# Brilliant Integration Contest - Season 3

Hi Brilliant! Here is the most awaited Integration contest! See Season 1 and Season 2.

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

• Direct problems on contour integration are not allowed. However, solutions which use contour integration are allowed.

• Also don't forget to upvote good problems and solutions!

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, 11 months ago

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Problem 6:

Prove that $\int_0^\infty \frac{\ln x}{(x+a)^2 + b^2}\,dx \; = \; \tfrac{1}{b}\,\tan^{-1}\tfrac{b}{a}\,\ln\sqrt{a^2+b^2}$ for all $$a,b > 0$$.

This problem has been solved by Aditya Sharma.

- 1 year, 11 months ago

Solution to problem 6 :

Let us denote the integral by $$I$$ .

$$\displaystyle I = \int_0^\infty \frac{\ln x}{(x+a)^2+b^2}dx = \int_0^\infty \frac{\ln x}{(x+a+ib)(x+a-ib)}$$

With the substitution $$xy=a^2+b^2$$ we have,

$$\displaystyle I=\ln(a^2+b^2)\int_0^\infty \frac{dy}{y^2+2ay+a^2+b^2} - \underbrace{\int_0^\infty \frac{\ln y}{(y+a)^2+b^2}dy}_I$$

$$\displaystyle I=\ln(\sqrt{a^2+b^2})\int_0^\infty \frac{dx}{(x+a)^2+b^2} = \frac{1}{b}\tan^{-1}\frac{b}{a}\ln\sqrt{a^2+b^2}$$.

- 1 year, 11 months ago

As an alternative to Aditya's proof...

Consider the cut plane $$\mathbb{C} \backslash [0,\infty)$$, so that $$0 < \mathrm{Arg}\,z < 2\pi$$ for all $$z$$, and consider the integral of $f(z) \; = \; \frac{(\log z)^2}{(z+a)^2 + b^2}$ around the keyhole contour $$\gamma_1 + \gamma_2 - \gamma_3 - \gamma_4$$, where

• $$\gamma_1$$ is the line segment $$z \,=\, x e^{0i}$$, for $$\varepsilon < x < R$$, just above the cut,
• $$\gamma_2$$ is the circular contour $$z \,=\, Re^{i\theta}$$ for $$0 < \theta < 2\pi$$,
• $$\gamma_3$$ the the line segment $$z \,=\, x e^{2\pi i}$$, for $$\varepsilon < x < R$$, just below the cut,
• $$\gamma_4$$ is the circular contour $$z \,=\, \varepsilon e^{i\theta}$$ for $$0 < \theta < 2\pi$$.

where we assume that $$0 < \varepsilon < \sqrt{a^2+b^2} < R$$. Now \begin{align} \int_{\gamma_1} f(z)\,dz & = \int_\varepsilon^R \frac{(\ln x)^2}{(x+a)^2 + b^2}\,dx \\ \int_{\gamma_3} f(z)\,dz & = \int_\varepsilon^R \frac{(\ln x + 2\pi i)^2}{(x+a)^2 + b^2}\,dx \\ \left(\int_{\gamma_1} - \int_{\gamma_3}\right)\,f(z)\,dz & = \int_{\varepsilon}^R \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \end{align} We also note that $\int_{\gamma_2} f(z)\,dz \; = \; O\big(\tfrac{\ln^2R}{R}\big) \hspace{2cm} \int_{\gamma_4} f(z)\,dz \; = \; O\big(\varepsilon \ln^2\varepsilon\big)$ as $$R \to \infty$$ and $$\varepsilon \to 0$$. Letting $$R \to \infty$$ and $$\varepsilon \to 0$$, we deduce that $\int_0^\infty \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \; = \; 2\pi i\Big[ \mathrm{Res}_{z=-a+ib} f(z) \,+\, \mathrm{Res}_{z=-a-ib} f(z)\Big]$ Now $$f(z)$$ has simple poles at each of $$-a \pm ib$$, and \begin{align} \mathrm{Res}_{z = -a\pm ib} f(z) & = \frac{\Big(\ln\sqrt{a^2+b^2} + i\pi \mp i\tan^{-1}\frac{b}{a}\Big)^2}{\pm 2ib} \\ \mathrm{Res}_{z = -a+ib} f(z) + \mathrm{Res}_{z = -a-ib} f(z) & = \frac{-4i(\ln\sqrt{a^2+b^2}+i\pi)\tan^{-1}\frac{b}{a}}{2ib} \\ & = - \frac{2}{b}\tan^{-1}\frac{b}{a}(\ln\sqrt{a^2+b^2} + i\pi) \end{align} and hence $\int_0^\infty \frac{4\pi^2 - 4\pi i \ln x}{(x+a)^2 + b^2}\,dx \; = \; \frac{4\pi}{b}\tan^{-1}\frac{b}{a}\big(\pi - i\ln\sqrt{a^2+b^2}\big)$ Taking imaginary parts of this equation gives the result.

- 1 year, 11 months ago

Problem 16:

Evaluate the integral $\int_0^1 \frac{x^{a-1}(1-x)^{b -1}}{(x+p)^{a+b}}\,dx$ for all $$a,b,p > 0$$.

This problem has been solved by Aditya Sharma.

- 1 year, 11 months ago

If we use the substitution, $$\displaystyle x=\frac{1}{u+1}$$ then

$$\displaystyle I =\int_0^\infty \frac{u^{b-1}}{(up+p+1)^{a+b}}\,du$$

Again using the substitution, $$\displaystyle u=y\left(\frac{p+1}{p}\right)$$ the integral transforms to,

$$\displaystyle I =\frac{1}{p^b(1+p)^a}\int_0^\infty \frac{y^{b-1}}{(1+y)^{a+b}}\,dy = \left(\frac{\beta(a,b)}{p^b(1+p)^a}\right)$$

- 1 year, 11 months ago

The substitution $z = \frac{(p+1)x}{p+x}$ Is even more direct. Not a hard problem, but a very elegant result.

- 1 year, 11 months ago

Yes, Although this substitution emerges by combining the chain of substitutions used, or it's tough to put it directly without knowing why it is the reqd. substitution.

- 1 year, 11 months ago

Problem 14 : Prove That

$\int_{0}^{\infty} \ln x\left[\ln \left( \dfrac{x+1}{2} \right) - \dfrac{1}{x+1} - \psi \left( \dfrac{x+1}{2} \right) \right] \mathrm{d}x = \dfrac{\ln^2 2}{2}+\ln2\cdot\ln\pi-1$

Notation : $$\psi(x)$$ denotes the Digamma Function.

This problem has been solved by Mark Hennings.

- 1 year, 11 months ago

This is a really nice use of previous results of this round of the contest!

Starting with the identity $\psi(x) \; = \; \ln x - \frac{1}{2x} - 2\int_0^\infty \frac{t}{(t^2+x^2)(e^{2\pi t}-1)}\,dt$ we deduce that \begin{align} \ln\big(\tfrac{x+1}{2}\big) + \tfrac{1}{x+1} - \psi\big(\tfrac{x+1}{2}\big) & = 2\int_0^\infty \frac{t}{\big(t^2 + \frac14(x+1)^2\big)(e^{2\pi t} - 1)}\,dt \\ & = 8\int_0^\infty \frac{t}{(4t^2 + (x+1)^2)(e^{2\pi t} - 1)}\,dt \\ & = 2\int_0^\infty \frac{t}{(t^2 + (x+1)^2)(e^{\pi t}-1)}\,dt \end{align} and hence \begin{align} \int_0^\infty \ln x & \left(\ln\big(\tfrac{x+1}{2}\big) + \tfrac{1}{x+1} - \psi\big(\tfrac{x+1}{2}\big)\right)\,dx \\ & = 2\int_0^\infty \left(\int_0^\infty \frac{\ln x}{(x+1)^2 + t^2}\,dx\right)\,\frac{t}{e^{\pi t}-1}\,dt \\ & = 2\int_0^\infty \left(\frac{1}{t} \tan^{-1}t \ln\sqrt{t^2 + 1}\right)\,\frac{t}{e^{\pi t}-1}\,dt \; = \; \int_0^\infty \frac{\tan^{-1}t \ln(t^2+1)}{e^{\pi t} - 1}\,dt \\ & = \tfrac12\ln^22 + \ln 2 \ln\pi - 1 \end{align} using the results of Problems $$5$$ and $$6$$.

- 1 year, 11 months ago

I have posted a question on Math Stack Exchange asking for alternate methods to evaluate the above integral (independently of Problems $$5$$ and $$6$$) here. If we can evaluate it using other methods, it'll also give us an alternate solution for Problem $$5$$.

- 1 year, 11 months ago

Yes indeed! This integral arose from my attempt to find an alternate solution of Problem 5 :) The first identity can be proved (using real analysis) by differentiating the identity $\int_{0}^{\infty} \dfrac{\log(1-e^{-2a\pi x})}{1+x^2} \mathrm{d}x = \pi \left[\dfrac{1}{2} \log (2a\pi ) + a(\log a - 1) - \log(\Gamma(a+1)) \right]$ wrt $$a$$. I have proved it here.

- 1 year, 11 months ago

Problem 13:

Evaluate $\int_0^{2\pi} e^{\cos\theta} \cos\big(n\theta - \sin\theta\big)\,d\theta$ for any integer $$n$$.

This problem has been solved by Ishan Singh.

- 1 year, 11 months ago

It's simplest using contour integration...

Note that $\int_0^{2\pi} e^{\cos\theta}\,\cos(\sin\theta - n\theta)\,d\theta \; = \; \mathfrak{Re}\left(\int_0^{2\pi} e^{\cos\theta} e^{i(\sin\theta - n\theta)}\,d\theta\right) \; = \; \mathfrak{Re}\left(\int_0^{2\pi} e^{e^{i\theta} - in\theta}\,d\theta\right)$ With the substitution $$z = e^{i\theta}$$, this becomes the contour integral $\mathfrak{Re}\left(\int_{|z|=1}e^z z^{-n} \frac{dz}{iz}\right) \; = \; \mathfrak{Re}\left(\frac{1}{i}\int_{|z|=1}\frac{e^z}{z^{n+1}}\,dz\right) \; = \; 2\pi\mathfrak{Re}\left(\mathrm{Res}_{z=0} \frac{e^z}{z^{n+1}}\right) \; = \; \left\{ \begin{array}{ll} \frac{2\pi}{n!} & n \ge 0 \\ 0 & n < 0 \end{array} \right.$

- 1 year, 11 months ago

$\text{I} = \int_{0}^{2 \pi} e^{\cos t} \cos (nt - \sin t) \ \mathrm{d}t = \Re{ \left(\int_{0}^{2 \pi} e^{i n t} \cdot e^{(\cos t - i \sin t)} \ \mathrm{d}t \right)}$

$= \Re{ \left(\int_{0}^{2 \pi} e^{i n t} \cdot e^{\Large e^{ -i t}} \ \mathrm{d}t \right)}$

$= \dfrac{2 \pi}{n!} + \Re{ \left(\sum_{r \geq 0 \ ; \ r \neq n} \left( \dfrac{1}{r!} \int_{0}^{2 \pi} e^{i t (n-r)} \ \mathrm{d}t \right) \right)}$

Since last integral is $$0$$, we have,

$\text{I} = \dfrac{2 \pi}{n!}$

If $$n$$ is negative, we can write $$n = -k \ ; \ k \in \mathbb{Z}^+$$, so that,

$\text{J} = \int_{0}^{2 \pi} e^{\cos t} \cos (kt + \sin t) \ \mathrm{d}t$

By the above method, we can similarly show that $$\text{J} = 0$$.

