Ethan's Notes

    Gaussian Integral

    #Math

    $\displaystyle \int_{-\infty}^{\infty} f_{X}(x) , \mathrm{d}x=1,,f_{X}(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left( \frac{x-\mu}{\sigma} \right)^{2}}$

    $\displaystyle \int_{-\infty}^{\infty} e^{-ax^{2}} , \mathrm{d}x=\sqrt{ \frac{\pi}{a} }$

    Spicy Integrals

    $\displaystyle \int_{0}^{\infty} x^{2n}e^{-x^{2}/a^{2}} , \mathrm{d}x=\sqrt{ \pi } \frac{(2n)!}{n!}\left( \frac{a}{2} \right)^{2n+1}$

    $\displaystyle \int_{0}^{\infty} x^{2n+1}e^{-x^{2}/a^{2}} , \mathrm{d}x=\frac{n!}{2}a^{2n+2}$

    $\displaystyle \int_{0}^{\infty} x^{2n}e^{-ax^{2}} , \mathrm{d}x=\frac{\Gamma\left( \frac{n+1}{2} \right)}{2a^{\left( \frac{n+1}{2} \right)}}$

    $\displaystyle \int_{0}^{\infty} \frac{x^{n}}{e^{x}-1} , \mathrm{d}x=n!\zeta(n+1)$

    Proof

    $$
    \begin{align}
    I&\equiv \int_{-\infty}^{\infty} e^{-x^{2}} , \mathrm{d}x \\\int_{-\infty}^{\infty} e^{-x^{2}}\mathrm{d}x ,\int_{-\infty}^{\infty} e^{-y^{2}} , \mathrm{d}y &=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^{2}}e^{-y^{2}} , \mathrm{d}x , \mathrm{d}y \\&=\int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^{2}+y^{2})} , \mathrm{d}x , \mathrm{d}y \\&= \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^{2}}r , \mathrm{d}r, \mathrm{d}\theta,,u\equiv r^{2} \\&=2\pi\int_{0}^{\infty} \frac{e^{-u}}{2} , \mathrm{d}u \\&=\pi(-e^{-u})^{\infty}_{0} \\&=\pi \\&=I^{2} \\\therefore I&=\sqrt{ \pi }
    \end{align}
    $$