Let $\Omega=[0,1]\times[0,1]$, $\lambda$ the restriction of the Lebesgue measure $\lambda^2$ on $\mathbb{R}$ to $\Omega$. Define
$$\begin{array}{rccc} f\colon& \Omega&\longrightarrow&\mathbb{R}\\& (x,y)&\mapsto& \begin{cases} (1-xy)^{-1} & \text{ if } xy\neq 1,\\ 0 & \text{ if } xy=1. \end{cases}\end{array}$$
I have already proven that $f$ is jointly measurable. Now I have to compute $\int_{\Omega}f\,\mathrm d\lambda$. There is a hint that I might need to apply Fubini's theorem twice, and that I can use $\sum_{n=1}^{\infty}\dfrac{1}{n^2}=\dfrac{\pi^2}{6}$ without proof.
I really don't know where to start, can somebody help me?
You have\begin{align}\int_0^1\int_0^1\frac1{1-xy}\,\mathrm dx\,\mathrm dy&=\int_0^1\int_0^1\sum_{n=1}^\infty(xy)^{n-1}\,\mathrm dx\,\mathrm dy\\&=\sum_{n=1}^\infty\frac1{n^2}\end{align}by the monotone convergence theorem and because$$(\forall n\in\mathbb{N}):\frac1{n^2}=\int_0^1\int_0^1\sum_{n=1}^\infty x^{n-1}y^{n-1}\,\mathrm dx\,\mathrm dy.$$