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

Evaluate
$$\int_{0}^{1} \arctan^{2}\left(\, x\,\right) \ln\left(\, 1 + x^{2} \over 2x^{2}\,\right)\,{\rm d}x$$

I substituted $x \equiv \tan\left(\,\theta\,\right)$ and got

$$-\int^{\pi/4}_{0}\theta^{2}\,{\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right) \over \cos^{2}\left(\,\theta\,\right)}\,{\rm d}\theta$$

After this, I thought of using the Taylor Expansion of $\ln\left(\, 2\sin^{2}\left(\,\theta\,\right)\,\right)$ near zero but that didn't do any good.

• you could reduce this to $\log$ integrals by expanding $\arctan(x)$ as $\frac{i}{2}(\log(1+i x)-\log(1-i x))$. The resulting expression should be (in principle) expressible in $\log$ ,$\text{Li}$ and $\zeta$ values. – tired Nov 21 '14 at 15:39
• The integral under consideration equals $$0.119548664034710922477515058213 .$$ – user64494 Nov 23 '14 at 17:38

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.

• @sos440 Coming from an expert in integration like yourself, this compliment means a lot. Thank you. – M.N.C.E. Nov 25 '14 at 4:11
• @M.N.C.E. Thanks! – Henry Nov 30 '14 at 13:38

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 $$I(a)=\int_0^{\pi/2} \frac{t\sin t}{1+\cos t}\ln\left(1-a\cos t\right)\,dt$$ 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 $$I(b)=\int_0^{\pi/2} \frac{t\sin t}{1+b\cos t}\,dt$$ 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 $$J(b)=\int_0^{\pi/2}\ln(1+b\cos t)\,dt$$ 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...

• I'm sorry, but the last result does not seem correct. We want to calculate $I(1) = \int_0^1 da I'(a)$, but this integral diverges due to the $\ln(-a)/(a(1+a))$ term. – user111187 Nov 23 '14 at 10:38
• @user111187 That's why I said, "I give up". I'm too tired. Sorry... – Anastasiya-Romanova 秀 Nov 23 '14 at 10:50
• @Anastasiya-Romanova秀 I bumped into this integral and noticed that your answer is very old so maybe you're no longer into that.. But I believe you might work out the solution of the indefinite integral in terms of polylogarithms buy substituting $\omega = \frac{1+ix}{2}$. In this way you get $\arctan x = (\log \omega + \mbox{Li}_1(\omega))/2i$. The logaritmic factor of the integrand function can also be easly written in terms of $\omega$... – dfnu Sep 28 '17 at 23:04

This is not an answer, but my approach suggests that the answer is

$$I := \int_{0}^{1} \arctan^{2} x \log \left( \frac{x^{2}+1}{2x^{2}} \right) \, dx = \frac{19\pi^{3}}{192} + \frac{5\pi}{16}\log^{2}2 - 6 \Im \mathrm{Li}_{3}\left( \frac{1+i}{2} \right).$$

My approach is to write

$$I = \int_{0}^{1} \arctan^{2} x \log \left( \frac{x^{2}+1}{2} \right) \, dx - 2\int_{0}^{1} \arctan^{2} x \log x \, dx,$$

introduce function $f(z) = \log \left( \frac{1+iz}{\sqrt{2}} \right)$ and write

$$\arctan^{2} x \log \left( \frac{x^{2}+1}{2} \right) = -\frac{1}{4} ( f(x) - f(-x))^{2}(f(x) + f(-x)).$$

This allows to write, with a bit help of complex analysis,

$$\int_{0}^{1} \arctan^{2} x \log \left( \frac{x^{2}+1}{2} \right) \, dx = - \frac{\sqrt{2}}{4} \int_{-\pi/4}^{\pi/4} \theta^{2}\log(\sqrt{2}e^{i\theta} - 1) e^{i\theta} \, d\theta,$$

which seems more tractable than the original one. But I was stuck here.

• So, $\mathrm{Im} \; \mathrm{Li}_3\left((1+i)/2\right)$, we meet again. – Gahawar Nov 23 '14 at 18:08
• @Gahawar, have you ever met this uncomfortable guy $\Im \mathrm{Li}_{3}(\sqrt{i/2})$ somewhere before? – Sangchul Lee Nov 23 '14 at 18:22
• Yes, a few months ago. I was not able to find a closed form for it, however Cleo found an expression involving a nasty hypergeometric series. math.stackexchange.com/questions/918680/… – Gahawar Nov 23 '14 at 18:26
• @Gahawar, Oh, certainly that is not a happy formula to me. – Sangchul Lee Nov 23 '14 at 18:31

