0}^{\infty} \dfrac{\sin(2nx) \cos(p\arctan(ax))}{(1+a^2x^2)^{\frac{p}{2}} \sin x} \ \mathrm{d}x \ ; \ n \in \mathbb{Z}, a,p>0$

- 1 year, 5 months ago

Let $$C_{n-1}(a)$$ be the previous integral. I suspect one can find a closed form for the ordinary generating function for the sequence $$\left(C_n(a)\right)$$ because $$U_{2n+1}(\cos x)=\dfrac{\sin((2n+2)x}{\sin x}$$, $$U_j$$ being the $$j$$-nth Chebyshev polynomial of the second kind.

- 1 year, 5 months ago

I looked hard on the web and could not find a closed expression for the Laplace transform of the $$U_n$$, which would have been really useful!

- 1 year, 5 months ago

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