How to evaluate $\int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin(x) \sin(y) \sin(x+y)}{x y(x+y)} ~{\rm d}x ~{\rm d}y$?

Question

I came across the problem of evaluating the double integral: $$\int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin(x) \sin(y) \sin(x+y)}{x y(x+y)} ~{\rm d}x ~{\rm d}y$$ My attempt at this was to substitute $\sin(x) \cos(y) + \cos(x) \sin(y)$ for $\sin(x+y)$, although I had no clue where to go from there. My second attempt was to use trigonometric identities to rewrite the integral as: $$\frac{1}{4} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin(2x) + \sin(2y) - \sin(2x + 2y)}{x^2 y + x y^2} ~{\rm d}x ~{\rm d}y$$ Which didn't seem to help, so I continued to get: $$\frac{1}{2} \int_{0}^{\infty} \int_{0}^{\infty} \frac{\sin(2x) \sin^2 (y) + \sin(2y) \sin^2 (x)}{x^2 y + x y^2} ~{\rm d}x ~{\rm d}y$$ I am sure I am missing something, but I am unsure how to integrate this. Any help or advice is appreciated.

Answer 1

This integral appeared as problem 11953 in the American Mathematical Monthly, Vol.124, January 2017 (proposed C. I. Valean (Romania)). The result is $\zeta(2)$.

This is the solution that I submitted a few months. For more solutions you can take a look at this page.

We have that $$\frac{|\sin(x) \sin(y) \sin(x+y)|}{xy(x+y)}\leq \left\{\begin{array}{ll} \frac{1}{xy(x+y)} & \text{for $(x,y)\in [1,+\infty)\times [1,+\infty)$}\\ \frac{1}{x(x+y)} & \text{for $(x,y)\in [1,+\infty)\times [0,1]$}\\ \frac{1}{y(x+y)} & \text{for $(x,y)\in [0,1]\times [1,+\infty)$}\\ 1 & \text{for $(x,y)\in [0,1]\times [0,1]$}\\ \end{array}\right. $$ Moreover $$\int_{1}^{\infty}\left(\int_{1}^{\infty}\frac{dx}{xy(x+y)}\right)dy=\int_{1}^{\infty}\frac{\ln(y+1)}{y^2}dy=2\ln(2),$$ and $$ \int_{0}^{1}\left(\int_{1}^{\infty}\frac{dx}{x(x+y)}\right)dy= \int_{0}^{1}\frac{\ln(y+1)}{y}dy= \frac{\pi^2}{12}.$$ Therefore, by Fubini-Tonelli Theorem, the given double integral can be written as iterated integrals \begin{align*} I&:=\int_{0}^{\infty}\int_{0}^{\infty}\frac{\sin(x) \sin(y) \sin(x+y)}{xy(x+y)}\,dx\,dy=\lim_{R\to +\infty} \int_{x=0}^{R}\left(\int_{y=0}^{R}\frac{\sin(x) \sin(y) \sin(x+y)}{xy(x+y)}\,dx\right)\,dy\\ &=\lim_{R\to +\infty} 2\int_{x=0}^{R}\left(\int_{y=0}^{x}\frac{\sin(x) \sin(y) \sin(x+y)}{xy(x+y)}\,dy\right)\,dx\\ &=\lim_{R\to +\infty}\frac{1}{2}\int_{x=0}^{R}\left(\int_{y=0}^{x}\frac{\sin(2x)+\sin(2y)-\sin(2x+2y)}{xy(x+y)}\,dy\right)\,dx. \end{align*} Then, after letting $u=2x$, $v=2y=tu$, \begin{align*} I&=\lim_{R\to +\infty}\int_{u=0}^{R}\left(\int_{v=0}^{u}\frac{\sin(u)+\sin(v)- \sin(u+v)}{uv(u+v)}\,dv\right)\,du\\ &=\lim_{R\to +\infty}\int_{u=0}^{R}\left(\int_{t=0}^{1}\frac{\sin(u)+\sin(tu)- \sin(u+tu)}{t(1+t)u^2}\,dt\right)\,du =\lim_{R\to +\infty}\int_{t=0}^{1}\frac{f_R(t)}{t(1+t)}\,dt, \end{align*} where \begin{align*} f_R(t):=\int_{u=0}^{R}&\frac{\sin(u)+\sin(tu)-\sin(u+tu)}{u^2}\,du =\left[-\frac{\sin(u)+\sin(tu)-\sin(u+tu)}{u}\right]_{u=0}^{R}\\ &\qquad+\int_{u=0}^{R}\frac{\cos(u)-\cos((1+t)u)}{u}\,du +t\int_{u=0}^{R}\frac{\cos(tu)-\cos((1+t)u)}{u}\,du. \end{align*} By using the Frullani's integral, $$\int_0^{+\infty}\frac{\cos(au)-\cos(bu)}{u}\,dt=\ln(b/a) \qquad\mbox{for $a,b>0$,}$$ it follows that $$\lim_{R\to +\infty} f_R(t)=f(t):=\ln(1+t)+t\ln((1+t)/t)=(1+t)\ln(1+t)-t\ln(t).$$ Note that for any $t\in [0,1]$, $$|f_R(t)-f(t)|\leq \int_{R}^{\infty}\frac{3}{u^2}\,du=\frac{3}{R}.$$ Hence, by the Dominated Convergence Theorem, we finally obtain \begin{align*} I&=\int_{0}^{1}\frac{f(t)}{t(1+t)}\,dt =\int_0^1 \frac{\ln(1+t)}{t}dt-\left[\ln(t)\ln(1+t)\right]_0^1+\int_0^1 \frac{\ln(1+t)}{t}dt\\ &=2\int_0^1 \frac{\ln(1+t)}{t}dt= 2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\int_0^1 t^{n-1}dt =2\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}=\zeta(2)=\frac{\pi^2}{6}. \end{align*}

