A Double Integral

Well, it is kinda obvious that $(e^{x^{2} + y^{2}})^{-1} = e^{-x^{2}}e^{-y^{2}}$ . Also observe that

$\int_{0}^{\infty} e^{-u^{2}} du = \left (\frac{1}{2} \sqrt{\pi} \; erf(x) \right )_{0}^{\infty} = \frac{\sqrt{\pi}}{2}$

Therefore

$4 \times \int_{0}^{\infty} \int_{0}^{\infty} e^{-x^{2}}e^{-y^{2}} \;dx\;dy = 4 \times \int_{0}^{\infty} e^{-x^{2}} \;dx \int_{0}^{\infty}e^{-y^{2}} \;dy = 4 \times \left ( \frac{\sqrt{\pi}}{2} \right )^{2} = \pi$

Another method is to observe that in polar coordinates $x^{2} + y^{2} = r^{2}$ and $dx \; dy = r \;dr \;d\theta$ where $0 < r < \infty$ and $0 < \theta < \pi /4$ . Then the integral becomes

$4 \times \int_{0}^{\infty} re^{-r^{2}} \;dr \int_{0}^{\frac{\pi}{4}} d\theta = 4 \times \frac{\pi}{4} \int_{0}^{\infty} re^{-r^{2}} \;dr$

which is easy to solve by making $r^{2} = u$ and the result is again $\pi$ .