<s>[INST] Review on double integral Forgot how to do these :
$$\displaystyle\int_0^1\int_0^1\frac{\text{d}x\text{d}y}{1-x^2y^2}$$
 [/INST] Consider the integral over $x$:
$$\int_0^1 dx \; \frac{1}{1-x^2 y^2} $$
Make the substitution $x = \sin{\theta}/y$, $dx = \cos{\theta}/y \: d \theta$:
$$\begin{align} &= \frac{1}{y} \int_0^{\arcsin{y}} d \theta \sec{\theta} \\ &=  \frac{1}{y} [\log{(\sec{\theta} + \tan{\theta})]_{0}^{\arcsin{y}}} \\ &= \frac{1}{2 y} \log{\left ( \frac{1+y}{1-y} \right )} \end{align} $$
Now you can do the integral over $y$:
$$\begin{align} \int_0^1 dy \: \int_0^1 dx \: \frac{1}{1-x^2 y^2} &= \frac{1}{2} \int_0^1 dy \: \frac{1}{y} \log{\left ( \frac{1+y}{1-y} \right )}\\ &= \sum_{n=0}^{\infty} \int_0^1 dy \: \frac{y^{2 n}}{2 n+1} \\ &= \sum_{n=0}^{\infty} \frac{1}{(2 n+1)^2}\\ &= \frac{\pi^2}{8} \\\\  \end{align} $$
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