ガウス積分の極座標変換を使わない計算

$$\begin{array}{rl}I& :={\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx\end{array}$$
とおく。ここで
$${\int }_0^{\mathrm{\infty }}\frac{dx}{1+x^2}={[\arctan x]}_0^{\mathrm{\infty }}=\frac{\pi }{2}$$
より
$$\begin{array}{rl}\frac{\pi }{2}& ={\int }_0^{\mathrm{\infty }}{\int }_0^{\mathrm{\infty }}{e}^{-(1+x^2)t}dtdx\\ & ={\int }_0^{\mathrm{\infty }}{e}^{-t}{\int }_0^{\mathrm{\infty }}{e}^{-t{x}^2}dxdt\\ & ={\int }_0^{\mathrm{\infty }}{e}^{-t}\frac{I}{\sqrt{t}}dt\\ & =2I{\int }_0^{\mathrm{\infty }}{e}^{-u^2}du(u=\sqrt{t})\\ & =2I^2\end{array}$$

$I$は正なので
$$I=\frac{\sqrt{\pi }}{2}.$$