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\ds{\int_{0}^{\infty}\int_{0}^{\infty}{\sin\pars{x}\sin\pars{y}\sin\pars{x + y} \over xy\pars{x + y}}\,\dd x\,\dd y}} \\[5mm] = &\ \int_{0}^{\infty}\int_{0}^{\infty}{\sin\pars{x}\sin\pars{y}\over xy}\ \overbrace{\bracks{{1 \over 2}\int_{-1}^{1}\expo{\ic k\pars{x + y}}\dd k}} ^{\ds{\sin\pars{x + y} \over x + y}}\ \,\dd x\,\dd y \\[5mm] = &\ {1 \over 2}\int_{-1}^{1}\bracks{\int_{0}^{\infty}{\sin\pars{x} \over x} \,\expo{\ic k x}\,\dd x}^{2}\dd k\label{1}\tag{1} = {1 \over 2}\int_{-1}^{1} \bracks{\pi + 2\ic\,\mrm{arctanh}\pars{k} \over 2}^{2}\dd k \\[5mm] = &\ {1 \over 4}\int_{0}^{1} \bracks{\pi^{2} - 4\,\mrm{arctanh}^{2}\pars{k}}\dd k = {\pi^{2} \over 4} - {1 \over 4}\int_{0}^{1}\ln^{2}\pars{1 - k \over 1 + k}\,\dd k \\[5mm] \stackrel{t\ =\ \pars{1 - k}/\pars{1 + k}}{=}\,\,\,& {\pi^{2} \over 4} - {1 \over 2}\ \underbrace{\int_{0}^{1}{\ln^{2}\pars{t} \over \pars{1 + t}^{2}}\,\dd t} _{\ds{\pi^{2} \over 6}} = {\pi^{2} \over 4} - {1 \over 2}\,{\pi^{2} \over 6} = \bbx{\pi^{2} \over 6}\label{2}\tag{2} \end{align}

Answer 3

In this answer, there are some gaps to be filled in by the reader (such as why the integration operators commute and why they commute with taking a limit). This answer is similar to Felix Marin's solution.