Therefore,

$\int_{0}^{2 \pi} e^{\cos t} \cos (nt - \sin t) \ \mathrm{d}t = \left\{ \begin{array}{ll} \frac{2\pi}{n!} & n \ge 0 \\ 0 & n < 0 \end{array} \right.$

- 1 year, 11 months ago

Problem 8:

Show that $\int_0^{\frac12\pi} \ln^2\big(a\sin\theta\big)\,d\theta \; = \; \tfrac{1}{24}\pi^3 + \tfrac{1}{6}\pi\Big[\ln^2\big(\tfrac{2}{a}\big) - 2\ln\big(\tfrac{2}{a}\big)\ln\big(\tfrac{a}{2}\big)\Big]$ for any $$a > 0$$.

This problem has been solved by Jasper Braun.

- 1 year, 11 months ago

Solution to Problem 8: $\displaystyle I=\int_0^\frac{\pi}{2}\ln^2{a}+2\ln{a}\ln{\sin{x}}+\ln^2{\sin{x}}dx=\frac{\pi\ln^2{a}}{2}-\ln{a}\pi\ln{2}+\int_0^\frac{\pi}{2}\ln^2{\sin{x}}dx$

$\displaystyle\int_0^\frac{\pi}{2}\ln^2{\sin{x}}dx = \int_0^1\frac{\ln^2{x}}{\sqrt{1-x^2}}dx\quad\text{(by }u= \sin{x})$

$\displaystyle\text{Now using }B(m,n) = 2\int_0^1x^{2m-1}(1-x^2)^{n-1}dx \text{ and thus }\int_0^1\frac{\ln^2{x}}{\sqrt{1-x^2}}dx = \frac{d^2}{dm^2}\frac{B(m,\frac{1}{2})}{2*4}\big|_\frac{1}{2}$

$\displaystyle\text{Using }\Gamma'(m) = \Gamma(m)\psi(m),\qquad\Gamma''(m) = \Gamma(m)\psi(m)^2+\Gamma(m)\psi^{(1)}(m)$

$\displaystyle\frac{\sqrt{\pi}}{8}\frac{d^2}{dm^2}\frac{\Gamma(m)}{\Gamma(m+\frac{1}{2})}\Big|_\frac{1}{2} = \frac{\pi\ln^2{2}}{2}+\frac{\pi\zeta(2)}{4}$

$\displaystyle I=\boxed{\frac{\pi^3}{24}+\frac{\pi\ln^2{2}}{2}+\frac{\pi\ln^2{a}}{2}-\pi\ln{a}\ln{2}}$

- 1 year, 11 months ago

Problem 5:

Prove that:

${\large\int}_0^\infty\frac{\ln\left(x^2+1\right)\,\arctan x}{e^{\pi x}-1}dx=\frac{\ln^22}2+\ln2\cdot\ln\pi-1.$

This problem has been solved by Mark Hennings.

- 1 year, 11 months ago

Solution of Problem 5

Starting with the Hermite formula for the generalized Riemann zeta function: $\zeta(s,a) \; = \; \tfrac12a^{-s} + a^{1-s}(s-1)^{-1} + 2\int_0^\infty \frac{(a^2+y^2)^{-\frac12s} \, \sin\big(s \tan^{-1}\frac{y}{a}\big)}{e^{2\pi y} - 1}\,dy$ we have \begin{align} \zeta(s,\tfrac12) & = 2^{s-1}\left(1 + \tfrac{1}{s-1}\right) + 2\int_0^\infty \frac{(\frac14 + y^2)^{-\frac12s}\,\sin\big(s \tan^{-1}2y\big)}{e^{2\pi y}-1}\,dy \\ & = 2^{s-1}\frac{s}{s-1} + 2^{s+1}\int_0^\infty \frac{(1 + 4y^2)^{-\frac12s} \, \sin\big(\tan^{-1}2y\big)}{e^{2\pi y} - 1}\,dy \\ & = 2^{s-1}\frac{s}{s-1} + 2^s \int_0^\infty \frac{(1+t^2)^{-\frac12s}\,\sin\big(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \end{align} so that $F(s) \; = \; \int_0^\infty \frac{(1+t^2)^{-\frac12s}\,\sin(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \; = \; 2^{-s}\zeta(s,\tfrac12) - \frac{s}{2(s-1)}$ Now since \begin{align} F''(s) & = \frac{\partial^2}{\partial s^2}\int_0^\infty \frac{(1 + t^2)^{-\frac12s} \, \sin(s \tan^{-1}t\big)}{e^{\pi t}-1}\,dt \\ & = \int_0^\infty \frac{(1+t^2)^{-\frac12s}}{e^{\pi t} - 1}\left\{ \begin{array}{l} \tfrac14\big(\ln(1+t^2)\big)^2 \sin(s\tan^{-1}t) - \ln(1 + t^2) \tan^{-1}t \cos\big(s \tan^{-1}t\big) \\ - \big(\tan^{-1}t\big)^2 \sin\big(s \tan^{-1}t\big)\end{array}\right\}\,dt \end{align} we deduce that $F''(0) \; = \; -\int_0^\infty \frac{\ln(1+t^2) \tan^{-1}t}{e^{\pi t}-1}\,dt$ and hence that the desired integral is $\int_0^\infty \frac{\ln(1+t^2) \tan^{-1}t}{e^{\pi t}-1}\,dt \; = \; -F''(0) \; = \; \frac{\partial^2}{\partial s^2}\Big[\tfrac{s}{2(s-1)} - 2^{-s}\zeta(s,\tfrac12)\Big] \Big|_{s=0}$ which simplifies to give $\int_0^\infty \frac{\ln(1+t^2) \tan^{-1}t}{e^{\pi t}-1}\,dt \; = \; -1 - \tfrac12\ln^22 + \ln2 \cdot \ln2\pi \; = \; -1 + \tfrac12\ln^22 + \ln2\ln\pi$

- 1 year, 11 months ago

Problem 2:

Evaluate $$\large \displaystyle \int_{0}^{1} \frac{(x^{2}+1)\tan^{-1}3x}{x} \, dx$$.

This problem has been solved by Aditya Sharma.

- 1 year, 11 months ago

The integral can be written as : $$\displaystyle I=\underbrace{\int_0^1 x\tan^{-1} (3x) dx}_{I_1}+\underbrace{\int_0^3 \frac{\tan^{-1}x}{x}dx}_{I_2}$$

A simple integration by parts would yield $$\displaystyle I_1 = \frac{5}{9}\tan^{-1}3 - \frac{1}{6}$$

$$I_2$$ is the Inverse Tangent Integral and we represent it by dilogarithms.

$$\displaystyle Ti_2(3)=\frac{1}{2i}(Li_2(3i)-Li_2(-3i))$$

So, $$\displaystyle I = \frac{5}{9}\tan^{-1}3 - \frac{1}{6} +\frac{1}{2i}(Li_2(3i)-Li_2(-3i))$$

- 1 year, 11 months ago

Problem 1:

Show that: $\int_{0}^{1}\frac{\cosh(a\log(x))\log(1+x)}{x}dx=\frac{1}{2a}(\pi \csc(\pi a)-\frac{1}{a}), \;\ a<1$

This problem has been solved by Aditya Sharma.

- 1 year, 11 months ago

If we denote the integral by $$I$$ & using $$2cosh(a\ln x)=x^a+x^{-a}$$ & $$\displaystyle \ln(1+x)=\sum_{n\ge 1}\frac{(-1)^{n-1}}{n}x^n$$ the integral can be transformed into,

$$\displaystyle I = \sum_{n\ge 1} \frac{(-1)^{n-1}}{n} \int_0^1 (x^{n+a-1}+x^{n-a-1})=\sum_{n\ge 1} \frac{(-1)^{n-1}}{n}(\frac{1}{n-a}+\frac{1}{n+a})=\frac{1}{2a}\sum_{n\ge 1} (-1)^{n-1} (\frac{1}{n-a}-\frac{1}{n+a})$$

For removing the alternating factor $$(-1)^{n-1}$$ this can be written as ,

$$\displaystyle 2a I = \sum_{n\ge 1} (\frac{1}{n-a}-\frac{1}{n+a})-2\sum_{n\ge 1} (\frac{1}{2n-a}-\frac{1}{2n+a})$$

which simplifies to ,

$$\displaystyle 2a I = \sum_{n\ge 1} (\frac{1}{n-a}-\frac{1}{n+a}-\frac{1}{n-\frac{a}{2}}+\frac{1}{n+\frac{a}{2}})$$

Using $$\displaystyle H_a = \sum_{n\ge 1}(\frac{1}{n}-\frac{1}{n+a})$$ we have,

$$\displaystyle I=\frac{1}{2a}(H_a-H_{-a}-H_{\frac{a}{2}}+H_{-\frac{a}{2}})$$

Again using reflection formula for harmonic numbers as $$a<1$$

i.e. $$\displaystyle H_{1-a}-H_a = \pi \cot(\pi a)-\frac{1}{a}+\frac{1}{1-a}$$ it simplifies to,

$$\displaystyle I = \frac{1}{2a} (-\pi\cot(\pi a)+\pi\cot(\pi\frac{a}{2})-\frac{1}{a}) = \frac{1}{2a}(\pi\csc(\pi a)-\frac{1}{a})$$

- 1 year, 11 months ago

Problem 20 :

Show that $$\displaystyle \int_0^1 \dfrac{x\ln x\arctan x}{1+x}\, dx=\dfrac{1}{2}\ln 2-\dfrac{1}{2}G\ln 2+\dfrac{\pi^3}{64}+\dfrac{\pi^2}{48}-\dfrac{\pi}{4}$$.

This problem has been solved by Mark Hennings and then by Fdp Dpf.

