# Double integral with strange Change of Variables

I am trying to compute the following integral, $$\iint_{\mathbb{R}^2} \left(\frac{1-e^{-xy}}{xy}\right)^2 e^{-x^2-y^2}dxdy$$

First I tried substituting $x=r\cos{\theta}, y=r\sin{\theta}$ but it didn't really give me anything.

For the second try, I tried $u=x^2+y^2, v=xy$ but after computing the Jacobian, it got really messy, I couldn't really continue further.

Would there be a way to change the variables so that I would be able to compute this integral?

Unleash the infinite power of symmetry! The value of the integral is twice the value of the integral over the region $0\leq y\leq x$, and by setting $y=x t, dy = x\,dt$ we get

$$I = 2 \iint_{(0,+\infty)^2} x\left(\frac{1-e^{-tx^2}}{tx^2}\right)^2 e^{-x^2-t^2 x^2}\,dx\,dt\stackrel{x\mapsto\sqrt{s}}{=} \iint_{(0,+\infty)^2}\left(\frac{1-e^{-ts}}{ts}\right)^2 e^{-s-t^2 s}\,ds\,dt$$ By Frullani's theorem we have $\int_{0}^{+\infty}\frac{e^{-au}-e^{-bu}}{u}\,du = \log\frac{b}{a}$ for any $a,b>0$, hence by integrating with respect to $s$ we have $$I = \int_{0}^{+\infty}\left[(1+t^2)\log\frac{1+t^2}{(1+t)^2}-2(1+t+t^2)\log\frac{1+t+t^2}{(1+t)^2}\right]\frac{dt}{t^2}$$ and by integration by parts $$I = \int_{0}^{+\infty}\left[2t\log\frac{1+t^2}{(1+t)^2}-2(1+2t)\log\frac{1+t+t^2}{(1+t)^2}\right]\frac{dt}{t}.$$ Now $1+t+t^2=\Phi_3(t)=\frac{1-t^3}{1-t}$ and both $\log(1\pm t^k)$ and $\frac{1}{t}\log(1\pm t^k)$ have manageable primitives in terms of $\log$ and $\text{Li}_2$. The final outcome is

$$\iint_{(0,+\infty)^2} \left(\frac{1-e^{-xy}}{xy}\right)^2 e^{-x^2-y^2}dxdy = \color{blue}{\pi -\frac{2 \pi }{\sqrt{3}}+\frac{\pi ^2}{9}}.$$

• Why do the answers vary? – zxcvber Apr 21 '18 at 6:35
• @zxcvber: my integral is over $(0,+\infty)^2$ and Felix Marin's integral is over $\mathbb{R}^2$. – Jack D'Aurizio Apr 21 '18 at 10:59
• Then would the final answer be 4 times of your answer? I actually don't see why the integral is not convergent for $\mathbb{R}^2$ – zxcvber Apr 22 '18 at 8:11
• @zxcvber: no, since the integrand function is not that symmetric. But the integral over $\mathbb{R}^+\times\mathbb{R}^-$ can be computed in the exact same way. – Jack D'Aurizio Apr 22 '18 at 16:49


• Why do the answers vary? – zxcvber Apr 21 '18 at 6:36
• @zxcvber My answer corresponds to the original question. Namely, $\texttt{over}\ \left(-\infty,\infty\right)^2$. Jack D'Aurizio followed another route. – Felix Marin Apr 21 '18 at 20:11