Write $$\text{sinc}(q)=\left\{ \begin{array}{ll} \frac{\sin(q)}{q}&\text{if }q\neq 0\,,\\ 1&\text{if }q=0\,. \end{array} \right.$$ Observe that $$\text{sinc}(q)=\frac{1}{2}\,\int_{-1}^{+1}\,\exp(\text{i}pq)\,\text{d}p\,.$$ Hence, the integral $\displaystyle I:=\int_0^\infty\,\int_0^\infty\,\text{sinc}(x)\,\text{sinc}(y)\,\text{sinc}(x+y)\,\text{d}x\,\text{d}y$ can be written as $$\begin{align} I&=\frac{1}{8}\,\int_0^\infty\,\int_0^\infty\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\exp(\text{i}rx)\,\exp(\text{i}sy)\,\exp\big(\text{i}t(x+y)\big)\,\text{d}r\,\text{d}s\,\text{d}t\,\text{d}x\,\text{d}y \\ &=\lim_{\epsilon\to 0^+}\,\small\frac{1}{8}\,\int_0^\infty\,\int_0^\infty\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\exp(\text{i}rx)\,\exp(\text{i}sy)\,\exp\big(\text{i}(t+\text{i}\epsilon)(x+y)\big)\,\text{d}r\,\text{d}s\,\text{d}t\,\text{d}x\,\text{d}y \\ &=\lim_{\epsilon\to 0^+}\,\small\frac{1}{8}\,\int_0^\infty\,\int_0^\infty\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\exp\big(\text{i}(r+t+\text{i}\epsilon)x\big)\,\exp\big(\text{i}(s+t+\text{i}\epsilon)y\big)\,\text{d}r\,\text{d}s\,\text{d}t\,\text{d}x\,\text{d}y\,. \end{align}$$ Ergo, $$ \begin{align} I&=\lim_{\epsilon\to0^+}\,\small\frac18\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_0^\infty\,\int_0^\infty\,\exp\big(\text{i}(r+t+\text{i}\epsilon)x\big)\,\exp(\text{i}(s+t+\text{i}\epsilon)y\big)\,\text{d}x\,\text{d}y\,\text{d}r\,\text{d}s\,\text{d}t \\ &=\lim_{\epsilon\to0^+}\,\frac18\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\left(\frac{\text{i}}{r+t+\text{i}\epsilon}\right)\,\left(\frac{\text{i}}{s+t+\text{i}\epsilon}\right)\,\text{d}r\,\text{d}s\,\text{d}t \\ &=-\lim_{\epsilon\to0^+}\,\frac18\,\int_{-1}^{+1}\,\Biggl(\ln\left(\frac{t+\text{i}\epsilon+1}{t+\text{i}\epsilon-1}\right)\Biggr)^2\,\text{d}t\\ &=-\lim_{\epsilon\to0^+}\,\frac18\,\int_{-1}^{+1}\,\Biggl(\ln\left(\frac{1+t+\text{i}\epsilon}{1-t-\text{i}\epsilon}\right)+\text{i}\pi\Biggr)^2\,\text{d}t \\ &=\lim_{\epsilon\to0^+}\,\frac18\,\int_{-1}^{+1}\,\Biggl(\pi^2-2\pi\text{i}\,\ln\left(\frac{1+t+\text{i}\epsilon}{1-t-\text{i}\epsilon}\right)-\Bigg(\ln\left(\frac{1+t+\text{i}\epsilon}{1-t-\text{i}\epsilon}\right)\Bigg)^2\Biggr)\,\text{d}t \\ &=\frac{\pi^2}{4}-\frac18\,\int_{-1}^{+1}\,\Bigg(\ln\left(\frac{1+t}{1-t}\right)\Bigg)^2\,\text{d}t\,, \end{align}$$ noting that $$\small\lim_{\epsilon\to0^+}\,\int_{-1}^{+1}\,\ln\left(\frac{1+t+\text{i}\epsilon}{1-t-\text{i}\epsilon}\right)\,\text{d}t=\lim_{\epsilon\to0^+}\,\Biggl((2+\text{i}\epsilon)\,\ln(2+\text{i}\epsilon)-(2-\text{i}\epsilon)\,\ln(2-\text{i}\epsilon)+2\pi\epsilon-2\text{i}\epsilon\,\ln(\epsilon)\Biggr)=0\,.$$ Now, by setting $u:=\frac{1}{2}\,\ln\left(\frac{1+t}{1-t}\right)$, we see that $t=\tanh(u)$, so $$\int_{-1}^{+1}\,\Bigg(\ln\left(\frac{1+t}{1-t}\right)\Bigg)^2\,\text{d}t=\int_{-\infty}^{+\infty}\,\frac{(2u)^2}{\big(\cosh(u)\big)^2}\,\text{d}u=4\,\int_{-\infty}^{+\infty}\,\left(\frac{u}{\cosh(u)}\right)^2\,\text{d}u\,.$$