- 1 year, 11 months ago

Note that \begin{align} I \; = \; \int_0^1 \frac{x \,\ln x\, \tan^{-1}x}{1+x}\,dx & = \int_0^1 \frac{x \ln x}{1+x}\left(\int_0^x \frac{du}{1+u^2}\right)\,dx \\ & = \int_0^1 \left(\int_u^1 \frac{x \ln x}{1+x}\,dx \right)\,\frac{du}{1+u^2} \\ & = \int_0^1 \Big[x\ln x - x - \ln x \ln(1+x) - \mathrm{Li}_2(-x)\Big]_{x=u}^1\,\frac{du}{1+u^2} \\ & = \int_0^1 \Big(-1 + \tfrac{1}{12}\pi^2 - u\ln u + u + \ln u \ln(1+u) +\mathrm{Li}_2(-u)\Big]\,\frac{du}{1+u^2} \\ & = -\tfrac14\pi + \tfrac{1}{48}\pi^2 + \tfrac{1}{48}\pi^3 + \tfrac12\ln2 + \int_0^1 \frac{\ln u \ln(1+u) + \mathrm{Li}_2(-u)}{1 + u^2}\,du \end{align} It is possible (integrating by parts and so forth) to calculate the indefinite integrals of both parts of the remaining integral in terms of polylogarithms, but the end result is extremely complicated! My expression for the indefinite integral $\int \frac{\ln u \ln(1+u)}{1 + u^2}\,du$ has over $$30$$ terms, and the indefinite integral for $\int \frac{\mathrm{Li}_2(-u)}{1+u^2}\,du$ is even more complicated. The definite integrals are \begin{align} \int_0^1 \frac{\ln u \ln(1+u)}{1 + u^2}\,du & = \tfrac{1}{128} \left(\begin{array}{l}11 \pi^3 - 256 G \ln2 + 10 i \pi^2 \ln2 + 12\pi (\ln2)^2 \\ - 8i(\ln2)^3 - 394i\mathrm{Li}_3(\tfrac12+\tfrac12i) + 210\zeta(3)\end{array}\right) \\ \int_0^1 \frac{\mathrm{Li}_2(-u)}{1+u^2}\,du & = \tfrac{1}{384}\left(\begin{array}{l}-35\pi^3 - 30 i \pi^2 \ln2 - 36 \pi (\ln2)^2 + 576G \ln2\\ + 24i(\ln2)^3 - 768i\mathrm{Li}_3(\tfrac12+\tfrac12i) + 630i\zeta(3)\end{array}\right) \end{align} so that $\int_0^1 \frac{\ln x \ln(1+x) + \mathrm{Li}_2(-x)}{1 + x^2}\,dx \; = \; -\tfrac{1}{192}\pi^3 - \tfrac12 G \ln2$ and hence $I \; = \; -\tfrac14\pi + \tfrac{1}{48}\pi^2 + \tfrac{1}{64}\pi^3 + \tfrac12\ln2 - \tfrac12G \ln2$ as required.

That was exhausting! Someone else can set the next one...

- 1 year, 10 months ago

SOLUTION OF PROBLEM 20:

Alternative solution.

Let $\displaystyle R(x)=\int_0^x \dfrac{t\ln t}{1+t}dt=\int_0^1 \dfrac{tx^2\ln(tx)}{1+xt}dt$

Note that,

$R(1)=\dfrac{\pi^2}{12}-1$

(Taylor expansion of integrand)

\begin{align} \displaystyle I=\int_0^1 \dfrac{x\ln x \arctan x}{1+x}dx \displaystyle &=\Big[R(x)\arctan x\Big]_0^1 -\int_0^1 \dfrac{R(x)}{1+x^2}dx\\ \displaystyle &= \dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\int_0^1\int_0^1 \dfrac{tx^2\ln(tx)}{(1+tx)(1+x^2)}dtdx\\ \displaystyle &= \dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\int_0^1\int_0^1 \dfrac{tx^2\ln t}{(1+tx)(1+x^2)}dtdx-\int_0^1\int_0^1 \dfrac{tx^2\ln x}{(1+tx)(1+x^2)}dtdx\\ &\displaystyle =\dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\int_0^1 \left[t\ln t\left(\dfrac{t\ln(1+x^2)}{2(1+t^2)}+\dfrac{\ln(1+tx)}{t(t^2+1)}-\dfrac{\arctan x}{1+t^2}\right)\right]_{x=0}^{x=1} dt-\\ &\displaystyle \int_0^1 \left[\dfrac{\ln x\Big(tx-\ln(1+tx)\Big)}{1+x^2}\right]_{t=0}^{t=1}dx\\ &=\displaystyle\dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\dfrac{\ln 2}{2}\int_0^1 \dfrac{t^2\ln t}{1+t^2}dt-\int_0^1 \dfrac{\ln t\ln(1+t)}{1+t^2}dt+\dfrac{\pi}{4}\int_0^1 \dfrac{t\ln t}{1+t^2}dt-\int_0^1 \dfrac{x\ln x}{1+x^2}dx+\\ &\int_0^1 \dfrac{\ln x\ln(1+x)}{1+x^2}dx\\ &=\displaystyle \dfrac{\pi^3}{48}-\dfrac{\pi}{4}-\dfrac{\ln 2}{2}\int_0^1 \dfrac{t^2\ln t}{1+t^2}dt+\left(\dfrac{\pi}{4}-1\right)\int_0^1 \dfrac{t\ln t}{1+t^2}dt\\ \end{align}

Using Taylor expansion of$\dfrac{x^2\ln x}{1+x^2}$ and $\dfrac{x\ln x}{1+x^2}$ one obtains,

$\displaystyle \int_0^1 \dfrac{x^2\ln x}{1+x^2}dx=G-1$

G being the Catalan constant.

$\displaystyle \int_0^1 \dfrac{x\ln x}{1+x^2}dx=-\dfrac{\pi^2}{48}$

Therefore,

$\boxed{\displaystyle I=\dfrac{\pi^3}{64}+\dfrac{\pi^2}{48}-\dfrac{\pi}{4}-\dfrac{1}{2}G\ln 2+\dfrac{1}{2}\ln 2}$

- 1 year, 10 months ago

(+1) Nice solution! You might be interested in Part 2 of the current season.

- 1 year, 10 months ago

Problem 19:

Show that $\int_0^\infty \frac{e^{-x^p} - e^{-x^q}}{x}\,dx \; = \; \frac{p-q}{pq}\gamma$ for $$p,q > 0$$, where $$\gamma$$ is the Euler-Mascheroni constant.

The problem has been solved first by Aditya Sharma and then by Ishan Singh.

- 1 year, 11 months ago

Solution to Problem 19

$$\displaystyle I=\int_0^\infty (e^{-x^p}-e^{-x^q})d(\ln x)$$

$$\displaystyle I= [(e^{-x^p}-e^{-x^q})\ln x]_0^\infty + p\int_0^\infty e^{-x^p}x^{p-1}\ln x-q\int_0^\infty e^{-x^q}x^{q-1}\ln x\,dx$$

The substitutions $$x^p=u\;,x^q=v$$ makes the last two integrals as,

$$\displaystyle p\int_0^\infty e^{-x^p}x^{p-1}\ln x-q\int_0^\infty e^{-x^q}x^{q-1}\ln x\,dx = \int_0^\infty e^{-x}x\ln x\, dx \left(\frac{q-p}{qp}\right)$$

$$\displaystyle = \int_0^\infty e^{-x}\ln x\, dx \left(\frac{q-p}{qp}\right) = \gamma\left(\frac{p-q}{qp}\right)$$

Now $$\displaystyle [(e^{-x^p}-e^{-x^q})\ln x]_0^\infty = \lim_{x\to\infty} \frac{\ln x}{\frac{1}{e^{-x^p}-e^{-x^q}}}- \lim_{x\to 0} \frac{\ln x}{\frac{1}{e^{-x^p}-e^{-x^q}}}$$

Both of these can be evaluated by applying L-Hospital's rule succesively and both of them are $$0$$.

Hence $$\displaystyle \int_0^\infty \frac{(e^{-x^p}-e^{-x^q})}{x}\, dx = \gamma\left(\frac{p-q}{pq}\right)$$

- 1 year, 11 months ago

Solution to Problem 19 :

Proposition 1 : $\int_{0}^{\infty} \left(e^{-x} - \dfrac{1}{x+1}\right)\dfrac{\mathrm{d}x}{x} = -\gamma$

Proof : Let,

$\text{J} = \int_{0}^{\infty} \left(e^{-x} - \dfrac{1}{x+1}\right)\dfrac{\mathrm{d}x}{x}$

$= \int_{0}^{\infty} \int_{0}^{\infty} e^{-x(a+1)} - \dfrac{e^{-ax}}{x+1} \mathrm{d}x \ \mathrm{d}a$

$\displaystyle = \int_{0}^{\infty} \dfrac{1}{a+1} \mathrm{d}a -\int_{0}^{\infty}\int_{0}^{\infty} \dfrac{e^{-ax}}{x+1}\mathrm{d}x \ \mathrm{d}a$

Substitute $$ax \mapsto x$$ in the second integral,

$\displaystyle \implies \text{J} = \int_{0}^{\infty} \dfrac{1}{a+1} \mathrm{d}a -\int_{0}^{\infty}\int_{0}^{\infty} \dfrac{e^{-x}}{x+a}\mathrm{d}x \ \mathrm{d}a$

Since $\displaystyle \int_{0}^{\infty} e^{-x}\mathrm{d}x = 1$

$\displaystyle \implies \text{J} = \int_{0}^{\infty} \int_{0}^{\infty} \dfrac{e^{-x}}{a+1} - \dfrac{e^{-x}}{x+a}\mathrm{d}x \ \mathrm{d}a$

$\displaystyle = \int_{0}^{\infty} e^{-x} \int_{0}^{\infty} \dfrac{1}{a+1} - \dfrac{1}{x+a}\mathrm{d}a \ \mathrm{d}x$

$\displaystyle = \int_{0}^{\infty} e^{-x} \left[\ln\left(\dfrac{a+1}{a+x}\right)\right]_{a=0}^{a \to \infty} \mathrm{d}x$

$\displaystyle =\int_{0}^{\infty} e^{-x} \ln x \ \mathrm{d}x$

$\displaystyle = -\gamma \quad \square$

Proposition 2 : $\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x = \gamma$

Proof : $\displaystyle \int_{0}^{1} \dfrac{1-e^{-x}}{x} \mathrm{d}x - \int_{1}^{\infty}\dfrac{e^{-x}}{x} \mathrm{d}x = \int_{0}^{\infty} \int_{0}^{1} (1-e^{-x})e^{-yx} \mathrm{d}x \ \mathrm{d}y - \int_{0}^{\infty} \int_{1}^{\infty} e^{-x(y+1)} \mathrm{d}x \ \mathrm{d}y$

$\displaystyle = \int_{0}^{\infty} \left(\dfrac{1-e^{-y}}{y} + \dfrac{e^{-(y+1)} - 1}{y+1} \right) \mathrm{d}y - \int_{0}^{\infty} \dfrac{e^{(y+1)}}{y+1} \mathrm{d}y$

$\displaystyle = -\int_{0}^{\infty} \left(e^{-y} - \dfrac{1}{y+1} \right)\dfrac{\mathrm{d}y}{y}$

