How to evaluate $\int_0^1 (\arctan x)^2 \ln(\frac{1+x^2}{2x^2}) dx$

Solution 1:

Results used

I will just state the following result as I do not wish to replicate Random Variable's brilliant work in this answer. $$\int^\frac{\pi}{4}_0\ln^2(\cos{x})\ {\rm d}x=\Im{\rm Li}_3(1-i)-\frac{\mathbf{G}}{2}\ln{2}+\frac{7\pi^3}{192}+\frac{5\pi}{16}\ln^2{2}$$ It is also quite easy to show that $$\sum^\infty_{n=1}\frac{H_n}{n^2}z^n={\rm Li}_3(z)-{\rm Li}_3(1-z)+{\rm Li}_2(1-z)\ln(1-z)+\frac{1}{2}\ln{z}\ln^2(1-z)+\zeta(3)$$

Splitting up the integral

We may split up the integral into 3 simpler integrals. $$\mathscr{I}=-\ln{2}\underbrace{\int^1_0\arctan^2{x}\ {\rm d}x}_{\mathscr{I}_1} -2\underbrace{\int^1_0\arctan^2{x}\ln{x}\ {\rm d}x}_{\mathscr{I}_2}+\underbrace{\int^1_0\arctan^2{x}\ln(1+x^2)\ {\rm d}x}_{\mathscr{I}_3}$$

Evaluation of $\mathscr{I}_1$

Integrate by parts. \begin{align} \mathscr{I}_1 =&x\arctan^2{x}\Bigg{|}^1_0-\int^1_0\frac{2x\arctan{x}}{1+x^2}{\rm d}x\\ =&\frac{\pi^2}{16}-\left[\ln(1+x^2)\arctan{x}\right]^1_0+\int^1_0\frac{\ln(1+x^2)}{1+x^2}{\rm d}x\\ =&\frac{\pi^2}{16}-\frac{\pi}{4}\ln{2}-2\int^\frac{\pi}{4}_0\ln(\cos{x})\ {\rm d}x\\ =&\frac{\pi^2}{16}-\frac{\pi}{4}\ln{2}+\frac{\pi}{2}\ln{2}+2\sum^\infty_{n=1}\frac{(-1)^n}{n}\int^\frac{\pi}{4}_0\cos(2nx)\ {\rm d}x\\ =&\frac{\pi^2}{16}+\frac{\pi}{4}\ln{2}+\sum^\infty_{n=1}\frac{(-1)^n\sin(n\pi/2)}{n^2}\\ =&\frac{\pi^2}{16}+\frac{\pi}{4}\ln{2}+\sum^\infty_{n=0}\frac{(-1)^{2n+1}(-1)^n}{(2n+1)^2}\\ =&\frac{\pi^2}{16}+\frac{\pi}{4}\ln{2}-\mathbf{G} \end{align}

Evaluation of $\mathscr{I}_2$ $\require{cancel}$ \begin{align} \mathscr{I}_2 =&\color{red}{\cancelto{0}{\color{grey}{x\arctan^2{x}\ln{x}\Bigg{|}^1_0}}}-\int^1_0\arctan^2{x}\ {\rm d}x-\int^1_0\frac{2x\arctan{x}\ln{x}}{1+x^2}{\rm d}x\\ =&-\frac{\pi^2}{16}-\frac{\pi}{4}\ln{2}+\mathbf{G}+2\sum^\infty_{n=0}\frac{(-1)^nH_{2n+1}}{(2n+3)^2}-\sum^\infty_{n=0}\frac{(-1)^nH_{n}}{(2n+3)^2}\\ =&\frac{\pi^3}{16}-\frac{\pi^2}{16}-\frac{\pi}{4}\ln{2}+\mathbf{G}-2\sum^\infty_{n=0}\frac{(-1)^nH_{2n+1}}{(2n+1)^2}+\sum^\infty_{n=1}\frac{(-1)^{n}H_n}{(2n+1)^2} \end{align} Since \begin{align} \sum^\infty_{n=0}\frac{(-1)^nH_{2n+1}}{(2n+1)^2} =&\Im\sum^\infty_{n=1}\frac{H_n}{n^2}i^n\\ =&-\Im{\rm Li}_3(1-i)-\frac{\mathbf{G}}{2}\ln{2}-\frac{\pi}{16}\ln^2{2} \end{align} and \begin{align} \sum^\infty_{n=1}\frac{(-1)^nH_n}{(2n+1)^2} =&\int^1_0\frac{\ln{x}\ln(1+x^2)}{1+x^2}{\rm d}x\\ =&-2\int^\frac{\pi}{4}_0\left(\ln(\sin{x})-\ln(\cos{x})\right)\ln(\cos{x})\ {\rm d}x\\ =&-\frac{1}{8}\frac{\partial^2{\rm B}}{\partial a\partial b}\left(\frac{1}{2},\frac{1}{2}\right)+2\int^\frac{\pi}{4}_0\ln^2(\cos{x})\ {\rm d}x\\ =&2\Im{\rm Li}_3(1-i)-\mathbf{G}\ln{2}+\frac{3\pi^3}{32}+\frac{\pi}{8}\ln^2{2} \end{align} We have $$\mathscr{I}_2=4\Im{\rm Li}_3(1-i)+\mathbf{G}+\frac{5\pi^3}{32}+\frac{\pi}{4}\ln^2{2}-\frac{\pi^2}{16}-\frac{\pi}{4}\ln{2}$$

