Brilliant Integration Contest - Season 1 (Part 2)

Thank you for your help. You're too kind to me. I really appreciate it \(\quad\) \(\ddot\smile\)

Solving this problem is really tedious job. Honestly, I'm unwilling to answer it too (even if I know how to solve it). I don't know what is the OP's motivation by posting two well-defined integrals in a single problem. If the OP has an elementary and a clever method than mine, please do share to us. Okay, here is an attempt using a cannon.

Let split the integral into two parts

\[I-J=\int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx-\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)\sqrt{x}\right)}{x}\,dx\]

Perform integration by parts for \(I\) by taking \(u={\rm{arcsinh}\,} x\) and \(dv=\dfrac{dx}{x}\).

\[\begin{align} \int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx&={\rm{arcsinh}\,} x\,\ln x\bigg|_0^1-\int_0^1\frac{\ln x}{\sqrt{1+x^2}}\,dx\qquad\Rightarrow\qquad x=\tan t\\ I&=\int_0^{\pi/4}\frac{\ln(\cos t)-\ln (\sin t)}{\cos t}\,dt\\ &=\int_0^{\pi/4}\frac{\ln(\cos t)}{\cos t}\,dt-\int_0^{\pi/4}\frac{\ln(\sin t)}{\cos t}\,dt\\ &=\frac{1}{2}\int_0^{\pi/4}\frac{\ln(1-\sin^2 t)}{\cos t}\,dt-\int_0^{\pi/4}\frac{\ln(\sin t)}{\cos t}\,dt\\ \end{align}\]

Putting \(y=\sin t\) and \(a=\dfrac{1}{\sqrt{2}}\), we get

\[\begin{align} \int_0^1\frac{{\rm{arcsinh}\,} x}{x}\,dx&=\int_0^{a}\frac{\ln (1-y^2)}{1-y^2}\,dy-\int_0^{a}\frac{\ln y}{1-y^2}\,dy\\ &=\int_0^{a}\frac{\ln (1-y)+\ln(1+y)}{(1-y)(1+y)}\,dy-\int_0^{a}\frac{\ln y}{(1-y)(1+y)}\,dy\\ &=I_1+I_2 \end{align}\]

Performing partial fractions decomposition we get

\[\begin{align} I_1&=\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1-y}\,dy\\ &=\frac{\ln^2 (1+a)}{4}+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy-\frac{\ln^2 (1-a)}{4}\\ &=\frac{1}{4}\ln^2\left(\frac{1+a}{1-a}\right)+\frac{1}{2}\int_0^{a}\frac{\ln (1-y)}{1+y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln (1+y)}{1-y}\,dy\qquad\Rightarrow\qquad 2z=1+y\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{1}{2}\int_c^{b}\frac{\ln (2-2z)}{z}\,dz+\frac{1}{2}\int_c^{b}\frac{\ln z}{1-z}\,dz\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{\ln2}{2}\int_c^{b}\frac{dz}{z}+\frac{1}{2}\int_c^{b}\frac{\ln (1-z)}{z}\,dz+\frac{1}{2}\int_{1-c}^{1-b}\frac{\ln (1-z)}{z}\,dz\\ &=\ln^2\left(1+\sqrt{2}\right)+\frac{\ln2}{2}\ln\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)+\frac{1}{2}\int_c^{b}\frac{\ln (1-z)}{z}\,dz+\frac{1}{2}\int_{1-c}^{1-b}\frac{\ln (1-z)}{z}\,dz\\ \end{align}\]

and

\[\begin{align} I_2&=\frac{1}{2}\int_0^{a}\frac{\ln y}{1-y}\,dy+\frac{1}{2}\int_0^{a}\frac{\ln y}{1+y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy+\frac{\ln y\ln(1+y)}{2}\bigg|_0^{a}-\int_0^{a}\frac{\ln (1+y)}{y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy+\frac{\ln a\ln(1+a)}{2}-\int_0^{a}\frac{\ln (1+y)}{y}\,dy\\ &=\frac{1}{2}\int_1^{1-a}\frac{\ln (1-y)}{y}\,dy-\int_0^{a}\frac{\ln (1+y)}{y}\,dy-\frac{\ln2}{4}\ln\left(\frac{1+\sqrt{2}}{\sqrt{2}}\right)\\ \end{align}\]

where \(b=\dfrac{1+\sqrt{2}}{2\sqrt{2}}\) and \(c=\dfrac{1}{2}\).