$= \gamma \ \left( \text{Using Proposition 1} \right) \ \square$

Now,

$\text{I} = \int_{0}^{\infty} \dfrac{e^{-x^p} - e^{-x^q}}{x} \ \mathrm{d}x$

$= \left( \int_{0}^{1} \dfrac{1-e^{-x^q}}{x} \ \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x^q}}{x} \ \mathrm{d}x \right) - \left( \int_{0}^{1} \dfrac{1-e^{-x^p}}{x} \ \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x^p}}{x} \ \mathrm{d}x \right)$

Substitute $$x^q \mapsto x$$ in the first integral and $$x^p \mapsto x$$ in the second integral to get,

$\text{I} = \dfrac{1}{q} \left( \int_{0}^{1} \dfrac{1-e^{-x}}{x} \ \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x}}{x} \ \mathrm{d}x \right) - \dfrac{1}{p} \left( \int_{0}^{1} \dfrac{1-e^{-x}}{x} \ \mathrm{d}x - \int_{1}^{\infty} \dfrac{e^{-x}}{x} \ \mathrm{d}x \right)$

Using Proposition 2, we get,

$\text{I} = \left(\dfrac{1}{q} - \dfrac{1}{p}\right) \gamma$

$\therefore \text{I} = \dfrac{p-q}{pq} \gamma \quad \square$

- 1 year, 10 months ago

Problem 18 : Prove That

$\int_{0}^{\infty} \log \left( \dfrac{a^2 + 2a \cos px + 1}{a^2 + 2a \cos qx + 1} \right) \ \dfrac{\mathrm{d}x}{x} = 2 \log \left( \dfrac{q}{p} \right) \log(1+a) \ \ ; \ -1 < a \leq 1$

This problem has been solved by Mark Hennings.

- 1 year, 11 months ago

There is some delicate interchanging of integrals and infinite sums needed here, so I will set this out in full. Fix $$p,q > 0$$. For $$0 < \varepsilon < X$$ we have $\int_{\varepsilon}^X \frac{\cos^k px - \cos^k qx}{x}\,dx \; =\; \left(\int_{p\varepsilon}^{pX} - \int_{q\varepsilon}^{qX}\right) \frac{\cos^kx}{x}\,dx \; =\; \int_{qX}^{pX} \frac{\cos^kx}{x}\,dx - \int_{q\varepsilon}^{p\varepsilon}\frac{\cos^kx}{x}\,dx$ which implies that $\left|\int_{\varepsilon}^X \frac{\cos^k pq - \cos^k qx}{x}\,dx \right| \; \le \; \frac{2|p-q|}{\mathrm{min}(p,q)} \hspace{1cm} 0 < \varepsilon < X$ Thus, if we define the functions $f_k(X) \; = \; \int_0^X \frac{\cos^k px - \cos^k qx}{x}\,dx \hspace{2cm} k \in \mathbb{N}\,,\, X > 0$ then we have shown that $\big|f_k(X)\big| \; \le \; \frac{2|p-q|}{\mathrm{min}(p,q)} \hspace{2cm} k \in \mathbb{N}\,,\, X > 0$ and hence the functions $$f_k(X)$$ are uniformly bounded in both $$k$$ and $$X$$. Moreover, the well-known Frullani integrals $\int_0^\infty \frac{\cos ax - \cos bx}{x}\,dx \; = \; \ln\big(\tfrac{b}{a}\big) \hspace{2cm} \int_0^\infty \frac{\sin ax - \sin bx}{x}\,dx \; = \; 0$ which hold whenever $$a$$ and $$b$$ are nonzero with the same sign, imply that $\int_0^\infty \frac{e^{irpx} - e^{irqx}}{x}\,dx \; =\; \ln\big(\tfrac{q}{p}\big) \hspace{2cm} r \neq 0$ and hence, since $f_k(X) \; = \; 2^{-k} \int_0^X \frac{(e^{ipx} + e^{ipx})^k - (e^{iqx} + e^{-iqx})^k}{x}\,dx \; = \; 2^{-k} \sum_{r=0}^k {k \choose r}\int_0^X \frac{e^{i(2r-k)px} - e^{i(2r-k)qx}}{x}\,dx$ we deduce that $\varphi_k \; = \; \lim_{X \to \infty} f_k(X) \; = \; \left\{ \begin{array}{ll} \displaystyle 2^{-k} \sum_{r=0}^k {k \choose r} \ln\big(\tfrac{q}{p}\big) & k \mbox{ odd} \\ \displaystyle 2^{-k} \sum_{0 \le r \le k \atop r \neq \frac12k} {k \choose r}\ln\big(\tfrac{q}{p}\big) & k \mbox{ even} \end{array} \right.$ and hence that $\varphi_k \; = \; \left\{ \begin{array}{ll} \ln\big(\tfrac{q}{p}\big) & k \mbox{ odd} \\ \ln\big(\tfrac{q}{p}\big)\left(1 - 2^{-k}{k \choose \frac12k}\right) & k \mbox{ even} \end{array} \right.$ Now, for $$-1 < a < 1$$, \begin{align} \int_0^X \ln\left(\frac{1 + 2a \cos px + a^2}{1 + 2a \cos qx + a^2}\right)\,\frac{dx}{x} & = \int_0^X \left\{ \ln\big(1 + \tfrac{2a}{1+a^2}\cos px\big) -\ln\big(1 + \tfrac{2a}{1+a^2}\cos qx\big)\right\}\,\frac{dx}{x} \\ & = \int_0^X \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\big(\tfrac{2a}{a^2+1}\big)^k\big[\cos^k px - \cos^k qx\big]\,\frac{dx}{x} \\ & = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\big(\tfrac{2a}{a^2+1}\big)^k f_k(X) \end{align} where the interchange of the sum and integral in the last line is valid because $$\big|\tfrac{2a}{a^2+1}\big| < 1$$, and hence the series is uniformly convergent in $$x$$. We can now use the uniform boundedness of the functions $$f_k(X)$$ to deduce that \begin{align} \int_0^\infty \ln\left(\frac{1 + 2a \cos px + a^2}{1 + 2a \cos qx + a^2}\right)\,\frac{dx}{x} & = \lim_{X \to \infty}\int_0^X \ln\left(\frac{1 + 2a \cos px + a^2}{1 + 2a \cos qx + a^2}\right)\,\frac{dx}{x} \\ & = \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\big(\tfrac{2a}{a^2+1}\big)^k \varphi_k \end{align} Thus the integral is equal to \begin{align} \ln\big(\tfrac{q}{p}\big)&\left[ \sum_{k=1}^\infty \frac{(-1)^{k-1}}{k}\big(\tfrac{2a}{a^2+1}\big)^k + \sum_{k=1}^\infty \frac{1}{2k}\big(\tfrac{2a}{a^2+1}\big)^{2k} 2^{-2k}{2k \choose k}\right] \\ & = \ln\big(\tfrac{q}{p}\big)\left[\ln\big(1 + \tfrac{2a}{a^2+1}\big) + \tfrac12\sum_{k=1}^\infty \frac{1}{k}{2k \choose k} \big(\tfrac{a^2}{(a^2+1)^2}\big)^{k}\right] \\ & = \ln\big(\tfrac{q}{p}\big)\left[\ln\big(\tfrac{(1+a)^2}{a^2+1}\big) - \ln\left(\tfrac12\left[1 + \sqrt{1 -4\tfrac{a^2}{(a^2+1)^2}}\right]\right)\right] \\ & = 2\ln\big(\tfrac{q}{p}\big)\ln(1+a) \end{align} as required.

- 1 year, 11 months ago

(+1) Nice solution! There is a slight typo in the Frullani Integral for cosine, the integral is for $$a$$ and $$b$$ whereas the logarithm is for $$p$$ and $$q$$.

- 1 year, 11 months ago

Problem 11:

Show that $\int_0^1 \frac{t^{\alpha-1} - t^{\beta-1}}{(1+t)\ln t}\,dt \; = \; \ln\left(\frac{\Gamma\big(\tfrac12+\tfrac12\alpha\big)\Gamma\big(\tfrac12\beta\big)}{\Gamma\big(\tfrac12+\tfrac12\beta\big)\Gamma\big(\tfrac12\alpha\big)}\right)$ for all $$\alpha,\beta > 0$$.

This problem has been solved by Aditya Sharma.

- 1 year, 11 months ago

Results used : $$\displaystyle \sum_{n\ge 0}\frac{1}{n+a} =-\psi(a)$$, $$\displaystyle \int \psi(ax) = \frac{1}{a}\ln(\Gamma(ax))$$

Let $$\displaystyle F(a)=\int_0^1 \frac{x^{a-1}}{(1+x)\ln x}dx$$ then by differentiating w.r.t to a,

$$\displaystyle F'(a)=\int_0^1 \frac{x^{a-1}}{1+x}dx = \sum_{n\ge 0}(-1)^n \frac{1}{n+a}$$

Now , $$\displaystyle \sum_{n\ge 0}(-1)^n \frac{1}{n+a} = \frac{1}{2}(\sum_{n\ge 0} \frac{1}{n+\frac{a}{2}}-\sum_{n\ge 0} \frac{1}{n+\frac{a+1}{2}}) = \frac{1}{2}(\psi(\frac{a+1}{2})-\psi(\frac{a}{2}))$$

So , $$\displaystyle F'(\alpha)-F'(\beta) = \frac{1}{2}(\psi(\frac{\alpha+1}{2})-\psi(\frac{\alpha}{2}))-\frac{1}{2}(\psi(\frac{\beta+1}{2})-\psi(\frac{\beta}{2}))$$

Integrating we have, $$\displaystyle F(\alpha)-F(\beta) = \left(\ln(\frac{\Gamma(\frac{1+\alpha}{2})\Gamma(\frac{\beta}{2})}{\Gamma(\frac{1+\beta}{2})\Gamma(\frac{\alpha}{2})})\right)+C$$

Now since, $$F(1)=0\implies C=0$$

So $$\displaystyle F(\alpha)-F(\beta) = \left(\ln\frac{\Gamma(\frac{1+\alpha}{2})\Gamma(\frac{\beta}{2})}{\Gamma(\frac{1+\beta}{2})\Gamma(\frac{\alpha}{2})}\right)$$

- 1 year, 11 months ago

@Aditya Kumar @Aditya Sharma @Ishan Singh @Mark Hennings @Rohith M.Athreya and all other participants:

I have made this to make seeing old problems easier. I will complete it by tomorrow.

- 1 year, 11 months ago

really nice :)

thank you

- 1 year, 11 months ago

That's very nice of you to compile the past problems! Thanks :)

- 1 year, 11 months ago

Sure! I hope it will make things easier!

- 1 year, 11 months ago

Problem 10:

Show that $\int_0^1 (x-1) e^{-x} \ln x\,dx \; = \; \frac{e-1}{e}$

This problem has been solved by Jasper Braun.

