$$ \begin{align} \int_{0}^{\infty}e^{-(1+t^2)x}dx&=\left[-\frac{e^{-(1+t^2)x}}{1+t^2}\right]_{0}^{\infty}=\frac{1}{1+t^2}\\ \int_{0}^{\infty}\int_{0}^{\infty}e^{-(1+t^2)x}dxdt&=\int_{0}^{\infty}\frac{1}{1+t^2}dt\\ \int_{0}^{\infty}e^{-x}\int_{0}^{\infty}e^{-xt^2}dtdx&=\int_{0}^{\frac{\pi}{2}}\frac{1}{1+\tan^2\theta}\frac{1}{\cos^2\theta}d\theta\\ \int_{0}^{\infty}e^{-x}\frac{1}{\sqrt{x}}\int_{0}^{\infty}e^{-\tau^2}d\tau dx&=\int_{0}^{\frac{\pi}{2}}d\theta\\ \int_{0}^{\infty}\frac{e^{-x}}{\sqrt{x}}dx\int_{0}^{\infty}e^{-\tau^2}d\tau &=\frac{\pi}{2}\\ &=\pi\int_{0}^{\infty}e^{-\rho}d\rho\\ &=\pi\\ I&=\sqrt{\pi} \end{align} $$ --- $$ \begin{align} I&=\int_{-\infty}^{\infty}e^{-x^2}dx\\ &=\int_{-\infty}^{\infty}e^{-y^2}dy\\ I^2&=\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy\\ &=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy\\ &=\int_{0}^{2\pi}\int_{0}^{\infty}e^{-r^2}rdrd\theta\\ &=2\pi\int_{0}^{\infty}e^{-r^2}rdr\\ &=\pi\int_{0}^{\infty}e^{-\rho}d\rho\\ &=\pi\\ I&=\sqrt{\pi} \end{align} $$