Now, let us evaluate \(J\). Set \(x=t^2\), we get

\[J=\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)\sqrt{x}\right)}{x}\,dx=2\int_0^1\frac{\ln\left(1+\left(\sqrt{2}-1\right)t\right)}{t}\,dt\]

Here is the cannon, recall a special function dilogarithm.

\[\operatorname{Li}_2(z)=-\int_0^z\frac{\ln(1 - t)}{t}\,dt\]

Hence, the rest integrals can be easily evaluated by using dilogarithm and an elementary substitution, i.e. \(t=kx\), where \(k\) is a constant. We may also utilize these identities \[\begin{align} \operatorname{Li}_2(z)+\operatorname{Li}_2(-z)&=\frac{1}{2}\operatorname{Li}_2(z^2)\\ \operatorname{Li}_2(1-z)+\operatorname{Li}_2\left(1-\frac{1}{z}\right)&=-\frac{\ln^2z}{2}\\ \operatorname{Li}_2(z)+\operatorname{Li}_2(1-z)&=\frac{{\pi}^2}{6}-\ln z \cdot\ln(1-z) \end{align}\] and special values \[\begin{align} \operatorname{Li}_2(-1)&=-\frac{{\pi}^2}{12}\\ \operatorname{Li}_2(0)&=0\\ \operatorname{Li}_2\left(\frac{1}{2}\right)&=\frac{{\pi}^2}{12}-\frac{\ln^2 2}{2}\\ \operatorname{Li}_2(1)&=\frac{{\pi}^2}{6} \end{align}\] Performing a cumbersome and a tedious calculation, we will get the announced result.

I hope you understand my feelings while trying to solve and to write it down. So, please do not ever post a problem like this again. LOL

Isn't this problem too difficult to high school students Anna? I decide to answer Problem 16 because I'm afraid if this continues till a week, this contest will lose its interest. IMHO, you should propose an easy problem so that this contest will be fun as the stated aims of it. So, here is a solution:

SOLUTION OF PROBLEM 16 :

