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 $$\sum_{n=0}^{\infty}{2n\choose n}x^{2n}=\frac{1}{\sqrt{1-4x^2}}$$
Now integrating $(*)$ twice and adjusting the constants we arrive at $$f(x)=\frac{-1+\sqrt{1-4x^2}+x\arcsin(2x)}{4x^2}$$
It remains to calculate $S$ through $f$ - $$S=\pi\left(4f\left(\frac{1}{2}\right)-2f\left(\frac{1}{4}\right)\right)=\frac{4}{3}\pi(3-3\sqrt{3}+\pi)$$
At the last point Abel's theorem for evaluating a function at the end of the interval of convergence of its series is needed. Also some other convergence issues should be made rigorous.
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