$$ \definecolor{rr}{rgb}{0.95,0.1,0} \definecolor{bb}{rgb}{0,0.4,1} \definecolor{gg}{rgb}{0,0.6,0} \definecolor{yy}{rgb}{0.8,0.5,0} \definecolor{pp}{rgb}{0.9,0,0.5} $$

가우스적분

~카테고리:
수학/나머지

가우스분포(정규분포)라는 함수 e-x2를 -∞~∞에서 적분한다.

가우스는 오만 데에 다 나오냐? 이것도 가우스 저것도 가우스. 망할 수학자 과학자들 작명센스 시발.

$$ \int_{-\infty}^\infty \operatorname{d}\!x \!\cdot\! e^{-x^2} \quad = \quad \sqrt{2\pi\over 2}$$
$$ \left( \textcolor{gg}{\int_{-\infty}^\infty \operatorname{d}\!x \!\cdot\! e^{-x^2}} \right)^2 $$
$$ \int_{-\infty}^\infty \operatorname{d}\!y \!\cdot\! e^{-y^2} = \int_{-\infty}^\infty \operatorname{d}\!x \!\cdot\! e^{-x^2} $$ $$ = \int_{-\infty}^\infty \int_{-\infty}^\infty \operatorname{d}\!x \!\cdot\! \operatorname{d}\!y \!\cdot\! \color{yy} e^{-x^2} \!\cdot\! e^{-y^2} $$
극좌표로 변환
x2 + y2 = r2
$$ = \int_{-\infty}^\infty \int_{-\infty}^\infty \operatorname{d}\!x \!\cdot\! \operatorname{d}\!y \!\cdot\! \color{yy} e^{-r^2} $$
dx·dy = r·dr·dθ $$ = \textcolor{rr}{\int_{0}^{2\pi} \operatorname{d}\!\theta} \textcolor{bb}{\int_{0}^\infty \operatorname{d}\!r \!\cdot\! r \!\cdot\! e^{-r^2}} $$
$$ \textcolor{bb}{\int_{0}^\infty \operatorname{d}\!r \!\cdot\! r \!\cdot\! e^{-r^2}} $$ $$ = \int_{0}^\infty \textcolor{pp}{\operatorname{d}\!r \!\cdot\! r} \!\cdot\! e^{-r^2} $$
$$ \color{pp}\operatorname{d}\!\left( r^2 \right) = 2 \cdot r \cdot dr $$
r=0,∞일 때 r2도 0,∞
$$ = \textcolor{pp}{\frac{1}{2}} \int_{\textcolor{pp}{r^2}=0}^\infty \textcolor{pp}{\operatorname{d}\!\left( r^2 \right)} \cdot e^{-r^2} $$
$$ = \frac{-1\;}{2} \int_{-r^2=0}^{-\infty} \operatorname{d}\!\left( -r^2 \right) \cdot e^{-r^2} $$
$$ = \frac{-1\;}{2} \int_{z=0}^{-\infty} \operatorname{d}\! z \cdot e^{z} $$
$$ = \frac{-1\;}{2} \textcolor{pp}{\left( e^{-\infty} - e^0 \right)} \quad = \quad \frac{-1\;}{2} \textcolor{pp}{\left( 0 - 1 \right)} $$
$$ = \frac{1}{2} $$
$$ = \textcolor{rr}{2\pi} \textcolor{bb}{\frac{1}{2}} $$ $$ \quad = \quad \textcolor{gg}{\frac{2\pi}{2}} $$

$$ \textcolor{gg}{\int_{-\infty}^\infty \operatorname{d}\!x \!\cdot\! e^{-x^2}} $$ $$ =\sqrt{\textcolor{gg}{\frac{2\pi}{2}}} $$