Gaussian integration

Gaussian integration

$\int_{-\infty}^{\infty}{e^{-x^2}}dx$

$I = \int_{-\infty}^{\infty}{e^{-x^2}}dx \\ I = \int_{-\infty}^{\infty}{e^{-y^2}}dy \\ I^2 = \int_{-\infty}^{\infty}{e^{-x^2}}dx \int_{-\infty}^{\infty}{e^{-y^2}}dy \\ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-x^2} e^{-y^2} dx dy \\ = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} e^{-(x^2+y^2)} dx dy$

극좌표계로 변경
$x = r cos\theta , y=r sin\theta \\ dx dy \rightarrow r dr d\theta$

$I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} rdr d\theta$
$u = r^2$ , $du=2rdr$
$\int_{0}^{\infty} e^{-r^2} rdr = \int_{0}^{\infty}e^{-u} \frac{1}{2} du\\ = \frac{1}{2}[-e^{-u}]_{0}^{\infty} = \frac{1}{2}$

$I^2 = \int_{0}^{2\pi} \int_{0}^{\infty} e^{-r^2} rdr d\theta \\ = \int_{0}^{2\pi}\frac{1}{2}d\theta \\ = \frac{1}{2} 2\pi = \pi$
$\therefore I = \int_{-\infty}^{\infty}{e^{-x^2}}dx = \sqrt{\pi}$

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