Evaluation of $\mathscr{I}_3$

$\mathscr{I}_3$ is rather straightforward to evaluate. \begin{align} \mathscr{I}_3 =&x\arctan^2{x}\ln(1+x^2)\Bigg{|}^1_0-\int^1_0\frac{2x\arctan{x}\ln(1+x^2)}{1+x^2}{\rm d}x-\int^1_0\frac{2x^2\arctan^2{x}}{1+x^2}{\rm d}x\\ =&\frac{\pi^2}{16}\ln{2}-\frac{1}{2}\ln^2(1+x^2)\arctan{x}\Bigg{|}^1_0+\frac{1}{2}\int^1_0\frac{\ln^2(1+x^2)}{1+x^2}{\rm d}x-2\int^1_0\arctan^2{x}\ {\rm d}x\\&+2\int^1_0\frac{\arctan^2{x}}{1+x^2}{\rm d}x\\ =&2\mathbf{G}+\frac{\pi^3}{96}+\frac{\pi^2}{16}\ln{2}-\frac{\pi}{8}\ln^2{2}-\frac{\pi^2}{8}-\frac{\pi}{2}\ln{2}+2\int^\frac{\pi}{4}_0\ln^2(\cos{x})\ {\rm d}x\\ =&2\Im{\rm Li}_3(1-i)-\mathbf{G}\ln{2}+2\mathbf{G}+\frac{\pi^3}{12}+\frac{\pi^2}{16}\ln{2}+\frac{\pi}{2}\ln^2{2}-\frac{\pi^2}{8}-\frac{\pi}{2}\ln{2} \end{align}

The closed form

Combining these results, we get \begin{align} \mathscr{I}=\Large{\boxed{\displaystyle \color{red}{-6\Im{\rm Li}_3(1-i)-\frac{11\pi^3}{48}-\frac{\pi}{4}\ln^2{2}}}} \end{align} as the closed form.

Solution 2:

Samurai, this is for the second time you post problems that relate to me. First you posted this question here which is exactly similar with my rated problem on Brilliant.org. I have raised objection to mods but they can do nothing since your post doesn't violate any rules here. Okay, fine. I can accept their reason. Now you post this question which I believe it's taken from one of proposed problems in Brilliant Integration Contest - Season 1 that I held on Brilliant.org. The original problem was proposed by Jatin Yadav as PROBLEM 7 but a day later he deleted this problem and changed to another problem after no-one can solve it included himself. According to him, it's taken from here, on Math S.E. You may want to take a look there.

I tried to solve this problem for hours but no success. Here is my attempt:

Set $x=\tan y$, we get \begin{align} I&=\int_0^1\arctan^2x\,\ln\left(\frac{1+x^2}{2x^2}\right)\,dx\\ &=-\int_0^{\pi/4} \frac{y^2\ln\left(2\sin^2y\right)}{\cos^2y}\,dy\\ &=-2\int_0^{\pi/4} \frac{y^2\ln\left(1-\cos2y\right)}{1+\cos2y}\,dy\\ &=-\frac{1}{4}\int_0^{\pi/2} \frac{t^2\ln\left(1-\cos t\right)}{1+\cos t}\,dt\qquad\Rightarrow\qquad t=2y\\ \end{align}

