Double Integral $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{-(x-y)^2}}{1 + (x + y)^2}\,dx\,dy$

Question

Evaluate $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{-(x-y)^2}}{1 + (x + y)^2}\,dx\,dy$$ by integrating over the square $\textit{S}_a:-a\leq x\leq a, -a\leq y \leq a$ and taking the limit as $a \rightarrow \infty $.

Answer 1

$$I=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} \frac{e^{-(x-y)^2}}{1+(x+y)^2} dx dy=$$ Let $x+y=u, x-y=v$, then $$\implies I= \frac{1}{2}\int_{\infty}^{\infty} \int_{-\infty}^{\infty} \frac{e^{-v^2}}{1+u^2} du dv$$ $$\implies I=\frac{1}{2}\int_{-\infty}^{\infty}e^{-v^2}dv \int_{-\infty}^{\infty} \frac{du}{1+u^2} du$$ $$I=\frac{1}{2} \sqrt{\pi} \pi =\frac{\pi \sqrt{\pi}}{2}$$

Answer 2

Let $u=\frac1{\sqrt2}(x-y)$ and $v=\frac1{\sqrt2}(x+y)$. Then, $dxdy=dudv$ and,

$$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\frac{e^{-(x-y)^2}}{1 + (x + y)^2}\,dx\,dy =\int_{-\infty}^{\infty}e^{-2u^2}du\int_{-\infty}^{\infty}\frac{dv}{1 + 2v^2}=\sqrt{\frac\pi2}\cdot \frac\pi{\sqrt2} = \frac{\pi^{3/2}}2$$