Rewrite the integral as follows \[\begin{align} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac12\int_{-\large\frac\pi2}^{\large\frac\pi2}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{v-1}\cos ax\ dx\\ &=\frac1{2^{v}}\int_{-\large\frac\pi2}^{\large\frac\pi2}\left(1+e^{2ix}\right)^{v-1}e^{-i(v-1)x}\cos ax\ dx\\ &=\frac1{2^{v}}\int_{-\large\frac\pi2}^{\large\frac\pi2}\sum_{n=0}^{v-1}\binom{v-1}{n} e^{2inx}\cdot e^{-i(v-1)x}\cos ax\ dx\\ &=\frac1{2^{v}}\sum_{n=0}^{v-1}\binom{v-1}{n} \int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i(2n-v+1)x}\cos ax\ dx.\qquad\qquad\qquad\tag1 \end{align}\] Consider \[ f(x)=\left\{ \begin{array}{l l} e^{i\omega x} & \quad \text{for}\ -\frac\pi2<x<\frac\pi2\\[12pt] 0 & \quad \text{otherwise} \end{array} \right. \] The Fourier transform of \(f(x)\) is \[\begin{align} \mathscr{F}\left[f(x)\right]&=\int_{-\infty}^\infty f(x)\ e^{-i\alpha x}\ dx\\ \int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i\omega x}\cos \alpha x\ dx-i\int_{-\large\frac\pi2}^{\large\frac\pi2}e^{i\omega x}\sin \alpha x\ dx&=\int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i\omega x}\ e^{-i\alpha x}\ dx\\ &=\int_{-\large\frac\pi2}^{\large\frac\pi2} e^{i(\omega-\alpha) x}\ dx\\ &=\left[\frac{e^{i(\omega-\alpha) x}}{i(\omega-\alpha)}\right]_{x=-\large\frac\pi2}^{\large\frac\pi2}\\ \Re\bigg(\mathscr{F}\left[f(x)\right]\bigg)&=\left[\frac{\sin(\omega-\alpha) x}{\omega-\alpha}\right]_{x=-\large\frac\pi2}^{\large\frac\pi2}\\ \int_{\large-\frac\pi2}^{\large\frac\pi2}\cos \alpha x\ dx&=\frac{2\sin(\omega-\alpha) \frac\pi2}{\omega-\alpha}.\qquad\qquad\qquad\tag2 \end{align}\] Using \((2)\), then \((1)\) turns out to be \[\begin{align} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac1{2^{v-1}}\sum_{n=0}^{v-1}\binom{v-1}{n} \frac{\sin(2n-v+1-a) \frac\pi2}{2n-v+1-a}\\ &=\frac1{2^{v}}\sum_{n=0}^{v-1}\binom{v-1}{n} \frac{\sin\left(n-\frac{v-1+a}2\right) \pi}{n-\frac{v-1+a}2}.\qquad\qquad\qquad\tag3\\ \end{align}\] Now, let us express \(\dbinom{y}{z}\) in term of beta function that can be related to \((3)\). \[\begin{align} \binom{y}{z}&=\frac{y!}{z!(y-z)!}\\ &=\frac{y!}{\Gamma(1+z)\Gamma(1+y-z)}\\ &=\frac{y!}{z\Gamma(z)\Gamma(1-z)(y-z)\cdots(1-z)}\\ &=\frac{\sin(\pi z)}{\pi z}\cdot\frac{y!}{(y-z)\cdots(1-z)}\\ &=\frac{\sin(\pi z)}{\pi z}\sum_{n=0}^{y}\binom{y}{n}(-1)^n\frac{n}{z-n}\\ &=\sum_{n=0}^{y}\binom{y}{n}\frac{\sin\pi(z-n)}{\pi(z-n)}.\qquad\qquad\qquad\tag4 \end{align}\] Using \((4)\), then \((3)\) turns out to be \[\begin{align} \int_0^{\Large\frac\pi2}\cos^{v-1}x\cos ax\ dx&=\frac{\pi}{2^{v}}\binom{v-1}{\frac{v-1+a}2}\\ &=\frac{\pi}{2^{v}}\frac{\Gamma(v)}{\Gamma\left(\frac{v+a+1}2\right)\Gamma\left(\frac{v-a+1}2\right)}\\ &=\frac{\pi}{2^{v}\ v\ \operatorname{B}\left(\frac{v+a+1}2,\frac{v-a+1}2\right)}\qquad\qquad\qquad\blacksquare \end{align}\]

There should be 3log(2) instead of 2log(2)

\[Solution\quad of\quad problem\quad 18:\\ I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { tan }^{ \frac { 1 }{ 3 } }x.dx } \\ put\quad { tan }^{ \frac { 1 }{ 3 } }x=t\\ dx\quad =\quad \frac { { 3t }^{ 2 } }{ 1+{ t }^{ 6 } } .dt\\ I\quad =\quad \int { \frac { { 3t }^{ 3 } }{ 1+{ t }^{ 6 } } .dt } \\ put\quad { t }^{ 2 }=u\\ 2t.dt=du\\ I\quad =\quad \frac { 3 }{ 2 } \int { \frac { u.du }{ 1+{ u }^{ 3 } } } \\ using\quad partial\quad fraction\quad ,\quad our\quad integration\\ turns\quad out\quad to\quad be\quad :\\ I\quad =\quad \frac { -1 }{ 2 } \int { \frac { du }{ 1+u } } \quad +\quad \frac { 1 }{ 2 } \int { \frac { (1+u)du }{ { u }^{ 2 }-u+1 } } \\ after\quad solving\quad and\quad applying\quad limits:\\ I\quad =\quad \int _{ 0 }^{ \frac { \pi }{ 4 } }{ { tan }^{ \frac { 1 }{ 3 } }x.dx } \quad =\quad \frac { 1 }{ 6 } (\pi \sqrt { 3 } -3\log { 2 } )\]

We have

\( \displaystyle \int_0^ {\infty} \frac{dx}{x^4 + a x^2 + b^2} = \frac{\pi}{2b \sqrt{2b+a}}, \)

which I proved on MSE. Plugging in \( a = 2 \cos2 \theta\) and using that \(2 \cos^2(\theta) = 1 + \cos 2\theta \) immediately gives the answer.

