integral from 0 to 1 of te^{-iwt}

解答

\int_{0}^{1} t e^{- i wt} d t

\frac{- 1 + cos ( w ) + w sin ( w )}{w ^{2}} + i \frac{- sin ( w ) + w cos ( w )}{w ^{2}}

求解步骤

\int_{0}^{1} t e^{- i wt} d t

使用换元积分法

= \int_{0}^{- i w} - \frac{e ^{u} u}{w ^{2}} d u

提出常数:

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

= - \frac{1}{w ^{2}} \cdot \int_{0}^{- i w} e^{u} u d u

使用分布积分法

= - \frac{1}{w ^{2}} [e^{u} u - \int e^{u} d u]_{0}^{- i w}

\int e^{u} d u = e^{u}

= - \frac{1}{w ^{2}} [e^{u} u - e^{u}]_{0}^{- i w}

化简

- \frac{1}{w ^{2}} [e^{u} u - e^{u}]_{0}^{- i w} : - \frac{[ e ^{u} u - e ^{u} ] _{0}^{- i w}}{w ^{2}}

= - \frac{[ e ^{u} u - e ^{u} ] _{0}^{- i w}}{w ^{2}}

带入上下限计算后得:

(- w sin (w) - cos (w) + 1) + i (sin (w) - w cos (w))

= - \frac{- w sin ( w ) - cos ( w ) + 1 + i ( sin ( w ) - w cos ( w ) )}{w ^{2}}

化简

= \frac{- 1 + cos ( w ) + w sin ( w )}{w ^{2}} + i \frac{- sin ( w ) + w cos ( w )}{w ^{2}}

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