Calculate the improper double integral

$$\iint_R e^{-x^2-y^2}\,\dd(x,y)$$ over the right half-plane $R$.

Solution

The double integral is easy to calculate in polar coordinates $(r,\varphi)$ with $r\ge 0$ and $-\tfrac{1}{2}\pi\le \varphi\le -\tfrac{1}{2}\pi$ as iterated integral:

$$\begin{aligned}\iint_R e^{-x^2-y^2}\,\,\dd(x,y)&= \int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\left(\int_{0}^{\infty}e^{-r^2}\,r\,\dd r\right)\,\dd\varphi\\[0.25cm] &=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\left(\lim_{\rho\to\infty}\biggl[-\frac{1}{2}e^{-r^2}\biggr]_{0}^{\rho}\right)\,\dd\varphi\\[0.25cm] &=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\left(\lim_{\rho\to\infty}\frac{1}{2}\Bigl(1-e^{-\rho^2}\Bigr)\right)\,\dd\varphi\\[0.25cm] &=\int_{-\frac{1}{2}\pi}^{\frac{1}{2}\pi}\frac{1}{2}\,\dd\varphi\\[0.25cm] &= \biggl[\frac{1}{2}\varphi\biggr]_{\varphi=-\frac{1}{2}\pi}^{\varphi=\frac{1}{2}\pi}\\[0.25cm] &= \frac{1}{2}\pi\end{aligned}$$
Due to symmetry of the integrand, the double integral over the left half-plane is also equal to $\tfrac{1}{2}\pi$ and so the improper double-integral of the integrand over the entire $xy$-plane is equal to $\pi$. Since $$\begin{aligned}\pi &=\iint_{\mathbb{R}^2} e^{-x^2+y^2}\,\dd(x,y)\\[0.25cm] &= \int_{x=-\infty}^{x=\infty}\int_{y=-\infty}^{y=\infty}e^{-x^2-y^2}\,\dd x \, \dd y\\[0.25cm]&=\int_{y=-\infty}^{y=\infty}e^{-y^2}\left( \int_{x=-\infty}^{x=\infty} e^{-x^2}\,\dd x\right) \dd y \\[0.25cm] &=\left(\int_{y=-\infty}^{y=\infty}e^{-y^2}\,\dd y\right)\cdot\left(\int_{x=-\infty}^{x=\infty}e^{-x^2}\,\dd x\right)\\[0.25cm] &=\left(\int_{-\infty}^{\infty}e^{-x^2}\,\dd x\right)^2\end{aligned}$$ we also determined the following improper integral in one variable: $$\int_{-\infty}^{\infty}e^{-x^2}\,\dd x=\sqrt{\pi}$$