- 1 year, 11 months ago

Solution to Problem 10:

$\displaystyle\int_0^1xe^{-x}\ln{x}-e^{-x}\ln{x}dx = \int_0^1e^{-x}\ln{x}dx+\int_0^1e^{-x}dx-\int_0^1e^{-x}\ln{x}dx = \boxed{1-\frac{1}{e}}\text{ (by IBP)}$

- 1 year, 11 months ago

You can also apply IBP this way (though almost same way),

$-\int_{0}^{1} (1-x)e^{-x} \ln x \ \mathrm{d}x = -\int_{0}^{1} \ln x \ \mathrm{d}(xe^{-x})$

- 1 year, 11 months ago

Problem 4:

Prove that $$\displaystyle \int_0^1 \frac{{\rm Li}_3^2(-x)}{x^2}\, dx = -\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{4}\zeta^2(2)+\frac{7}{4}\zeta(4)-3\zeta(3)\ln 2+6\zeta(3)-6\zeta(2)\ln 2+6\zeta(2)-12\ln^2 2$$

This problem has been solved by Aditya Kumar.

- 1 year, 11 months ago

Solution to Problem 4:

$$\displaystyle I=\int^1_0\frac{Li_3^2(-x)}{x^2}{\rm d}x$$

$I=-\frac{Li_3^2(-x)}{x}\Bigg |^1_0+\int^1_0\frac{2Li_2(-x)Li_3(-x)}{x^2}{\rm d}x=-\frac{9}{16}\zeta^2(3)-\frac{2Li_2(-x)Li_3(-x)}{x}\Bigg |^1_0+\int^1_0\frac{2Li_2^2(-x)}{x^2}{\rm d}x-\int^1_0\frac{2Li_3(-x)\ln(1+x)}{x^2}{\rm d}x$

$I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{2Li_2^2(-x)}{x}\Bigg |^1_0-\int^1_0\frac{4Li_2(-x)\ln(1+x)}{x^2}{\rm d}x+\frac{2Li_3(-x)\ln(1+x)}{x}\Bigg |^1_0-\int^1_0\frac{2Li_3(-x)}{x(1+x)}{\rm d}x-\int^1_0\frac{2Li_2(-x)\ln(1+x)}{x^2}{\rm d}x$

$I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\int^1_0\frac{2Li_3(-x)}{1+x}{\rm d}x+\frac{6Li_2(-x) \ \ln(1+x)}{x}\Bigg |^1_0-\int^1_0\frac{6Li_2(-x)}{x(1+x)}{\rm d}x+\int^1_0\frac{6\ln^2(1+x)}{x^2}{\rm d}x$

$I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\frac{9} \ {2}\zeta(3)-3\zeta(2)\ln{2}+\int^1_0\frac{6Li_2(-x)}{1+x}{\rm d}x+\int^1_0\frac{2Li_3(-x)}{1+x}{\rm d}x-\frac{6\ln^2(1+x)}{x}\Bigg |^1_0+\int^1_0\frac{12\ln(1+x)}{x(1+x)}{\rm d}x$

$I=-\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{2}\zeta^2(2)+\frac{7}{4}\zeta(4)-\frac{3}{2}\zeta(3)\ln{2}+\frac{9}{2}\zeta(3)-3\zeta(2)\ln{2}+6\zeta(2)-12\ln^2{2}+\int^1_0\frac{6Li_2(-x)}{1+x}{\rm d}x+\int^1_0\frac{2Li_3(-x)}{1+x}{\rm d}x$

Now,

$I_1=\int^1_0\frac{6Li_2(-x)}{1+x}{\rm d}x=6Li_2(-x)\ln(1+x)\Bigg |^1_0+\int^1_0\frac{6\ln^2(1+x)}{x}{\rm d}x$

$$I_1=\frac{3}{2}\zeta(3)-3\zeta(2)\ln{2}$$

$I_2=\int^1_0\frac{2Li_3(-x)}{1+x}{\rm d}x=2Li_3(-x)\ln(1+x)\Bigg |^1_0-\int^1_0\underbrace{\frac{2Li_2(-x) \ln(1+x)}{x} {\rm d}x}_{\displaystyle\small{2Li_2(-x){\rm d}Li_2(-x)}}$

$$I_2=\frac{1}{4}\zeta^2(2)-\frac{3}{2}\zeta(3)\ln{2}$$

Therefore,

$\boxed{\displaystyle \int_0^1 \frac{{\rm Li}_3^2(-x)}{x^2}\, dx = -\frac{9}{16}\zeta^2(3)-\frac{3}{4}\zeta(2)\zeta(3)-\frac{1}{4}\zeta^2(2)+\frac{7}{4}\zeta(4)-3\zeta(3)\ln 2+6\zeta(3)-6\zeta(2)\ln 2+6\zeta(2)-12\ln^2 2}$

- 1 year, 11 months ago

Problem 03:

Show that $$\displaystyle \int_0^1 \frac{\ln x \ln^2(1+x)}{x}\, dx = \dfrac{\pi^4}{24}-4{\rm Li}_4\left(\frac{1}{2}\right)-\frac{7\zeta(3)\ln 2}{2}+\frac{\pi^2\ln^2 2}{6}-\frac{\ln^4 2}{6}$$.

This problem has been solved by Jasper Braun.

- 1 year, 11 months ago

Solution to Problem 3:

$$\displaystyle H_n^{(k)} = \sum_{j=1}^{\infty}\frac{H_n^{(k)}x^i}{i^n}\quad H(x) = \frac{-\ln{(1-x)}}{1-x}$$

$\displaystyle I = \int_{0}^{\infty}\frac{\ln^2{(1+x)}\ln{x}}{x}dx = -\int_{0}^{1}\frac{\ln{(1+x)}\ln^2{x}}{1+x}dx=\quad\text{by parts}$

$\displaystyle -\int_{1}^{2}\frac{\ln^2{(x-1)}\ln{x}}{x}dx=\quad\text{(by } u = 1+x)$

$\displaystyle -\int_{\frac{1}{2}}^{1}\frac{\ln^2{(\frac{1}{x}-1)}\ln{\frac{1}{x}}}{x}dx=\int_{\frac{1}{2}}^{1}\frac{\ln{x}\ln^2{(1-x)}}{x}-\frac{2\ln^2{x}\ln{(1-t)}}{x}+\frac{\ln^3{x}}{x}dx\quad\text{(by }u = 1/x)$

$\displaystyle\int\frac{\ln{x}\ln^2{(1-x)}}{x} = \frac{\ln^2{x}}{2}\ln^2{(1-x)}+\int\frac{\ln^2{x}\ln{(1-x)}}{1-x}dx$

$\displaystyle \int\frac{\ln^2{x}\ln{(1-x)}}{1-x}dx = -\sum_{n=1}^{\infty}H_n\frac{d^2}{dn^2}\int x^ndx = -\sum_{n=1}^{\infty}H_n\big(\frac{x^{n+1}\ln^2{x}}{n+1}-\frac{2x^{n+1}\ln{x}}{(1+n)^2}+\frac{2x^{n+1}}{(1+n)^3}\big)=$

Now use $$H_n = H_{n+1} - \frac{1}{n+1}$$ to obtain:

$\displaystyle -\Big(H_1(x)\ln^2{x}-Li_2(x)\ln^2{x}-2H_2(x)\ln{x}+2Li_3(x)\ln{x}+2H_3(x)-2Li_4(x)\Big)$

$\displaystyle\textbf{Next }-2\int\frac{\ln^2{x}\ln{(1-x)}}{x}dx = 2\sum_{n=1}^{\infty}\frac{1}{n}\frac{d^2}{dn^2}\int x^{n-1} =$

$\displaystyle 2\sum_{n=1}^{\infty}\frac{x^n\ln^2{x}}{n^2}-\frac{2x^n\ln{x}}{n^3}+\frac{2x^n}{n^4} = 2\big(Li_2(x)\ln^2{x}-2Li_3(x)\ln{x}+2Li_4(x)\big)$

$\displaystyle I = \frac{\ln^4{x}}{4}\Big|^1_{\frac{1}{2}} + \frac{\ln^2{x}}{2}\ln^2{(1-x)}\Big|^1_{\frac{1}{2}}-2H_3(1)+2Li_4(1)+$

$\displaystyle +H_1(\frac{1}{2})\ln^2{2}-Li_2(\frac{1}{2})\ln^2{2}+2H_2(\frac{1}{2})\ln{2}-2Li_3(\frac{1}{2})\ln{2}+2H_3(\frac{1}{2})-2Li_4(\frac{1}{2})+$

$\displaystyle4Li_4(1)-2Li_2(\frac{1}{2})\ln^2{2}-4Li_3(\frac{1}{2})\ln{2}-4Li_4(\frac{1}{2}) =$

$\displaystyle\frac{-3ln^4{2}}{4}-2H_3(1)+H_1(\frac{1}{2})\ln^2{2}+2H_2(\frac{1}{2})\ln{2}+2H_3(\frac{1}{2})-6Li_4(\frac{1}{2})+6Li_4(1)-3Li_2(\frac{1}{2})\ln^2{2}-6Li_3(\frac{1}{2})\ln{2}$

Now using this for the evaluation of $$H_n(x)$$

$\displaystyle -2H_3(1) = \frac{-\pi^4}{36},\ln^2{2}H_1(\frac{1}{2})=\frac{\ln^4{2}}{2}+\ln^2{2}Li_2{\frac{1}{2}},2\ln{2}H_2{\frac{1}{2}} = 2\ln{2}\zeta(3)-\zeta(2)\ln^2{2},2H_3(\frac{1}{2})= \frac{\pi^4}{360}+\frac{\ln^4{2}}{12}-\frac{\zeta(3)\ln{2}}{4}+2Li_4(\frac{1}{2})$

Adding all this together and using formulas for $$Li_3$$ and $$Li_2$$(I have checked on a calculator) gets the desired answer. Anyone may post the next problem.

- 1 year, 11 months ago

Problem 23:

Evaluate the indefinite integral, $$\displaystyle \large \int \frac{x}{((1+x^{2})^{1012} (2+x^2)^{3012})^{\frac{1}{2012}}} \, dx$$.

This problem has been solved by Mark Hennings.

- 1 year, 10 months ago

Start with the substitution $$y = x^2 + 1$$. Then \begin{align} I_{p}(x) & = \int \frac{x\,dx}{(1 + x^2)^{1-p}(2 + x^2)^{1 + p}} \; = \; \tfrac12\int \frac{dy}{y^{1-p}(1 + y)^{1+p}} \\ & = \tfrac12 \int \left(\frac{y}{1+y}\right)^{p-1}\,\frac{dy}{(1+y)^2} \end{align} Then the substitution $$z = \frac{y}{1+y}$$ gives $I_p(x) \; = \; \tfrac12\int z^{p-1}\,dz \; = \; \tfrac{1}{2p}z^p + c \; = \; \tfrac{1}{2p}\left(\tfrac{x^2+1}{x^2+2}\right)^p + c$ In this case, we have $$p = \tfrac{1000}{2012} = \tfrac{250}{503}$$, and hence $I_{\frac{250}{503}}(x) \; = \; \tfrac{503}{500} \left(\tfrac{x^2+1}{x^2+2}\right)^{\frac{250}{503}} + c$

- 1 year, 10 months ago

Problem 22:

Evaluate $\int_0^{\frac14\pi} \ln(1 + \tan x)\,dx$

This problem has been solved by Rohinth M.Athreya.