We now need to show that $$\int_{-\infty}^\infty\,\left(\frac{u}{\cosh(u)}\right)^2\,\text{d}u=\frac{\pi^2}{6}\,.$$ Use the closed contour $$\gamma_L:=[-L,+L]\cup[+L,+L+2\pi\text{i}]\cup[+L+2\pi\text{i},-L+2\pi\text{i}]\cup[-L+2\pi\text{i},-L]\,,$$ oriented in the counterclockwise direction, to evaluate the integral $$\lim_{L\to\infty}\,\oint_{\gamma_L}\,\frac{u^3}{\big(\cosh(u)\big)^2}\,\text{d}u=-6\pi\text{i}\,\int_{-\infty}^{+\infty}\,\left(\frac{u}{\cosh(u)}\right)^2\,\text{d}u+8\pi^3\text{i}\,\int_{-\infty}^{+\infty}\,\frac{\text{d}u}{\big(\cosh(u)\big)^2}\,.$$ Since $$\int_{-\infty}^{+\infty}\,\frac{\text{d}u}{\big(\cosh(u)\big)^2}=\big(\tanh(u)\big)\Big|^{u=+\infty}_{u=-\infty}=2\,,$$ we conclude that $$\begin{align}\int_{-\infty}^{+\infty}\,\left(\frac{u}{\cosh(u)}\right)^2\,\text{d}u &=\frac{16\pi^2}{6}-\frac{1}{6\pi\text{i}}\,\left(\lim_{L\to\infty}\,\oint_{\gamma_L}\,\frac{u^3}{\big(\cosh(u)\big)^2}\,\text{d}u\right) \\&=\small\frac{16\pi^2}{6}-\frac{1}{3}\,\text{Res}_{u=\frac{\pi\text{i}}{2}}\left(\frac{u^3}{\big(\cosh(u)\big)^2}\right)-\frac{1}{3}\,\text{Res}_{u=\frac{3\pi\text{i}}{2}}\left(\frac{u^3}{\big(\cosh(u)\big)^2}\right)\,. \end{align}$$ Note that $$\text{Res}_{u=\frac{\pi\text{i}}{2}}\left(\frac{u^3}{\big(\cosh(u)\big)^2}\right)=\frac{3\pi^2}{4}\text{ and }\text{Res}_{u=\frac{3\pi\text{i}}{2}}\left(\frac{u^3}{\big(\cosh(u)\big)^2}\right)=\frac{27\pi^2}{4}\,.$$ That is, $$\int_{-\infty}^{+\infty}\,\left(\frac{u}{\cosh(u)}\right)^2\,\text{d}u=\frac{16\pi^2}{6}-\frac{3\pi^2}{12}-\frac{27\pi^2}{12}=\frac{\pi^2}{6}\,.$$ Consequently, $$I=\frac{\pi^2}{4}-\frac{1}{8}\Biggl(4\left(\frac{\pi^2}{6}\right)\Biggr)=\frac{\pi^2}{6}=\zeta(2)\,,$$ where $\zeta$ is the Riemann zeta-function.

