> [!NOTE] Theorem > Let $\mu\in \mathbb{R},$ $\sigma>0$ and $f_{X}(x)=\frac{1}{\sqrt{ 2 \pi \sigma^{2} }} \cdot e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}.$Then $f_{X}$ is [[Riemann integration|integrable]] with $\int_{-\infty}^{\infty} f_{X}(x) \, dx =1$ **Proof**: Substituting $u=\frac{x-\mu}{\sigma}$ gives, $\int_{-\infty}^{+\infty}\frac1{\sqrt{2\pi\sigma^2}}\cdot e^{-\frac{(x-\mu)^2}{2\sigma^2}}\mathrm{d}x=\int_{-\infty}^{+\infty}\frac1{\sqrt{\pi}}\cdot e^{-u^2}\mathrm{d}u.$Thus STS $\int_{\infty }^{\infty} e^{-x^{2}}= \sqrt{ \pi } \, dx.$ # Applications See [[Normal Distribution]].