- 1 year, 10 months ago

$$\displaystyle \large \int_{0}^{\frac{\pi}{4}} ln(1+\tan(\frac{\pi}{4} - x)\, dx$$

$$\displaystyle \large \int_{0}^{\frac{\pi}{4}} ln(\frac{2}{1+\tan x}\,dx$$

if we denote the integral by $$I$$. it is also $$\frac{\pi ln2}{4} -I$$

and so the integral is $$\frac{\pi ln2}{8}$$.

- 1 year, 10 months ago

Problem 17 :

Find a closed form for $$\displaystyle \int_0^\infty \frac{\sin^{2n+1}x}{x}\, dx$$ for $$\, n\ge 0$$

This problem has been solved by Ishan SIngh

- 1 year, 11 months ago

Since,

$\sin x = \dfrac{e^{ix} - e^{-ix}}{2i}$

We have,

$\sin^{2n+1} x = \dfrac{(-1)^n}{2^{2n}} \sum_{r=0}^{n} (-1)^r \dbinom{2n+1}{r} \sin (2r+1)x$

$\implies \int_{0}^{\infty} \dfrac{\sin^{2n+1} x}{x} \ \mathrm{d}x = \dfrac{(-1)^n}{2^{2n}} \sum_{r=0}^{n} (-1)^r \dbinom{2n+1}{r} \int_{0}^{\infty} \dfrac{\sin (2r+1)x}{x} \ \mathrm{d}x$

$= \dfrac{(-1)^n \pi}{2^{2n+1}} \sum_{r=0}^{n} (-1)^r \dbinom{2n+1}{r}$

$= \dfrac{(-1)^n \pi}{2^{2n+1}} \sum_{r=0}^{n} \left( (-1)^r \dbinom{2n}{r} - (-1)^{r-1} \dbinom{2n}{r-1} \right)$

Clearly, the above sum telescopes. Evaluating it, we have,

$\int_{0}^{\infty} \dfrac{\sin^{2n+1} x}{x} \ \mathrm{d}x = \dfrac{(-1)^n \pi}{2^{2n+1}} (-1)^n \dbinom{2n}{n}$

$\therefore \int_{0}^{\infty} \dfrac{\sin^{2n+1} x}{x} \ \mathrm{d}x = \dfrac{\pi}{2^{2n+1}} \dbinom{2n}{n} \ \forall \ n \in \mathbb{Z}^+$

- 1 year, 11 months ago

As an alternative,

$$\displaystyle I=\int_0^\infty \frac{\sin^{2n+1}x}{x}\,dx = \int_0^\infty\int_0^\infty e^{-xy}\sin^{2n+1}x\,dx\,dy$$

Now by IBP two times we can create a recurrence relation and thus evaluate,

$$\displaystyle J_n = \int_0^\infty e^{-ax}\sin^{2n+1}x\,dx = \frac{(2n+1)!}{\prod_{r=1}^{n}(a^2 +(2r+1)^2)}$$

Thus, $$\displaystyle I=(2n+1)!\int_0^\infty \frac{dy}{\prod_{r=1}^{n}(a^2 +(2r+1)^2)}$$

The expression $$\displaystyle \frac{1}{\prod_{r=1}^{n}(a^2 +(2r+1)^2)} = \sum_{r=0}^n \frac{A_r}{y^2+(2r+1)^2}$$

where the coefficients can be determined by the cover-up rule as $$\displaystyle A_k = \frac{(-1)^k}{2^{2n}}\frac{2k+1}{(n-k)!(n+k+1)!}$$

So, $$\displaystyle I=(2n+1)!\sum_{k=0}^n \int_0^\infty \frac{A_k}{y^2+(2k+1)^2}\,dy = \frac{\pi}{2^{2n+1}} \sum_{k=0}^{n} (-1)^k \binom{2n+1}{n-k} = \frac{\pi}{2^{2n+1}}\binom{2n}{n}$$

- 1 year, 11 months ago

That technique looks familiar!

- 1 year, 11 months ago

(+1) Nice ! I have another method so all total we could have atleast 2 real methods to evaluate this.

- 1 year, 11 months ago

It is interesting to note that the sum $\sum_{r=0}^m (-1)^r {n \choose r} \; = \; (-1)^m{n-1 \choose m} \hspace{2cm} 0 \le m < n$ can be proved by considering the coefficients of $$x^m$$ on both sides of the identity $(1 + x)^{n-1} \; = \; (1 + x)^{-1} \times (1+x)^n \hspace{1cm} |x| < 1 \;.$

- 1 year, 11 months ago

Problem 15 : Prove That

$\int_{0}^{\infty} \dfrac{\cos (2nx)}{\cosh^{2m} x} \mathrm{d}x = \dfrac{\sqrt{\pi}}{2} \dfrac{|\Gamma(m+in)|^2}{\Gamma(m) \Gamma \left(m + \dfrac{1}{2} \right)} \ ; \ (m,n) \in \mathbb{R}^+$

This problem has been solved by Mark Hennings.

- 1 year, 11 months ago

Note that \begin{align} B(z,w) & = \int_0^\infty \frac{t^{z-1}}{(1+t)^{z+w}}\,dt \; = \; \left(\int_0^1 + \int_1^\infty\right) \frac{t^{z-1}}{(1+t)^{z+w}}\,dt \\ & = \int_0^1 \frac{t^{z-1}}{(1+t)^{z+w}}\,dt + \int_1^0 \frac{t^{1-z}}{(1+t^{-1})^{z+w}} \big(-t^{-2}\,dt\big) \\ & = \int_0^1 \frac{t^{z-1} + t^{w-1}}{(1+t)^{z+w}}\,dt \end{align} for all $$\mathfrak{Re}\,z\,,\,\mathfrak{Re}\,w \,>\, 0$$, so that (with the substitution $$t = e^{-2x}$$) \begin{align} B(m+in,m-in) & = \int_0^1 \frac{t^{m+in-1} + t^{m-in-1}}{(1+t)^{2m}}\,dt \; = \; \int_0^1 \frac{t^{m-1}(t^{in} + t^{-in})}{(1 + t)^{2m}}\,dt \\ & = \int_\infty^0 \frac{e^{2(1-m)x} \big(e^{-2inx} + e^{2inx}\big)}{(1 + e^{-2x})^{2m}}\,(-2e^{-2x})\,dx \\ & = 4\int_0^\infty \frac{e^{-2mx} \cos2nx}{(1 + e^{-2x})^{2m}}\,dx \; = \; 4\int_0^\infty \frac{\cos 2nx}{(e^x + e^{-x})^{2m}}\,dx \\ & = 4^{1-m}\int_0^\infty \frac{\cos 2nx}{\cosh^{2m}x}\,dx \end{align} for any real $$n$$ and $$m > 0$$, so that \begin{align} \int_0^\infty \frac{\cos 2nx}{\cosh^{2m}x}\,dx & = 4^{m-1} B(m+in,m-in) \; = \; \frac{4^{m-1}}{\Gamma(2m)} \Gamma(m+in) \Gamma(m-in) \\ & = \frac{\sqrt{\pi}}{2\Gamma(m)\Gamma(m+\frac12)} \big|\Gamma(m+in)\big|^2 \end{align} for any real $$n$$ and any $$m > 0$$, using the duplication formula $\Gamma(2m) \;= \; \tfrac{1}{\sqrt{2\pi}} 2^{2m - \frac12} \Gamma(m) \Gamma(m+\tfrac12)$

- 1 year, 11 months ago

Problem 12

Show that $$\displaystyle \int_0^\infty \left(\frac{\ln x\arctan x}{x}\right)^2 dx = \pi^3\frac{\ln 2}{4}-\frac{\pi^3}{12}+2\pi\ln 2-\frac{\pi}{4}\zeta(3)$$

This problem has been solved by Mark Hennings

- 1 year, 11 months ago

Consider the principal branch of the logarithm, so that the plane is cut along the negative real axis. Suppose that $$a > 0$$. If we integrate $f(z) \; = \; \frac{(\log z)^2}{z^2 + a^2}$ around the contour $$\gamma_1 + \gamma_2 - \gamma_3 - \gamma_4$$, where

• $$\gamma_1$$ is the straight line $$z = x$$ for $$\varepsilon < x < R$$,
• $$\gamma_2$$ is the semicircular arc $$z = Re^{i\theta}$$ for $$0 < \theta < \pi$$,
• $$\gamma_3$$ is the straight line $$z = xe^{i\pi}$$ for $$\varepsilon < x < R$$, just above the cut,
• $$\gamma_4$$ is the semicircular arc $$z = \varepsilon e^{i\theta}$$ for $$0 < \theta < \pi$$,

where $$0 < \varepsilon < a < R$$. Then $\int_{\gamma_1} f(z)\,dz \; = \; \int_\varepsilon^R \frac{\ln^2x}{x^2 + a^2}\,dx \hspace{2cm} \int_{\gamma_3} f(z)\,dz \; = \; -\int_\varepsilon^R \frac{(\ln x + i\pi)^2}{x^2 + a^2}\,dx$ while $\lim_{\varepsilon \to 0}\int_{\gamma_4} f(z)\,dz \; = \; \lim_{R \to \infty}\int_{\gamma_2} f(z)\,dz \; = \; 0$ and hence we deduce that $\int_0^\infty \frac{\ln^2x + (\ln x + i\pi)^2}{x^2+a^2}\,dx \; = \; 2\pi i \mathrm{Res}_{z=ia} f(z) \; = \; \tfrac{\pi}{a}(\ln a + \tfrac12\pi i)^2$ so that $\int_0^\infty \frac{\ln^2x}{x^2 + a^2}\,dx \; = \; \tfrac{\pi}{2a}\,\ln^2a + \tfrac{\pi^3}{8a} \hspace{2cm} \int_0^\infty \frac{\ln x}{x^2 + a^2}\,dx \; = \; \tfrac{\pi}{2a}\ln a$

