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

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.

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*}

• Your maths are very good. In the past I tried to use changes of variables in the integral $\frac{4}{\pi}\int_0^{\pi/2}z\sin(z)I(z)dz$ (now I believe that there is a typo in the formula by user75829) from this comment of a post in MathOverflow. I don't know if there are (in the literature) attempts of interesting changes of variable when one evokes the calculation of $\zeta(3)$ in closed-form. I add this comment if you want to explore it in your home. Thus isn't required a response of this message, good week. – user243301 Apr 22 '18 at 12:28
• (+1) Nicely done. And Hi to Cornel. – Jack D'Aurizio Apr 22 '18 at 16:32

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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.

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, $$\int_0^\infty\,\int_0^\infty\,\text{sinc}(x)\,\text{sinc}(y)\,\text{sinc}(x-y)\,\text{d}x\,\text{d}y=\frac{\pi^2}{2}-\frac{\pi^2}{6}=\frac{\pi^2}{3}=2\,\zeta(2)\,.$$