Evaluation of double integral
Evaluate $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\left(\frac{1-e^{-xy}}{xy}\right)^2e^{-x^2-y^2}\text{d}x\text{d}y$$ Solution. One trivial observation is that $$\left(\frac{1-e^{-xy}}{xy}\right)^2=\sum_{n=0}^{\infty}\frac{2(-1)^n(2^{n+1}-1)}{(n+2)!}(xy)^n$$ Moreover $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}(xy)^{2n}e^{-x^2-y^2}\text{d}x\text{d}y=\left(\int_{-\infty}^{\infty}x^{2n}e^{-x^2}\text{d}x\right)^2=\frac{(2n)!^2}{2^{4n}(n!)^2}\pi$$ and $0$ when the power of $xy$ is odd. Combining the previous two we arrive at the following series $$S=\sum_{n=0}^{\infty}{2n\choose n}\frac{2^{2-2n}-2^{1-4n}}{(2n+1)(2n+2)}\pi$$ Now it is natural to consider the series $$f(x):= \sum_{n=0}^{\infty}{2n\choose n}\frac{x^{2n}}{(2n+1)(2n+2)}$$ Observe that $$(x^2f(x))''= \sum_{n=0}^{\infty}{2n\choose n}x^{2n}\ \ \ (*)$$ Recall that similar central binomials occur at the expansion of $(1-x)^{-1/2}$ and it indeed turns out that $$\sum_{n=0}^{\infty}{2n\choose n}x^{2n}=\fra