Now $\frac{\tan^{-1}x}{x} \; =\; \int_0^1 \frac{du}{1 + x^2u^2}$ and hence \begin{align} \int_0^\infty \left(\frac{\tan^{-1}x \ln x}{x}\right)^2\,dx & = \int_0^1 \int_0^1 \left(\int_0^\infty \frac{\ln^2x}{(1 + x^2u^2)(1 + x^2v^2)}\,dx\right)\,du\,dv \\ & = \int_0^1 \int_0^1 \left(\int_0^\infty \ln^2x\left\{ \frac{1}{x^2 + u^{-2}} - \frac{1}{x^2 + v^{-2}}\right\}\,dx \right)\,\frac{du\,dv}{u^2-v^2} \\ & = \int_0^1 \int_0^1 \left\{ \tfrac12\pi u\,\ln^2u + \tfrac18\pi^3u - \tfrac12\pi v\, \ln^2v - \tfrac18\pi^3v\right\} \frac{du\,dv}{u^2 - v^2} \\ & = \tfrac12\pi \int_0^1 \int_0^1 \frac{u \ln^2u - v\ln^2v}{u^2 - v^2}\,du\,dv + \tfrac18\pi^3 \int_0^1 \int_0^1 \frac{du\,dv}{u+v} \\ & = \tfrac12\pi\left\{ -\tfrac16\pi^2 + 4\ln2 - \tfrac12\zeta(3)\right\} + \tfrac18\pi^3 \times 2\ln2 \\ & = \pi\left(\tfrac14\pi^2\ln2 - \tfrac{1}{12}\pi^2 + 2\ln2 - \tfrac14\zeta(3)\right) \end{align} I initially relied on Mathematica to evaluate the double integral $\int_0^1\int_0^1 \frac{u\ln^2u - v\ln^2v}{u^2 - v^2}\,du\,dv$ but here is a proper derivation.

\begin{align} \int_0^1 \int_0^1 \frac{u\ln^2u - v\ln^2v}{u^2 - v^2}\,du\,dv & = 2\int_0^1 \,du \int_0^u\,dv \frac{u\ln^2u - v\ln^2v}{u^2-v^2} \\ & = 2\int_0^1\,du \int_0^u\,dv \frac{1}{u^2-v^2}\int_v^u (2\ln w + \ln^2w)\,dw \\ & = \int\int\int_{0 < v < w < u < 1}\frac{2\ln w + \ln^2w}{u}\left(\frac{1}{u+v} + \frac{1}{u-v}\right) \\ & = \iint_{0 < w < u < 1} \frac{2\ln w + \ln^2w}{u}\ln\left(\tfrac{u+w}{u-w}\right)\,dw\,du \end{align} Since it is easy to check (by differentiating back again) that $\int (2\ln w + \ln^2w)\ln\left(\tfrac{u+w}{u-w}\right)\,dw \; = \; \begin{array}{l} \displaystyle u \ln^2w \ln\big(1 + \tfrac{w}{u}\big) + w\ln^2w \ln\big(\tfrac{u+w}{u-w}\big) \\ \displaystyle + u\ln^2w\ln\big(1 - \tfrac{w}{u}\big) + 2u\ln w \mathrm{Li}_2\big(-\tfrac{w}{u}\big) + 2u \ln w \mathrm{Li}_2\big(\tfrac{w}{u}\big) \\ \displaystyle - 2u\mathrm{Li}_3\big(-\tfrac{w}{u}\big) - 2u\mathrm{Li}_3\big(\tfrac{w}{u}\big) + c\end{array}$ it follows that $\int_0^u (2 \ln w + \ln^2w)\ln\left(\tfrac{u+w}{u-w}\right)\,dw \; = \; \tfrac16 u \left(\pi^2 \ln u + 12 \ln2 \ln^2u - 3 \zeta(3)\right)$ and so $\int_0^1 \int_0^1 \frac{u\ln^2u - v\ln^2v}{u^2 - v^2}\,du\,dv \; = \; \tfrac16\int_0^1\left(\pi^2 \ln u + 12 \ln2 \ln^2u - 3 \zeta(3)\right)\,du \; = \; -\tfrac16\pi^2 + 4\ln2 - \tfrac12\zeta(3)$ as required.

- 1 year, 11 months ago

Comment deleted Dec 18, 2016

Out of interest - what was your technique?

- 1 year, 11 months ago

It was almost same with yours except for the complex analysis part & the evaluation of the double integral. I did it like,

With a substitution $$xy=a^2$$ the integral $$\displaystyle \int_0^\infty \frac{\ln x}{x^2+a^2}dx$$ transforms into,

$$\displaystyle \int_0^\infty \frac{\ln x}{x^2+a^2}dx = \int_0^\infty \frac{\ln a}{x^2+a^2}dx-\int_0^\infty \frac{\ln x}{x^2+a^2}dx$$ and so $$\displaystyle \int_0^\infty \frac{\ln x}{x^2+a^2}dx = \frac{\pi}{2a}\ln a$$

With the same substitution the integral $$\displaystyle \int_0^\infty \frac{\ln^2 x}{x^2+a^2}dx$$ can be calculated.

The double integral encountered $$\displaystyle J=\int_0^1\int_0^1 \frac{x\ln^2 x-y\ln^2 y}{x^2-y^2}\,dx\,dy$$ here was symmetric and thus it equals,

$$\displaystyle J=2\int_0^1\int_0^1 \frac{x\ln^2 x}{x^2-y^2} \,dx \,dy\ = \int_0^1 \ln^2 x \ln\left(\frac{1+x}{1-x}\right) \space dx = \int_0^1 \ln^2 x\ln(1+x) dx - \int_0^1 \ln^2 x\ln(1-x)$$

Both of $$\displaystyle \int_0^1 \ln^2 x \ln(1+x) dx$$ & $$\displaystyle \int_0^1 \ln^2 x \ln(1-x)dx$$ can be evaluated by simple techniques and that gives the result .

- 1 year, 11 months ago

I am not clear about your statement that the substitution $$x = a^2/y$$ works for $$\ln^2x$$, since we get $\int_0^\infty \frac{\ln^2x}{x^2+a^2}\,dx \; = \; \int_0^\infty \frac{(2\ln a - \ln y)^2}{y^2 + a^2}\,dy$ and the integrals involving $$\ln^2x$$ just cancel out, and we retrieve the previous formula for the $$\ln x$$ integral.

There is also a problem with the way you calculate $$J$$, since the integral $\int_0^1 \int_0^1 \frac{x\ln^2x}{x^2-y^2}\,dy\,dx$ does not converge. When you try to calculate the integral $\int_0^1 \frac{2x}{x^2 - y^2}\,dy$ as $\ln\tfrac{1+x}{1-x}$ you are integrating right through the singularity this function of $$y$$ has at $$y=x$$. Perhaps this integral exists as a principal value integral, i.e. as $\lim_{\varepsilon \to 0} \left(\int_0^{x-\varepsilon} + \int_{x + \varepsilon}^1\right) \frac{2x}{x^2 - y^2}\,dy$ but it is not at all obvious why this should be the right thing to do.

It is weird, but we need the two singular terms together, so that their separate singularities cancel out, leaving an integral we can evaluate.

- 1 year, 11 months ago

Yes I see what's wrong with my approach however producing the desired result. I will try to focus on these deeper arguments , I really wasn't enough attentive to note that singularity and convergence part here. However I learnt something new here and I'm happy about it :)

- 1 year, 11 months ago

Problem 9:

Prove $\displaystyle\int_0^1Li_2^2(x)dx = \frac{\pi^4}{36}-4\zeta(3)-2\zeta(2)+6$

This problem has been solved by Mark Hennings.

- 1 year, 11 months ago

I will be using the Euler result that $\sum_{u=1}^\infty\frac{H_u}{u^2} \; = \; 2\zeta(3)$ Integrating by parts, \begin{align} \int_0^1 \mathrm{Li}_2^2(x)\,dx & = \zeta(2)^2 - 2\int_0^1 \mathrm{Li}_1(x)\mathrm{Li}_2(x)\,dx \; = \; \zeta(2)^2 - 2\sum_{r,s=1}^\infty \frac{1}{rs^2}\int_0^1 x^{r+s}\,dx \\ & = \zeta(2)^2 - 2\sum_{r,s=1}^\infty \frac{1}{rs^2(r+s+1)} \; = \; \zeta(2)^2 - 2\sum_{r,s=1}^\infty\frac{1}{s^2(s+1)}\left(\frac{1}{r} - \frac{1}{r+s+1}\right) \\ &= \zeta(2)^2 - 2\sum_{s=1}^\infty \frac{H_{s+1}}{s^2(s+1)} \; = \; \zeta(2)^2 - 2\sum_{s=1}^\infty H_{s+1}\left(\frac{1}{s^2} - \frac{1}{s(s+1)}\right) \\ &= \zeta(2)^2 - 2\sum_{s=1}^\infty \frac{H_{s+1}}{s^2} + 2\sum_{s=1}^\infty \frac{H_{s+1}}{s(s+1)} \\ & = \zeta(2)^2 - 2\sum_{s=1}^\infty \frac{H_s}{s^2} - 2\sum_{s=1}^\infty \frac{1}{s^2(s+1)} + 2\sum_{s=1}^\infty H_{s+1}\left(\frac{1}{s} - \frac{1}{s+1}\right) \\ & = \zeta(2)^2 - 4\zeta(3) - 2\sum_{s=1}^\infty \left(\frac{1}{s^2} - \frac{1}{s(s+1)}\right) + 2\lim_{S \to \infty}\left(\sum_{s=1}^S \frac{H_{s+1}}{s} - \sum_{s=2}^{S+1} \frac{H_s}{s}\right) \\ &= \zeta(2)^2 - 4\zeta(3) - 2\zeta(2) + 2 + 2\lim_{S \to \infty}\left(\tfrac32 + \sum_{s=2}^S \frac{1}{s(s+1)} - \frac{H_{S+1}}{S}\right) \\ &= \zeta(2)^2 - 4\zeta(3) - 2\zeta(2) + 6 \end{align} as required.

- 1 year, 11 months ago

Problem 7:

Show that for $$n\ge2$$, $$\displaystyle \int_0^\infty \frac{\ln x}{x^n+1}dx = -\frac{\pi^2}{n^2}\cot\left(\frac{\pi}{n}\right)\csc\left(\frac{\pi}{n}\right)$$

This problem has been solved by Mark Hennings.