Solution 23

A well-known problem. Sub \( \tan x = u \) to get

\( \displaystyle \int_0^1 du \frac{du \ln u}{1+u^2} = \sum_{k \geq 0} (-1)^k \int_0^1 du \ln u \, u^{2k} = \sum_{k \geq 0} (-1)^k \frac{1}{(2k+1)^2} = G. \)

The penultimate equality follows from integration by parts.

@Ronak Agarwal @Ruben Doornenbal The answer given is wrong!!! It should be -G!!! You must have forgotten the negative sign.......

\[ I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} \int_{r}^{r+1} \frac{x-r-\frac{1}{2}}{x} dx \] \[ I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 1-(r+\frac{1}{2})\ln(\frac{r+1}{r}) \] \[ 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+1)\ln(\frac{r+1}{r}) \] \[ 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(\frac{r+1}{r}) + \ln(\frac{r+1}{r}) \] \[ 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-(2r+2)\ln(r+1) +(2r+2)\ln(r)+ \ln(\frac{r+1}{r}) \] \[ 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r) \] \[ 2I = \lim_{n \rightarrow \infty} \sum_{r=1}^{r=n-1} 2-2((r+1)\ln(r+1)-r\ln(r))+\ln(r+1)+\ln(r) \] \[ 2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+2\ln(n!)-\ln(n) \] \[ 2I = \lim_{n \rightarrow \infty} 2(n-1)-2n\ln(n)+(2n+1)\ln(n)-2n+\ln(2\pi)-\ln(n) \] \[ 2I = \ln(2\pi)-2 \] \[ I =\frac{\ln(2\pi)}{2}-1 \]

This one is easy too :) \[\frac{\ln2\pi}{2}-1\]

@Shivang Jindal : Sorry, I was kidding & I am busy right now so I have no time to write down my answer. Could you elaborate yours then you're good to go (propose your problem). Sorry for the inconvenience...

Trick , is to break the integral from \( (1,2),(2,3)...(n-1,n) \). Then, we compute the sum in terms of \( n \) . and then use Stirling approximation :) .

\(Solution\quad of\quad Problem\quad 22\)

\(I=\displaystyle \int _{ 0 }^{ a }{ \frac { xdx }{ cos(x)cos(a-x) } } =\int _{ 0 }^{ a }{ \frac { (a-x)dx }{ cos(x)cos(a-x) } } \)

Adding these two forms we get :

\(I=\displaystyle \frac { a }{ 2 } \int _{ 0 }^{ a }{ \frac { dx }{ cos(x)cos(a-x) } } \)

Multiplying and dividing by \(sin(a)\) we get :

\(I=\displaystyle \frac { a }{ 2sin(a) } \int _{ 0 }^{ a }{ \frac { sin(x+(a-x))dx }{ cos(x)cos(a-x) } } \)

\(I=\displaystyle \frac { a }{ 2sin(a) } \int _{ 0 }^{ a }{ (tan(x)+tan(a-x))dx } \)

Also since \(\displaystyle \int _{ 0 }^{ a }{ tan(x)dx } =\int _{ 0 }^{ a }{ tan(a-x)dx } \)

We get \(I=\displaystyle \frac{a}{sin(a)}\int _{ 0 }^{ a }{ tan(x)dx } \)

\(I=\frac { a }{ sin(a) } ln(sec(a))\)

\(Solution\quad of\quad problem\quad 24\)

First note that :

\(arcsech(x)=ln(\frac{1+\sqrt{1-{x}^{2}}}{x})\)

In our integral put \(x=sin(\theta)\)

\(\displaystyle I=\int _{ 0 }^{ \frac { \pi }{ 2 } }{ \theta cos\theta ln(\frac { 1+cos\theta }{ sin\theta } ) } d\theta \)

Applying integration by parts we get , \(u=ln(\frac{1+cos\theta }{sin\theta}),dv=\theta cos\theta d\theta \)

