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