- 1 year, 11 months ago

Solution to Problem 7:

Consider the integrals of $f(z) \; = \; \frac{1}{z^n + 1} \hspace{1cm} g(z) \; = \; \frac{\log z}{z^n + 1}$ around the contour $$\gamma_1 + \gamma_2 - \gamma_3$$, where

• $$\gamma_1$$ is the line segment $$z \,=\, x$$, for $$0 < x < R$$,
• $$\gamma_2$$ is the circular contour $$z \,=\, Re^{i\theta}$$ for $$0 < \theta < \frac{2\pi}{n}$$,
• $$\gamma_3$$ the the line segment $$z \,=\, x e^{\frac{2\pi i}{n}}$$, for $$0 < x < R$$.

where $$R > 1$$. Now \begin{align} \int_{\gamma_1} f(z)\,dz & = \int_0^R \frac{1}{x^n+1}\,dx \\ \int_{\gamma_3} f(z)\,dz & = e^{\frac{2 \pi i}{n}}\int_0^R \frac{1}{x^n + 1}\,dx \\ \left(\int_{\gamma_1} - \int_{\gamma_3}\right)\,f(z)\,dz & = -2ie^{\frac{\pi i}{n}}\sin\frac{\pi}{n}\int_0^R \frac{1}{x^n + 1}\,dx \\ \int_{\gamma_1} g(z)\,dz & = \int_0^R \frac{\ln x}{x^n+1}\,dx \\ \int_{\gamma_3} g(z)\,dz & = e^{\frac{2 \pi i}{n}}\int_0^R \frac{\ln x + i\frac{\pi}{n}}{x^n + 1}\,dx \\ \left(\int_{\gamma_1} - \int_{\gamma_3}\right)\,g(z)\,dz & = -2ie^{\frac{\pi i}{n}}\sin\frac{\pi}{n}\int_0^R \frac{\ln x}{x^n + 1}\,dx - i\frac{\pi}{n}e^{\frac{2\pi i}{n}}\int_0^R \frac{1}{x^n + 1}\,dx \end{align} Since $\int_{\gamma_2} f(z)\,dz \; = \; O\big(R^{1-n}\big) \hspace{2cm} \int_{\gamma_2} g(z)\,dz \; = \; O\big(\ln R \,R^{1-n}\big)$ as $$R \to \infty$$, we deduce that (since both $$f(z)$$ and $$g(z)$$ just have simple poles at $$e^{\frac{\pi i}{n}}$$ inside this contour) \begin{align} \displaystyle -2ie^{\frac{\pi i}{n}}\sin\frac{\pi}{n}\int_0^\infty \frac{1}{x^n + 1}\,dx & = \displaystyle 2\pi i \mathrm{Res}_{z = e^{\frac{\pi i}{n}}} f(z) \; = \; -\frac{2\pi i}{n}e^{\frac{\pi i}{n}} \\ \displaystyle -2ie^{\frac{\pi i}{n}}\sin\frac{\pi}{n}\int_0^R \frac{\ln x}{x^n + 1}\,dx - i\frac{\pi}{n}e^{\frac{2\pi i}{n}}\int_0^R \frac{1}{x^n + 1}\,dx & =\displaystyle 2\pi i \mathrm{Res}_{z = e^{\frac{\pi i}{n}}} g(z) \; = \; \frac{2\pi ^2}{n^2}e^{\frac{\pi i}{n}} \end{align} and so $\int_0^\infty \frac{1}{x^n + 1}\,dx \; = \; \frac{\pi}{n}\mathrm{cosec}\,\frac{\pi}{n} \hspace{2cm} \int_0^\infty \frac{\ln x}{x^n + 1}\,dx \; = \; -\frac{\pi^2}{n^2}\, \cot\frac{\pi}{n}\,\mathrm{cosec}\frac{\pi}{n}$

- 1 year, 11 months ago

PROBLEM 24:

Show that $\int_0^\infty e^{-ax^2 - bx^{-2}}\,dx \; = \; \tfrac12\sqrt{\tfrac{\pi}{a}} e^{-2\sqrt{ab}}$ for any $$a,b > 0$$.

- 1 year, 10 months ago

Let,

$\text{I} = \int_{0}^{\infty} e^{-ax^2 - \frac{b}{x^{2}}} \ \mathrm{d}x$

$= \dfrac{1}{2} \int_{-\infty}^{\infty} e^{-ax^2 - \frac{b}{x^{2}}} \ \mathrm{d}x$

Sunstituting $$x \mapsto \dfrac{x}{\sqrt{a}}$$, we have,

$\text{I}= \dfrac{1}{2\sqrt{a}} \int_{-\infty}^{\infty} e^{-x^2 - \frac{ab}{x^{2}}} \ \mathrm{d}x$

$= \dfrac{e^{-2\sqrt{ab}}}{2\sqrt{a}} \int_{-\infty}^{\infty} e^{- \left(x - \frac{\sqrt{ab}}{x} \right)^2 } \ \mathrm{d}x$

Using Glasser's Master Theorem, we have,

$\text{I} = \dfrac{e^{-2\sqrt{ab}}}{2\sqrt{a}} \int_{-\infty}^{\infty} e^{-x^2 } \ \mathrm{d}x$

$= \dfrac{1}{2} \sqrt{\dfrac{\pi}{a}} e^{-2\sqrt{ab}} \quad \square$

- 1 year, 10 months ago

You don't need a master theorem!

Completing the square $\int_0^\infty e^{-a x^2 - b x^{-2}}\,dx \; = \; e^{-2\sqrt{ab}}\int_0^\infty e^{-(\sqrt{a} x - \sqrt{b} x^{-1})^2}\,dx$ Now the substitution $$y \,=\, \sqrt{\tfrac{b}{a}} x^{-1}$$ yields \begin{align} \int_0^\infty e^{-(\sqrt{a}x - \sqrt{b} x^{-1})^2}\,dx & = \int_\infty^0 e^{-(\sqrt{b}y - \sqrt{a}y)^2} \times \big(-\sqrt{\tfrac{b}{a}}y^{-2}\big)\,dy \\ & = \sqrt{\tfrac{b}{a}}\int_0^\infty x^{-2} e^{-(\sqrt{a}x - \sqrt{b}x^{-1})^2}\,dy \\ & = \frac12\int_0^\infty \left(1 + \sqrt{\tfrac{b}{a}}x^{-2}\right) e^{-(\sqrt{a}x - \sqrt{b}x^{-1})^2}\,dx \\ & = \frac{1}{2\sqrt{a}}\int_0^\infty \left(\sqrt{a} + \sqrt{b}x^{-2}\right) e^{-(\sqrt{a} x -\sqrt{b}x^{-1})^2}\,dx \end{align} The further substitution $$u = \sqrt{a}x - \sqrt{b}x^{-1}$$ gives $\int_0^\infty e^{-(\sqrt{a}x - \sqrt{b} x^{-1})^2}\,dx \; = \; \frac{1}{2\sqrt{a}}\int_{\mathbb{R}} e^{-u^2}\,du \; = \; \tfrac12\sqrt{\tfrac{\pi}{a}}$ which implies that $\int_0^\infty e^{-a x^2 - b x^{-2}}\,dx \; = \; \tfrac12\sqrt{\tfrac{\pi}{a}}e^{-2\sqrt{ab}}$

- 1 year, 10 months ago

I know. I have solved it that way also, but I thought solving it this way gives it a new perspective and also introduces to a new Theorem.

- 1 year, 10 months ago

Problem 21:

Evaluate $$\displaystyle\int_0^\frac{\pi}{2}\ln^2{\sin{x}}\ln^2{\cos{x}}\,dx$$

This problem has been solved by Mark Hennings.

- 1 year, 10 months ago

Start with $f(a,b) \; = \; \int_0^{\frac12\pi} \sin^{a-1}x \cos^{b-1}x\,dx \; = \; \tfrac12B\big(\tfrac12a,\tfrac12b\big) \; = \; \frac{\Gamma(\frac12a)\Gamma(\frac12b)}{2\Gamma(\frac12a+\frac12b)}$ for $$a,b > 0$$. Then \begin{align} \frac{\partial f}{\partial a} & = \tfrac12 f(a,b)\Big[\psi\big(\tfrac12a\big) - \psi\big(\tfrac12a+\tfrac12b\big)\Big] \\ \frac{\partial^2 f}{\partial a^2} & = \tfrac14 f(a,b) \Big[ \psi'\big(\tfrac12a\big) - \psi'\big(\tfrac12a+\tfrac12b\big) + \big[\psi\big(\tfrac12a\big) - \psi\big(\tfrac12a+\tfrac12b\big)\big]^2\Big] \end{align} and so $g(b) \; = \; \frac{\partial^2 f}{\partial a^2}(1,b) \; = \; \tfrac18 A(b)B(b)$ where $A(b) \; = \; \frac{\Gamma(\frac12)\Gamma(\frac12b)}{\Gamma(\frac12+\frac12b)} \; = \; \frac{\sqrt{\pi}\Gamma(\frac12b)}{\Gamma(\frac12+\frac12b)} \hspace{1cm} B(b) \; = \; \psi'\big(\tfrac12\big) - \psi'\big(\tfrac12+\tfrac12b\big) + \big[\psi\big(\tfrac12\big) - \psi\big(\tfrac12+\tfrac12b\big)\big]^2$ Thus \begin{align} A'(b) & = \tfrac12A(b)\big[\psi\big(\tfrac12b\big) - \psi\big(\tfrac12+\tfrac12b\big)\big] \\ A''(b) & = \tfrac14A(b) \Big[ \psi'\big(\tfrac12b\big) - \psi'\big(\tfrac12 + \tfrac12b\big) + \big[\psi\big(\tfrac12b\big) - \psi\big(\tfrac12+\tfrac12b\big)\big]^2\Big] \end{align} and hence $A(1) \; = \; \pi \hspace{1.5cm} A'(1) \; = \; -\pi \ln2 \hspace{1.5cm} A''(1) \; = \; \tfrac{1}{12}\pi^3 + \pi(\ln2)^2$ Also \begin{align} B'(b) & = -\tfrac12\psi''\big(\tfrac12+\tfrac12b\big) - \big[\psi\big(\tfrac12\big) - \psi\big(\tfrac12+\tfrac12b\big)\big]\psi'\big(\tfrac12+\tfrac12b\big) \\ B''(b) & = -\tfrac14\psi'''\big(\tfrac12+\tfrac12b\big) + \tfrac12\psi'\big(\tfrac12+\tfrac12b\big)^2 - \tfrac12\big[\psi\big(\tfrac12\big) - \psi\big(\tfrac12+\tfrac12b\big)\big]\psi''\big(\tfrac12+\tfrac12b\big) \end{align} and hence $B(1) \; =\; \tfrac13\pi^2 + 4(\ln2)^2 \hspace{1.5cm} B'(1) \; = \; \tfrac13\pi^2\ln2 + \zeta(3) \hspace{1.5cm} B''(1) \; = \; -\tfrac{1}{360}\pi^4 - 2 \zeta(3)\ln2$ and hence \begin{align} \int_0^{\frac12\pi} \ln^2(\sin x) \ln^2(\cos x)\,dx & = \frac{\partial^4 f}{\partial a^2 \partial b^2} \Big|_{a=b=1} \; = \; g''(1) \\ & = \tfrac18\big[A''(1)B(1) + 2A'(1)B'(1) + A(1)B''(1)\big] \\ & = \tfrac{1}{320}\pi^5 + \tfrac12\pi(\ln2)^4 - \tfrac12\pi\zeta(3)\ln2 \end{align}

- 1 year, 10 months ago

Solve LaTex: $\displaysize \large \int _((cos \phi)^-1)^((cos \phi)^-1)\((e^(\delta*x))/(\sqrt (1-(x^2)))dx$