$$$x$$$에 대한 $$$e^{- \frac{x^{2}}{2 \sigma^{2}}}$$$의 적분

$$$\int e^{- \frac{x^{2}}{2 \sigma^{2}}}\, dx$$$을(를) 구하시오.

$$$u=\frac{\sqrt{2} x}{2 \left|{\sigma}\right|}$$$라 하자.

그러면 $$$du=\left(\frac{\sqrt{2} x}{2 \left|{\sigma}\right|}\right)^{\prime }dx = \frac{\sqrt{2}}{2 \left|{\sigma}\right|} dx$$$ (단계는 »에서 볼 수 있습니다), 그리고 $$$dx = \sqrt{2} \left|{\sigma}\right| du$$$임을 얻습니다.

따라서,

$${\color{red}{\int{e^{- \frac{x^{2}}{2 \sigma^{2}}} d x}}} = {\color{red}{\int{\sqrt{2} e^{- u^{2}} \left|{\sigma}\right| d u}}}$$

상수배 법칙 $$$\int c f{\left(u \right)}\, du = c \int f{\left(u \right)}\, du$$$을 $$$c=\sqrt{2} \left|{\sigma}\right|$$$와 $$$f{\left(u \right)} = e^{- u^{2}}$$$에 적용하세요:

$${\color{red}{\int{\sqrt{2} e^{- u^{2}} \left|{\sigma}\right| d u}}} = {\color{red}{\sqrt{2} \left|{\sigma}\right| \int{e^{- u^{2}} d u}}}$$

이 적분(오차 함수)은 닫힌형 표현이 없습니다:

$$\sqrt{2} \left|{\sigma}\right| {\color{red}{\int{e^{- u^{2}} d u}}} = \sqrt{2} \left|{\sigma}\right| {\color{red}{\left(\frac{\sqrt{\pi} \operatorname{erf}{\left(u \right)}}{2}\right)}}$$

다음 $$$u=\frac{\sqrt{2} x}{2 \left|{\sigma}\right|}$$$을 기억하라:

$$\frac{\sqrt{2} \sqrt{\pi} \left|{\sigma}\right| \operatorname{erf}{\left({\color{red}{u}} \right)}}{2} = \frac{\sqrt{2} \sqrt{\pi} \left|{\sigma}\right| \operatorname{erf}{\left({\color{red}{\left(\frac{\sqrt{2} x}{2 \left|{\sigma}\right|}\right)}} \right)}}{2}$$

따라서,

$$\int{e^{- \frac{x^{2}}{2 \sigma^{2}}} d x} = \frac{\sqrt{2} \sqrt{\pi} \left|{\sigma}\right| \operatorname{erf}{\left(\frac{\sqrt{2} x}{2 \left|{\sigma}\right|} \right)}}{2}$$

적분 상수를 추가하세요:

$$\int{e^{- \frac{x^{2}}{2 \sigma^{2}}} d x} = \frac{\sqrt{2} \sqrt{\pi} \left|{\sigma}\right| \operatorname{erf}{\left(\frac{\sqrt{2} x}{2 \left|{\sigma}\right|} \right)}}{2}+C$$

$$$\int e^{- \frac{x^{2}}{2 \sigma^{2}}}\, dx = \frac{\sqrt{2} \sqrt{\pi} \left|{\sigma}\right| \operatorname{erf}{\left(\frac{\sqrt{2} x}{2 \left|{\sigma}\right|} \right)}}{2} + C$$$A