How to compute $\parallel f \parallel_{L_2(\mathbb{R}^2)}$ for $f(x,y)=\frac{1}{1+(x-y)^2}$?

Question

So I want to compute $$\int\limits_\mathbb{R} \int\limits_\mathbb{R} \frac{1}{(1+(x-y)^2)^2} dxdy.$$

As I understand, I cannot reduce it to $1$-dimensional integrals, since Fubini's theorem requires measure of whole space to be finite.

Thank you.

Answer 1

The integrand is positive. Thus, you can use Fubini's theorem: \begin{align} \int_{\mathbb{R}^2} \dfrac{1}{\left(1+(x-y)^2\right)^2} \mathrm{d}\lambda &= \int_{\mathbb{R}} \left(\int_\mathbb{R}\dfrac{1}{\left(1+(x-y)^2\right)^2} \mathrm{d}x \right)\mathrm{d}y \\ &= \int_{\mathbb{R}} \left(\int_\mathbb{R}\dfrac{1}{\left(1+u^2\right)^2} \mathrm{d}u \right)\mathrm{d}y \text{ by substitution } u=x-y \\ &= \int_{\mathbb{R}} C\mathrm{d}y \\ &= +\infty \end{align} where $C >0$ is a constant

Answer 2

$\int_{\mathbb R} \frac 1{(1+(x-y)^{2})^{2}} dx=\int_{\mathbb R} \frac 1{(1+t^{2})^{2}} dt $ which is independent of $y$ hence the integral w.r.t $y$ is $\infty$. The function does not belong to $L^{2}$.