What is the integral from 0 to 1 of te^{-iwt} ?

The integral from 0 to 1 of te^{-iwt} is (-1+cos(w)+wsin(w))/(w^2)+i(-sin(w)+wcos(w))/(w^2)

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

\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}}

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

\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

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{0}^{- i w}

= - \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}

Simplify

- \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}}

Compute the boundaries:

(- 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}}

Simplify

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

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What is the integral from 0 to 1 of te^{-iwt} ?
The integral from 0 to 1 of te^{-iwt} is (-1+cos(w)+wsin(w))/(w^2)+i(-sin(w)+wcos(w))/(w^2)