Consider the double integral $\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\mathrm{e}^{-\left( x^2 + y^2 \right) }\,\mathrm{d}x}\,\mathrm{d}y} \end{align*}$. This can be evaluated by converting to polars. Since we are integrating over the entire real line, that means in polars $\displaystyle \begin{align*} 0 \leq r \leq \infty \end{align*}$ and $\displaystyle \begin{align*} 0 \leq \theta \leq 2\pi \end{align*}$. We have

$\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\mathrm{e}^{-\left( x^2 + y^2 \right) } \, \mathrm{d}x}\,\mathrm{d}y} &= \int_0^{2\pi}{\int_0^{\infty}{\mathrm{e}^{-r^2}\,r\,\mathrm{d}r}\,\mathrm{d}\theta} \\ &= -\frac{1}{2} \int_0^{2\pi}{\int_0^{\infty}{-2r\,\mathrm{e}^{-r^2} \,\mathrm{d}r}\,\mathrm{d}\theta} \\ &= -\frac{1}{2} \int_0^{2\pi}{\int_0^{-\infty}{\mathrm{e}^u\,\mathrm{d}u}\, \mathrm{d} \theta} \textrm{ after substituting } u = -r^2 \implies \mathrm{d}u = -2r\,\mathrm{d}r \\ &= \frac{1}{2} \int_0^{2\pi}{\int_{-\infty}^0{\mathrm{e}^u\,\mathrm{d}u }\,\mathrm{d}\theta} \\ &= \frac{1}{2} \int_0^{2\pi}{\lim_{\epsilon \to -\infty}\int_{\epsilon}^0{\mathrm{e}^u\,\mathrm{d}u }\,\mathrm{d}\theta} \\ &= \frac{1}{2} \int_0^{2\pi}{\lim_{\epsilon \to -\infty}{ \left[ \mathrm{e}^u \right] _{\epsilon}^0 }\mathrm{d}\theta} \\ &= \frac{1}{2} \int_0^{2\pi}{\lim_{\epsilon \to -\infty}{ \left( \mathrm{e}^0 - \mathrm{e}^{\epsilon} \right) }\,\mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\pi}{ \left( 1 - 0 \right) \, \mathrm{d}\theta } \\ &= \frac{1}{2} \int_0^{2\pi}{1\,\mathrm{d}\theta} \\ &= \frac{1}{2} \left[ \theta \right] _0^{2\pi} \\ &= \frac{1}{2} \left( 2\pi - 0 \right) \\ &= \pi \end{align*}$

Now let's try to evaluate the double integral in another way...

$\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\mathrm{e}^{-\left( x^2 + y^2 \right) } \,\mathrm{d}x}\,\mathrm{d}y} &= \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{e}^{-y^2}\, \mathrm{d}x}\,\mathrm{d}y} \\ &= \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \int_{-\infty}^{\infty}{\mathrm{e}^{-y^2}\,\mathrm{d}y} \end{align*}$

Notice that these two integrals are numerically identical, and thus $\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} = \int_{-\infty}^{\infty}{\mathrm{e}^{-y^2}\,\mathrm{d}y} \end{align*}$, giving

$\displaystyle \begin{align*} \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \int_{-\infty}^{\infty}{\mathrm{e}^{-y^2}\,\mathrm{d}y} &= \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x}\int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \\ &= \left( \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \right) ^2 \end{align*}$

But we already know what this integral's value is, it's $\displaystyle \begin{align*} \pi \end{align*}$, thus

$\displaystyle \begin{align*} \left( \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} \right) ^2 &= \pi \\ \int_{-\infty}^{\infty}{\mathrm{e}^{-x^2}\,\mathrm{d}x} &= \sqrt{\pi} \end{align*}$