Use integration by parts by taking $u=t^2$ and $dv=\dfrac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt$, then \begin{align} v&=\int\frac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt \end{align} Use integration by parts by taking $u=\ln\left(1-\cos t\right)$ and $dv=\dfrac{dt}{1+\cos t}$, by Weierstrass substitution: $x=\tan\left(\dfrac{t}{2}\right)$ then \begin{align} v=\int\frac{dt}{1+\cos t}=\int \,dx=\tan\left(\frac{t}{2}\right)=\frac{\sin t}{1+\cos t} \end{align} Hence \begin{align} \int\frac{\ln\left(1-\cos t\right)}{1+\cos t}\,dt &=\frac{\sin t}{1+\cos t}\ln\left(1-\cos t\right)-\int\frac{\sin t}{1+\cos t}\cdot\frac{\sin t}{1-\cos t}\,dt\\ &=\frac{\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t \end{align} and \begin{align} I&=-\frac{1}{4}\left[\frac{t^2\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t^3\right]_0^{\pi/2}+\frac{1}{2}\int_0^{\pi/2}\left[\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)-t^2\right]\,dt\\ &=\frac{\pi^3}{32}+\frac{1}{2}\int_0^{\pi/2}\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)\,dt-\frac{\pi^3}{48}\\ &=\frac{\pi^3}{96}+\frac{1}{2}\int_0^{\pi/2}\frac{t\sin t}{1+\cos t}\ln\left(1-\cos t\right)\,dt \end{align} Consider \begin{equation} I(a)=\int_0^{\pi/2} \frac{t\sin t}{1+\cos t}\ln\left(1-a\cos t\right)\,dt \end{equation} so that $I(0)=0$ and \begin{align} I'(a)&=-\int_0^{\pi/2} \frac{t\sin t\cos t}{(1-a\cos t)(1+\cos t)}\,dt\\ &=\frac{1}{1+a}\int_0^{\pi/2} \left(\frac{t\sin t}{1+\cos t}-\frac{t\sin t}{1-a\cos t}\right)\,dt\\ \end{align} Now consider \begin{equation} I(b)=\int_0^{\pi/2} \frac{t\sin t}{1+b\cos t}\,dt \end{equation} Use integration by parts by taking $u=t$ and $dv=\dfrac{\sin t}{1+b\cos t}\,dt$, then \begin{align} I(b)&= \frac{t\ln(1+b\cos t)}{b}\bigg|_0^{\pi/2}-\frac{1}{b}\int_0^{\pi/2}\ln(1+b\cos t)\,dt\\ &=-\frac{1}{b}\int_0^{\pi/2}\ln(1+b\cos t)\,dt \end{align} Consider \begin{equation} J(b)=\int_0^{\pi/2}\ln(1+b\cos t)\,dt \end{equation} so that $J(0)=0$ and \begin{align} J'(b)&=\int_0^{\pi/2} \frac{\cos t}{1+b\cos t}\,dt\\ &=\frac{1}{b}\int_0^{\pi/2} \left(1-\frac{1}{1+b\cos t}\right)\,dt\\ &=\frac{\pi}{2b}-\int_0^{\pi/2} \frac{dt}{1+b\cos t}\qquad\Rightarrow\qquad x=\tan\left(\frac{t}{2}\right)\\ &=\frac{\pi}{2b}-\int_0^{1} \frac{2}{1+b+(1-b)x^2}\,dx\qquad\Rightarrow\qquad x=\sqrt{\frac{1+b}{1-b}}\tan z\\ &=\frac{\pi}{2b}-\frac{2}{\sqrt{1-b^2}}\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\\ J(b)&=\frac{\pi}{2}\ln b-\int\frac{2}{\sqrt{1-b^2}}\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\,db\\ \end{align} Again we use integration by parts by taking $u=\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)$ and $dv=\dfrac{2}{\sqrt{1-b^2}}$, we have \begin{align} J(b)&=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\int\frac{\arcsin b}{1-b}\sqrt{\frac{1-b}{1+b}}\,db\\ &=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\int\frac{\arcsin b}{\sqrt{1-b^2}}\,db\\ &=\frac{\pi}{2}\ln b-2\arctan\left(\sqrt{\frac{1-b}{1+b}}\right)\arcsin b-\frac{\arcsin^2 b}{2} \end{align} Therefore \begin{align} I'(a)&=\frac{\pi^2}{8(1+a)}-\frac{1}{a(1+a)}\left[\frac{\pi}{2}\ln (-a)-2\arctan\left(\sqrt{\frac{1+a}{1-a}}\right)\arcsin (-a)-\frac{\arcsin^2(-a)}{2}\right]\\ \end{align} From this step, I give up. Perhaps someone else want to continue it. Be my guest...