$\int_{0}^{\frac{\pi}{2}} ln(sin(x))dx$

We know,
$sin x = cos( \frac{\pi}{2}-x)$
$I = \int_{0}^{\frac{\pi}{2}} ln(sin(x))dx \\ = \int_{0}^{\frac{\pi}{2}} ln( cos( \frac{\pi}{2}-x))dx \\ (u = \frac{\pi}{2}-x, du = -dx) \\ = \int_\frac{\pi}{2}^{0}ln(cos(u))-du\\ = \int_{0}^{\frac{\pi}{2}}ln(cos(u))du$
So,
$I = \int_{0}^{\frac{\pi}{2}} ln(sin(x))dx = \int_{0}^{\frac{\pi}{2}}ln(cos(u))du$
$2I = \int_{0}^{\frac{\pi}{2}} ln(sin(x))dx + \int_{0}^{\frac{\pi}{2}}ln(cos(x))dx\\ =\int_{0}^{\frac{\pi}{2}} ln(sin(x))+ln(cos(x))dx\\ =\int_{0}^{\frac{\pi}{2}} ln(sin(x)cos(x))dx \\ =\int_{0}^{\frac{\pi}{2}} ln(\frac{1}{2} sin(2x))dx \\ = \int_{0}^{\frac{\pi}{2}} ln(\frac{1}{2})dx + \int_{0}^{\frac{\pi}{2}} ln(sin(2x))dx \\ =\frac{\pi}{2}ln(\frac{1}{2})+ \int_{0}^{\frac{\pi}{2}} ln(sin(2x))dx \\ =-\frac{\pi}{2}ln(2)+ \int_{0}^{\frac{\pi}{2}} ln(sin(2x))dx \\ (u=2x, du=2dx)$

$=-\frac{\pi}{2}ln(2)+ \frac{1}{2}\int_{0}^{\pi} ln(sin(u))du$
The ln area of Sin from 0 to pi is 2 times of the area of sin from 0 to pi/2.
$=-\frac{\pi}{2}ln(2)+ \int_{0}^{\frac{\pi}{2}} ln(sin(u))du \\ 2I=-\frac{\pi}{2}ln(2)+ I \\ \therefore I = \int_{0}^{\frac{\pi}{2}} ln(sin(x))dx = -\frac{\pi}{2}ln(2)$

Author
crazyj7@gmail.com
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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
$x = r cos\theta , y=r sin\theta \\ dx dy \rightarrow r dr d\theta$

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

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

Author
crazyj7@gmail.com
Written with StackEdit.