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Interestingly, we have $$J:=\int_{-\infty}^{+\infty}\,\int_{-\infty}^{+\infty}\,\text{sinc}(x)\,\text{sinc}(y)\,\text{sinc}(x+y)\,\text{d}x\,\text{d}y=\pi^2\,.$$ To show this, we proceed as before: $$\begin{align} J&=\frac{1}{8}\,\int_{-\infty}^{+\infty}\,\int_{-\infty}^{+\infty}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\exp(\text{i}rx)\,\exp(\text{i}sy)\,\exp\big(\text{i}t(x+y)\big)\,\text{d}r\,\text{d}s\,\text{d}t\,\text{d}x\,\text{d}y \\&=\frac{1}{8}\,\int_{-\infty}^{+\infty}\,\int_{-\infty}^{+\infty}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\exp\big(\text{i}(r+t)x\big)\,\exp(\text{i}(s+t)y\big)\,\text{d}r\,\text{d}s\,\text{d}t\,\text{d}x\,\text{d}y \\&=\frac{1}{8}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-\infty}^{+\infty}\,\int_{-\infty}^{+\infty}\,\exp\big(\text{i}(r+t)x\big)\,\exp\big(\text{i}(s+t)y\big)\,\text{d}x\,\text{d}y\,\text{d}r\,\text{d}s\,\text{d}t\,. \end{align}$$ Let $\delta$ denote the Dirac delta distribution (noting that $\displaystyle\delta(p)=\frac{1}{2\pi}\,\int_{-\infty}^{+\infty}\,\exp(\text{i}pq)\,\text{d}q$). This gives $$\begin{align} J&=\frac{1}{8}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\int_{-1}^{+1}\,\big(2\pi\,\delta(r+t)\big)\,\big(2\pi\,\delta(s+t)\big)\text{d}r\,\text{d}s\,\text{d}t\\ &=\frac{\pi^2}{2}\,\int_{-1}^{+1}\,\text{d}t=\pi^2=6\,\zeta(2)\,. \end{align}$$ As a consequence, $$\int_0^\infty\,\int_0^\infty\,\text{sinc}(x)\,\text{sinc}(y)\,\text{sinc}(x-y)\,\text{d}x\,\text{d}y=\frac{\pi^2}{2}-\frac{\pi^2}{6}=\frac{\pi^2}{3}=2\,\zeta(2)\,.$$

Answer 4

Let $$I=\int_0^\infty\int_0^\infty\frac{\sin(x)\cos(x)\ln(x+y)}{xy(x+y)}\ dx\ dy$$

and

$$I(a)=\int_0^\infty\int_0^\infty\frac{\sin(x)\cos(y)\sin(a(x+y))}{xy(x+y)}\ dx\ dy,\quad I(0)=0,\quad I(1)=I$$

$$I'(a)=\int_0^\infty\int_0^\infty\frac{\sin(x)\cos(y)\cos(a(x+y))}{xy}\ dx\ dy$$ $$=\Re\int_0^\infty\int_0^\infty\frac{\sin(x)\cos(y)e^{ia(x+y)}}{xy}\ dx\ dy$$

$$=\Re\int_0^\infty\int_0^\infty\frac{\sin(x)e^{iax}}{x}\cdot\frac{\sin(y)e^{iay}}{y}\ dx\ dy=\Re\left(\color{blue}{\int_0^\infty\frac{\sin(x)e^{iax}}{x}\ dx}\right)^2$$ $$=\Re\left(i\ \text{arctanh}(1/a)\right)^2=-\Re\ \text{arctanh}^2{(1/a)}$$

