What is the integral of xe^{-iwx} ?

The integral of xe^{-iwx} is 1/(w^2)(cos(wx)+wxsin(wx))+i 1/(w^2)(-sin(wx)+wxcos(wx))+C

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integral of xe^{-iwx}

\int x e^{- i w x} d x

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

Solution steps

\int x e^{- i w x} d x

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 x

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

Simplify

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

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

Add a constant to the solution

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

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What is the integral of xe^{-iwx} ?
The integral of xe^{-iwx} is 1/(w^2)(cos(wx)+wxsin(wx))+i 1/(w^2)(-sin(wx)+wxcos(wx))+C