\(\displaystyle I=(\theta sin\theta +cos\theta )ln(\frac { 1+cos\theta }{ sin\theta } )\overset { \frac { \pi }{ 2 } }{ \underset { 0 }{ | } } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ (\theta sin\theta +cos\theta )cosec\theta d\theta } \)

\(\displaystyle I=(\theta sin\theta +cos\theta )ln(\frac { 1+cos\theta }{ sin\theta } )+\frac { { \theta }^{ 2 } }{ 2 } +ln(sin\theta )\overset { \frac { \pi }{ 2 } }{ \underset { 0 }{ | } } \)

Which on evaluating we get :

\(I=\frac { { \pi }^{ 2 } }{ 8 } -ln(2)\)

SOLUTION 26

Consider

\( \displaystyle I(a) = \int_0^\infty dx \frac{x^{a-1}}{\cosh x} = 2 \sum \limits_{k \geq 0} (-1)^k \int_0^\infty dx x^{a-1} e^{-(2k+1)x} = 2 \Gamma(a) \beta(a). \)

Our integral is

\( \displaystyle I'(1) = 2 \Gamma'(1)\beta(1) + 2 \Gamma(1) \beta'(1) = 2(-\gamma)(\pi/4) + 2 \frac \pi 4 \left[\gamma + 2 \ln 2 + 3 \ln \pi - 4 \ln \Gamma \frac 1 4 \right]. \)

Here we used a result from Mathworld. Using the Euler reflection formula,

\( \displaystyle \Gamma(1/4) = \pi \sqrt 2 (\Gamma(3/4))^{-1}. \)

Collecting all the terms gives \( \displaystyle I'(1) = -\frac \pi 2 \ln \pi + 2\pi \ln \Gamma(3/4), \)

which equals the stated result.

Solution of Problem 27

Set \(t=\tanh x\), we have

\[\begin{align} \int_0^\infty\ln^2(\tanh x)\,dx&=\int_0^1\frac{\ln^2t}{1-t^2}\,dt\\ &=\int_0^1\sum_{n=0}^\infty t^{2n}\ln^2t\,dt\\ &=\sum_{n=0}^\infty\int_0^1 t^{2n}\ln^2t\,dt\\ &=2\sum_{n=0}^\infty\frac{1}{(2n+1)^3}\qquad\Rightarrow\qquad\text{see solution of Problem 13}\\ &=2\left[\sum_{n=1}^\infty\frac{1}{n^3}-\sum_{n=1}^\infty\frac{1}{(2n)^3}\right]\\ &=\frac{7}{4}\sum_{n=1}^\infty\frac{1}{n^3}\\ &=\frac{7}{4}\zeta(3) \end{align}\]

Solution 28

The integral equals

\(\displaystyle I = \Re \int_{-\infty}^\infty dx \frac{\left(e^{2x} - e^{- 2x} \right) e^{\pi x} e^{2 i x}}{e^{2 \pi x} - 1}. \)

Substitute \( e^{\pi x} = u \). We get

\( \displaystyle I = J_+ - J_-, \)

where

\(\displaystyle \begin{align} J_\pm &= \Re \frac 1 \pi \int_0^\infty dx \frac{u^{2( i\pm1)/\pi}}{u^2 - 1} \\&= -\Re \frac 1 2 \cot\left[\frac \pi 2 \left(2(i\pm1)/\pi + 1\right) \right] \\&= \Re\frac 1 2 \tan(i \pm 1). \end{align} \)

Here we made use of the well-known integral

\(\displaystyle PV \int_0^\infty \frac{x^{a-1}}{1-x^b} = \frac \pi b \cot \frac{\pi a}{b}. \)

Now using the identity

\(\displaystyle \tan\frac{A+B}{2} = \frac{\sin A + \sin B}{\cos A + \cos B} \)

gives

\(\displaystyle J_\pm = \pm \frac 1 2 \frac{\sin 2}{\cosh 2 + \cos 2}, \)

which gives the desired result.

