Gaussian Integral

${\int }_{-\infty }^{\infty }{f}_X(x)\mathrm{d}x=1,f_X(x)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{-\frac{1}{2}{\left(\frac{x-\mu }{\sigma }\right)}^2}$

• $X$ is a random variable identically distributed to the Normal Distribution

${\int }_{-\infty }^{\infty }{e}^{-a{x}^2}\mathrm{d}x=\sqrt{\frac{\pi }{a}}$

Spicy Integrals

${\int }_0^{\infty }{x}^{2n}{e}^{-x^2/a^2}\mathrm{d}x=\sqrt{\pi }\frac{(2n)!}{n!}{\left(\frac{a}{2}\right)}^{2n+1}$

• Derived by using Gamma Function

${\int }_0^{\infty }{x}^{2n+1}{e}^{-x^2/a^2}\mathrm{d}x=\frac{n!}{2}{a}^{2n+2}$

• Derived by using Gamma Function

${\int }_0^{\infty }{x}^{2n}{e}^{-a{x}^2}\mathrm{d}x=\frac{\Gamma \left(\frac{n+1}{2}\right)}{2a^{\left(\frac{n+1}{2}\right)}}$

• Derived by using Gamma Function

${\int }_0^{\infty }\frac{x^n}{e^x-1}\mathrm{d}x=n!\zeta (n+1)$

• $\zeta$ is the Riemann Zeta function

Proof

$$\begin{array}{rl}I& \equiv {\int }_{-\infty }^{\infty }{e}^{-x^2}\mathrm{d}x\\ {\int }_{-\infty }^{\infty }{e}^{-x^2}\mathrm{d}x{\int }_{-\infty }^{\infty }{e}^{-y^2}\mathrm{d}y& ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-x^2}{e}^{-y^2}\mathrm{d}x\mathrm{d}y\\ & ={\int }_{-\infty }^{\infty }{\int }_{-\infty }^{\infty }{e}^{-(x^2+y^2)}\mathrm{d}x\mathrm{d}y\\ & ={\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-r^2}r\mathrm{d}r\mathrm{d}\theta ,u\equiv r^2\\ & =2\pi {\int }_0^{\infty }\frac{e^{-u}}{2}\mathrm{d}u\\ & =\pi (-e^{-u}{)}_0^{\infty }\\ & =\pi \\ & =I^2\\ \therefore I& =\sqrt{\pi }\end{array}$$