$$\Longrightarrow I=-\Re\int_0^1\text{arctanh}(1/a)^2\ da=-\frac14\Re\int_0^1\ln^2\left(-\frac{1-a}{1+a}\right)\ da=-\frac12\Re\int_0^1\frac{\ln^2(-x)}{(1+x)^2}\ dx$$

$$=-\frac12\int_0^1\frac{\ln^2(x)-\pi^2}{(1+x)^2}\ dx=-\frac12\int_0^1\frac{\ln^2(x)}{(1+x)^2}\ dx+\frac{\pi^2}2\underbrace{\int_0^1\frac{1}{(1+x)^2}\ dx}_{1/2}$$ $$=\frac12\sum_{n=1}^\infty(-1)^nn\int_0^1 x^{n-1}\ln^2(x)\ dx+\frac{\pi^2}{4}=\sum_{n=1}^\infty\frac{(-1)^n}{n^2}+\frac{\pi^2}{4}=-\frac{\pi^2}{12}+\frac{\pi^2}{4}=\frac{\pi^2}{6}$$

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Proving the blue integral;

Let $$A=\int_0^\infty\frac{\sin(x)e^{iax}}{x}\ dx=\int_0^\infty\frac{\sin(x)e^{-\frac{a}{i}x}}{x}\ dx$$

and $$I(r)=\int_0^\infty\frac{\sin(r x)e^{-\frac{a}{i}x}}{x}\ dx, \quad I(0)=0, \quad I(1)=A$$

$$\Longrightarrow I'(r)=\int_0^\infty \cos(r x)e^{-\frac{a}{i}x}\ dx=\frac{a/i}{r^2+(a/i)^2}=-\frac{ia}{r^2-a^2}$$

$$\Longrightarrow A=i\int_0^1\frac{a}{a^2-r^2}\ dr=i\ \text{arccoth}(a)=i\ \text{arctanh}(1/a)$$

Answer 5

Write the integral as $$ I = \mathrm{Im} \int_0^\infty dx \int_0^\infty dy \frac{\sin x \sin y}{xy(x+y)}e^{i(x+y)} $$ Define $$ J(a) \equiv \int_0^\infty dx \int_0^\infty dy \frac{\sin x \sin y}{xy(x+y)}e^{ia(x+y)} $$ so $$ J'(a) = i \int_0^\infty dx \int_0^\infty dy \frac{\sin x \sin y}{xy}e^{ia(x+y)} \equiv i (K(a))^2, $$ where $$ K(a) = \int_0^\infty dx \frac{\sin x}{x}e^{i a x} $$ We have, for $\mathrm{Im}\ a = \epsilon > 0$, $$ K'(a) = i\ \int_0^\infty dx \sin x\ e^{i a x} = \frac i 2 \left[\frac{1}{a+1} - \frac{1}{a-1} \right] $$ so $$ K(a) = \int_{\infty + \epsilon i}^{a} dz\ K'(z) = i\ \mathrm{arctanh}(1/a) $$

We have, for example from Riemann-Lebesgue, $$ J(\infty+\epsilon i) = 0 $$ whence $$ J(1) = \int_\infty^1 da\ J'(a) $$ Taking the imaginary part, we get $$ I = \int_1^\infty dx\ \mathrm{arctanh}^2 \frac 1 x = \int_0^1 \frac{dx}{x^2}\ \mathrm{arctanh}^2 x \\= \frac 1 4 \int_0^1 dx\ \frac{\ln^2(1+x)+\ln^2(1-x)-2 \ln(1+x)\ln(1-x)}{x^2} $$ The integral can be evaluated by finding the antiderivative expressed in terms of the polylogarithm, or by a series expansion. I don't have time to write down all the steps, which are standard. The result is $$ I = \frac{\pi^2}{6} $$