SOLUTION OF PROBLEM 17

Substitute \( x = 2t - 1\) to get

\( I = \displaystyle{\begin{equation} \int_0^1 \frac{dt}{\sqrt[3]{(1-t)t^2} } = B(1/3,2/3) = \frac{\Gamma(1/3)\Gamma(2/3)}{\Gamma(1)} = \frac{\pi}{\sin \pi/3} = \frac{2 \pi}{\sqrt 3}\end{equation}} \)

\(Solution\quad to\quad problem\quad 29\)

This one's easy.

Apply integration by parts \(u={atan(x)}^{2}\) and \(dv={x}^{2}dx\)

\(\displaystyle I=\frac { { x }^{ 3 }{ { tan }^{ -1 }(x) }^{ 2 } }{ 3 } \overset { 1 }{ \underset { 0 }{ | } } -\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ \frac { { x }^{ 3 }{ tan }^{ -1 }x }{ 1+{ x }^{ 2 } } dx } \)

\(\displaystyle I=\frac { { \pi }^{ 2 } }{ 48 } -\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ \frac { ({ x }^{ 3 }+x-x){ tan }^{ -1 }x }{ 1+{ x }^{ 2 } } dx }\)

\(\displaystyle I=\frac { { \pi }^{ 2 } }{ 48 } -\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ x{ tan }^{ -1 }xdx } +\frac { 2 }{ 3 } \int _{ 0 }^{ 1 }{ \frac { x{ tan }^{ -1 }x }{ 1+{ x }^{ 2 } } dx } \)

\(\displaystyle J=\int _{ 0 }^{ 1 }{ x{ tan }^{ -1 }xdx } \)

Integration by parts \(u=tan^{-1}(x),dv=xdx\)

\(\displaystyle J=\frac { { x }^{ 2 }{ tan }^{ -1 }x }{ 2 } \overset { 1 }{ \underset { 0 }{ | } } -\frac { 1 }{ 2 } \int _{ 0 }^{ 1 }{ \frac { { x }^{ 2 }dx }{ 1+{ x }^{ 2 } } } \)

Rest is trivial and it evaluates out to be :

\(J=\frac{\pi}{4}-\frac{1}{2}\)

\(\displaystyle K=\int _{ 0 }^{ 1 }{ \frac { x{ tan }^{ -1 }xdx }{ 1+{ x }^{ 2 } } } \)

Put \(x=tan(\theta)\) to get the integral as :

\(\displaystyle K=\int _{ 0 }^{ \frac{\pi}{4} }{ \theta tan\theta d\theta } \)

Integration by parts \(dv=tan\theta d\theta , u=\theta\)

\(\displaystyle K=\theta ln(sec\theta )\overset { \frac { \pi }{ 4 } }{ \underset { 0 }{ | } } +\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cos\theta )d\theta } \)

\(\displaystyle K=\frac { \pi ln(2) }{ 8 } +\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cos\theta )d\theta } \)

Now \(\displaystyle M=\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cos\theta )d\theta } =\int _{ \frac { \pi }{ 4 } }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta } -\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(sin\theta )d\theta } \)

Adding both these forms we get :

\(\displaystyle M=\frac{1}{2}\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cot\theta )d\theta } +\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta }\)

\(M=\frac { G }{ 2 } -\frac { \pi ln(2) }{ 4 }\)

Here I have two general results :

\(\displaystyle \frac { -\pi ln(2) }{ 2 } =\int _{ 0 }^{ \frac { \pi }{ 2 } }{ ln(sin\theta )d\theta },\int _{ 0 }^{ \frac { \pi }{ 4 } }{ ln(cot\theta )d\theta } =G \)

Where \(G\) is the catalan's constant,.

Finally we get \(K= \frac{G}{2}-\frac{\pi ln(2)}{8}\)

Using all these values we finally get \(I\) as :

\(\large I=\frac{{\pi}^{2}}{48}+\frac{G}{3}+\frac{1}{3}-\frac{\pi}{6}-\frac{\pi ln2}{12}\)

\(\Rightarrow \large \boxed{I=\frac { { (4-\pi ) }^{ 2 } }{ 48 } +\frac { G }{ 3 } -\frac { \pi ln(2) }{ 12 } }\)

SOLUTION OF PROBLEM 31

\[I(a)=\int_0^\pi\,\ln\left(1-2a\cos x+a^2\right)\;dx=\begin{cases} 0 &,\quad|a| < 1\\[12pt] 2\pi\ln|a| &,\quad |a| > 1 \end{cases}\]

For a complete proof, you may refer to my answer on Math S.E.

