integral of 1/(sqrt(2pi))e^{(x^2)/2}

解

\int \frac{1}{2 π} e^{\frac{x ^{2}}{2}} d x

\frac{1}{2} erfi (\frac{x}{2}) + C

解答ステップ

\int \frac{1}{2 π} e^{\frac{x ^{2}}{2}} d x

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2 π} \cdot \int e^{\frac{x ^{2}}{2}} d x

2 π = 2^{\frac{1}{2}} π^{\frac{1}{2}}

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} \cdot \int e^{\frac{x ^{2}}{2}} d x

指数の規則を適用する:

\frac{a ^{n}}{b ^{n}} = (\frac{a}{b})^{n}

\frac{x ^{2}}{2} = (\frac{x}{2})^{2}

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} \cdot \int e^{(\frac{x}{2})^{2}} d x

u置換積分法を適用する

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} \cdot \int e^{u^{2}} 2 d u

定数を除く:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} 2 \cdot \int e^{u^{2}} d u

非初等積分を使用する:

\int e^{u^{2}} d u = \frac{π}{2} erfi (u)

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} 2 \frac{π}{2} erfi (u)

代用を戻す

u = \frac{x}{2}

= \frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} 2 \frac{π}{2} erfi (\frac{x}{2})

簡素化

\frac{1}{2 ^{\frac{1}{2}} π ^{\frac{1}{2}}} 2 \frac{π}{2} erfi (\frac{x}{2}) : \frac{1}{2} erfi (\frac{x}{2})

= \frac{1}{2} erfi (\frac{x}{2})

定数を解答に追加する

= \frac{1}{2} erfi (\frac{x}{2}) + C