What is the integral of te^{-iwt} ?

The integral of te^{-iwt} is 1/(w^2)(cos(wt)+wtsin(wt))+i 1/(w^2)(-sin(wt)+wtcos(wt))+C

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integral of te^{-iwt}

\int t e^{- i wt} d t

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

Solution steps

\int t e^{- i wt} d t

Apply u-substitution

: \int - \frac{e ^{u} u}{w ^{2}} d u

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

Apply Integration By Parts

: e^{u} u - \int e^{u} d u

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

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

= - \frac{1}{w ^{2}} (e^{u} u - e^{u})

Substitute back

u = - w i t

= - \frac{1}{w ^{2}} (e^{- w i t} (- w i t) - e^{- w i t})

Simplify

- \frac{1}{w ^{2}} (e^{- w i t} (- w i t) - e^{- w i t}) : \frac{1}{w ^{2}} (cos (wt) + wt sin (wt)) + i \frac{1}{w ^{2}} (- sin (wt) + wt cos (wt))

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

Add a constant to the solution

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

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What is the integral of te^{-iwt} ?
The integral of te^{-iwt} is 1/(w^2)(cos(wt)+wtsin(wt))+i 1/(w^2)(-sin(wt)+wtcos(wt))+C