4

Find the integral $$I=\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin (x+y)}{xy(x+y)}\,dx\,dy$$

My try: define $$I(b):=\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin (x+y)}{xy(x+y)}e^{-bxy(x+y)}\,dx\,dy$$ then $$I'(b)=-\int_0^\infty\int_0^{\infty}\sin x \sin y \sin (x+y) e^{-bxy(x+y)}\,dx\,dy$$

But I can't find $I(b)$.

Thank you

Quanto
  • 97,352
athos
  • 5,177
  • 2
    I have answered a more general question a while back. – metamorphy Apr 10 '21 at 04:43
  • 1
    A very nice general solution from @metamorphy. It is surprisingly concise: $I(n)=\pi^nB_n$, where $B_n$ is Bernoulli number. For $n=2$ it gives immediately $\pi^2/6$. I saw several such problems solved (for given $n=1,2,3$) and even evaluated one of them (for example, https://math.stackexchange.com/questions/3513089/unexpected-appearances-of-pi2-6?noredirect=1 or https://artofproblemsolving.com/community/c7h2465658p20815855) , but could never have imagined that there were such solution for the general case. The beauty of mathematics is infinite :) – Svyatoslav Apr 10 '21 at 20:01

1 Answers1

6

Note that $I=J(1)$, where

$$J(a) =\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \sin [a(x+y)]}{xy(x+y)} dx dy$$ and \begin{align} J’(a) &=\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y \cos [a(x+y)]}{xy} dx dy\\ & =\int_0^\infty\int_0^{\infty}\frac{\sin x \sin y [\cos (ax)\cos(ay)- \sin (ax)\sin(ay)]}{xy} dx dy\\ & =\left(\int_0^\infty\frac{\sin x \cos (ax)}xdx \right)^2 - \left(\int_0^\infty\frac{\sin x \sin (ax)}xdx \right)^2\\ & =\frac14\left(\int_0^\infty\frac{\sin (1-a)x + \sin (1+a)x }{x} dx\right)^2\\ & \>\>\>\>\>\>\>-\frac14\left(\int_0^\infty\frac{\cos (1-a)x -\cos (1+a)x }{x} dx\right)^2\\ &=\frac14\left(\pi^2 - \ln^2 \frac{1+a}{1-a} \right) \end{align} Thus $$I=\int_0^1 J’(a)da=\frac14 \int_0^1 \left(\pi^2 - \ln^2 \frac{1+a}{1-a} \right) da=\frac14 \left(\pi^2 - \frac{\pi^2}3 \right) =\frac{\pi^2}{6} $$

Quanto
  • 97,352