I posted this problem, because there are numerous elegant ways to solve this problem. There exist a simple solution, by using Riemann integral[which i not commonly used], making this problem extremely beautiful. Also, this problem can be done elegantly using Physics[gravitation]. I will post both the solutions, if anyone wants.

SOLUTION 32

We have

\( \displaystyle I = \Im \int_0^\infty \frac{x e^{ix + e^{ix}} dx}{1+x^2} = \Im \sum\limits_{k \geq 0} \frac{1}{k!} \int_0^\infty \frac{x e^{i(k+1)x} dx}{1+x^2} = \Im \sum\limits_{k \geq 0} \frac{1}{k!} \pi i e^{-(k+1)} = \pi e^{-1+e^{-1}}. \)

Here we made use of the fact that for positive \(a \),

\( \displaystyle \int_0^\infty \frac{x e^{i a x} dx}{1+x^2} = \frac 1 i \frac{d}{da} \int_0^\infty \frac{e^{i a x} dx}{1+x^2} = \frac 1 i \frac{d}{da} \pi e^{-a} = i \pi e^{-a}. \)

Solution to problem 19

We will integrate by parts to get rid of the arccos. The remaining integral will be trivial by symmetry. Observe that (at least for \( 0 < x < 1\))

\( \displaystyle \partial_x \cos^{-1} \left( \frac{2x}{1+x^2}\right) = \frac{2(1+x^2) - (2x)(2x)}{1+x^2} \frac{-1}{\sqrt{1 - \left(\frac{2x}{1+x^2} \right)^2}} = \frac{-2}{1+x^2}. \)

Also,

\( \displaystyle \frac{x^4}{1-x^4} = -1 + \frac{1}{2} \left[\frac{1}{1+x^2} + \frac{1}{1-x^2} \right]. \)

Integrating by parts now yields

\( \displaystyle \int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} dx \frac{x^{4}}{1-x^{4}}\cos^{-1}\left(\frac{2x}{1+x^{2}}\right) = \left( -x + \frac{1}{2} \tan^{-1} x + \frac{1}{2}\tanh^{-1} x \right) \cos^{-1}\left(\frac{2x}{1+x^{2}}\right) \left. \right\lvert _{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} - \int_{-\frac{1}{\sqrt{3}}}^{\frac{1}{\sqrt{3}}} dx \left( -x + \frac{1}{2} \tan^{-1} x + \frac{1}{2} \tanh^{-1} x \right) \frac{-2}{1+x^2} \\= \pi \left[\frac{\pi}{12} - \frac{1}{\sqrt 3} + \frac{1}{4} \log \left( 2 + \sqrt{3} \right) \right], \) because the integrand is odd. Here we used the easy fact that \( \displaystyle 2 + \sqrt 3 = \frac{\sqrt 3 + 1}{\sqrt 3 - 1}. \)

A well-known problem. Set \(t=\ln\left(\frac{1}{x}\right)\), we have

\[\int_0^1 \ln \ln \left(\frac{1}{x}\right)\,dx=\int_0^\infty e^{-t}\ln t\, dt\]

The last expression is derivative of gamma function where its parameter \(a=1\), we have

\[\lim_{a\to1}\partial_a\,\Gamma(a)=\lim_{a\to1}\partial_a\,\int_0^\infty t^{a-1}e^{-t}\, dt=\lim_{a\to1}\psi(a)\Gamma(a)=-\gamma\]

where \(\psi(x)\) is the digamma function and \(\gamma\) is Euler–Mascheroni constant.

SOLUTION OF PROBLEM 33 :

I am surprised that no one answers this question.

