integral of xe^{-(ax+b)^2}

解答

\int x e^{- (a x + b)^{2}} d x

\frac{1}{a ^{2}} (- \frac{1}{2} e^{- a^{2} x^{2} - 2 ab x - b^{2}} - \frac{π}{2} b erf (a x + b)) + C

求解步骤

\int x e^{- (a x + b)^{2}} d x

使用换元积分法

= \int \frac{e ^{- u^{2}} ( u - b )}{a ^{2}} d u

提出常数:

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

= \frac{1}{a ^{2}} \cdot \int e^{- u^{2}} (u - b) d u

乘开

e^{- u^{2}} (u - b) : e^{- u^{2}} u - b e^{- u^{2}}

= \frac{1}{a ^{2}} \cdot \int e^{- u^{2}} u - b e^{- u^{2}} d u

使用积分加法定则:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \frac{1}{a ^{2}} (\int e^{- u^{2}} u d u - \int b e^{- u^{2}} d u)

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

\int b e^{- u^{2}} d u = b \frac{π}{2} erf (u)

= \frac{1}{a ^{2}} (- \frac{1}{2} e^{- u^{2}} - b \frac{π}{2} erf (u))

u = a x + b

代回

= \frac{1}{a ^{2}} (- \frac{1}{2} e^{- (a x + b)^{2}} - b \frac{π}{2} erf (a x + b))

乘开

- (a x + b)^{2} : - a^{2} x^{2} - 2 ab x - b^{2}

= \frac{1}{a ^{2}} (- \frac{1}{2} e^{- a^{2} x^{2} - 2 ab x - b^{2}} - \frac{π}{2} b erf (a x + b))

解答补常数

= \frac{1}{a ^{2}} (- \frac{1}{2} e^{- a^{2} x^{2} - 2 ab x - b^{2}} - \frac{π}{2} b erf (a x + b)) + C

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