By the way, there is a closed-form antiderivative (that could be proved by differentiation): $$\int\arctan^2x\cdot \ln\left(\frac{1+x^2}{2x^2}\right)\,dx=\\ \frac16\left[3 i \left\{\left(2 \operatorname{Li}_2(i x)-2 \operatorname{Li}_2(-i x)+\operatorname{Li}_2\left(\frac{2 x}{x+i}\right)-\operatorname{Li}_2\left(\frac{2 x}{x-i}\right)\right)\cdot \ln \left(\frac{1+x^2}{x^2}\right)\\ +2 \left(2 \operatorname{Li}_3\left(\frac{x}{x-i}\right)-2 \operatorname{Li}_3\left(\frac{x}{x+i}\right)+\operatorname{Li}_3\left(\frac{2 x}{x+i}\right)-\operatorname{Li}_3\left(\frac{2 x}{x-i}\right)\right)\\ +\left(\operatorname{Li}_2\left(\frac{1}{2}-\frac{i x}{2}\right)-\operatorname{Li}_2\left(\frac{i x}{2}+\frac{1}{2}\right)\right)\cdot\ln2\right\}\\ +3 \left(2 \operatorname{Li}_2\left(-x^2\right)+\ln ^2\left(1+x^2\right)+\ln \left(1+x^2\right)\cdot\ln2-2 \ln ^22\right)\cdot\arctan x\\ +6 x \ln \left(\frac{1+x^2}{2 x^2}\right)\cdot\arctan^2x+4 \arctan^3x\right]\color{gray}{+C}$$ This enables us to evaluate a definite integral over any region.

• This result can be obtained with Mathematica or WolframAlpha by evaluating Integrate[ (1/2 I Log[1 - I x] - 1/2 I Log[1 + I x])^2 (Log[1 + I x] + Log[1 - I x] - Log[2] - 2 Log[x]), x] and subsequent simplification. – Vladimir Reshetnikov Dec 28 '15 at 23:31
• (+1) Marvellous! The most powerful solution for this question. – Henry Jan 13 '16 at 3:16

I dont'have (yet) a complete solution for now.

Let $I=\displaystyle \int_0^1 (\arctan(x))^2 \ln\Big(\dfrac{1+x^2}{2x^2}\Big)dx$

$I=\displaystyle \int_0^1 (\arctan(x))^2\ln(1+x^2)dx-2\int_0^1 (\arctan(x))^2\ln(x)dx-\ln(2)\int_0^1 (\arctan(x))^2dx$

I know how to compute the last one ;)

Let $J=\displaystyle \int_0^{\frac{\pi}{2}} \log(\sin x)dx$

The value of $J$ is well known to be $-\dfrac{\pi}{2}\ln 2$

Perform integration by parts:

$J=\displaystyle \Big[x\log(\sin x)\Big]_0^{\frac{\pi}{2}}-\int_0^{\frac{\pi}{2}}\dfrac{x}{\tan x}dx=-\int_0^{\frac{\pi}{2}}\dfrac{x}{\tan x}dx$

Let $K=-J$

Perform the change of variable $u=\tan x$

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

$K\displaystyle =\int_0^{1}\dfrac{\arctan x}{x(1+x^2)}dx+\int_1^{+\infty}\dfrac{\arctan x}{x(1+x^2)}dx$

In the second integral, in the right member perform the change of variable $u=\dfrac{1}{x}$:

$K= \displaystyle\int_0^1 \dfrac{\arctan x }{x(1+x^2)}dx+ \int_0^1 \dfrac{x\arctan \Big(\dfrac{1}{x}\Big) }{1+x^2}dx$

For $x>0$, $\arctan \Big(\dfrac{1}{x}\Big)+\arctan x=\dfrac{\pi}{2}$

Therefore:

$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x(1+x^2)}dx+\dfrac{\pi}{2}\int_0^1\dfrac{x}{1+x^2}dx-\int_0^1\dfrac{x\arctan x}{1+x^2}dx$

For $x\neq 0$ ,$\dfrac{1}{x(1+x^2)}=\dfrac{1}{x}-\dfrac{x}{1+x^2}$

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

The derivative of $(\arctan x)^2$ is $\dfrac{2\arctan x}{1+x^2}$

Hence:

$\displaystyle \int_0^1 \dfrac{2x\arctan x }{1+x^2}dx=\Big[x(\arctan x)^2\Big]_0^1-\int_0^1 (\arctan x)^2dx=\dfrac{\pi^2}{16}-\int_0^1 (\arctan x)^2dx$

Therefore:

$K=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx-\dfrac{\pi^2}{16}+\int_0^1 (\arctan x)^2dx+\dfrac{\pi}{4}\Big[\log(1+x^2)\Big]_0^1$

Recall $K=\dfrac{\pi}{2}\ln 2$

Therefore:

$\displaystyle\int_0^1 (\arctan x)^2 dx=\dfrac{\pi^2}{16}-G+\dfrac{\pi}{4}\ln 2$

Where $G=\displaystyle\int_0^1 \dfrac{\arctan x }{x}dx$ is the Catalan's constant.