Okay, here is the solution. Setting \(t=x^2\) and followed by partial fractions decomposition, we have \[\begin{align} \int_0^\infty\frac{\sqrt{t}\ln(t+q)}{(t+1)(t+q)}\,dt&=2\int_0^\infty \frac{x^2\ln\left(x^2+q\right)}{\left(x^2+1\right)\left(x^2+q\right)}\,dx\\ &=\frac{2q}{q-1}\int_0^\infty \frac{\ln\left(x^2+q\right)}{x^2+q}\,dx+\frac{2}{1-q}\int_0^\infty \frac{\ln\left(x^2+q\right)}{x^2+1}\,dx\\ &=\frac{2q}{q-1}I+\frac{2}{1-q}J \end{align}\] Setting \(x=\sqrt{q}\tan\theta\) to \(I\), we have \[\begin{align} I&=\frac{1}{\sqrt{q}}\int_0^{\pi/2}\ln q\,\,d\theta+\frac{1}{\sqrt{q}}\int_0^{\pi/2}\ln\left(1+\tan^2\theta\right)\,d\theta\\ &=\frac{\pi\ln q}{2\sqrt{q}}-\frac{2}{\sqrt{q}}\int_0^{\pi/2}\ln\left(\cos\theta\right)\,d\theta\\ &=\frac{\pi\ln q}{2\sqrt{q}}+\frac{\pi\ln2}{\sqrt{q}}\\ &=\frac{\pi\ln \left(2\sqrt{q}\right)}{\sqrt{q}} \end{align}\] Differentiating \(J\) w.r.t. \(q\), we have \[\begin{align} J'(q)&=\int_0^\infty \frac{dx}{\left(x^2+q\right)\left(x^2+1\right)}\\ &=\frac{1}{1-q}\underbrace{\int_0^\infty \frac{dx}{x^2+q}}_{\color{red}{\text{set}\,x=\sqrt{q}\tan\theta}}-\frac{1}{1-q}\int_0^\infty \frac{dx}{x^2+1}\\ &=\frac{\pi}{2\sqrt{q}\,(1-q)}-\frac{\pi}{2(1-q)}\\ J(q)&=\pi\underbrace{\int \frac{dq}{2\sqrt{q}\,(1-q)}}_{\color{red}{\text{set}\,x=\sqrt{q}}}+\frac{\pi}{2}\ln(1-q)\\ &=\frac{\pi}{2}\ln\left(\frac{1+\sqrt{q}}{1-\sqrt{q}}\right)+\frac{\pi}{2}\ln(1-q)+C\\ &=\frac{\pi}{2}\ln\left(\frac{\left(1+\sqrt{q}\right)(1-q)}{1-\sqrt{q}}\right)+C\\ &=\pi\ln\left(1+\sqrt{q}\right)+C \end{align}\] Now, to determine the value of \(C\), let us evaluate \(J(0)\). \[\begin{align} J(0)&=2\int_0^\infty \frac{\ln x}{x^2+1}\,dx\qquad\Rightarrow\qquad x=\tan\theta\\ &=2\int_0^{\pi/2}\ln(\tan\theta)\,d\theta\\ &=2\int_0^{\pi/2}\ln(\sin\theta)\,d\theta-2\int_0^{\pi/2}\ln(\cos\theta)\,d\theta\\ &=0 \end{align}\] Since \(J(0)=0\), then \(C=0\). Hence \[\begin{align} \int_0^\infty\frac{\sqrt{t}\ln(t+q)}{(t+1)(t+q)}\,dt &=\frac{2\pi\sqrt{q}}{q-1}\ln \left(2\sqrt{q}\right)+\frac{2\pi}{1-q}\ln\left(1+\sqrt{q}\right)\\ &=\frac{\pi}{1-q}\left[2\ln\left(1+\sqrt{q}\right)-\sqrt{q}\ln \left(4q\right)\right] \end{align}\] It holds for \(q>0\).

Although I know how to evaluate the second integral, i.e. use \(q=\frac{1}{2}\), but I refuse to answer it since it violates the rules. Sorry... \(\quad\ddot\smile\)

P.S. PROBLEM 34 will be posted here: Brilliant Integration Contest - Season 1 (Part 3)

Hint: the value of the first integral is

\( \displaystyle \pi \frac{2 \log \left(\sqrt{q}+1\right)-\sqrt{q}\log(4q) }{1-q}. \)