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

The integral of xe^{-ikx} is 1/(k^2)(cos(kx)+kxsin(kx))+i 1/(k^2)(-sin(kx)+kxcos(kx))+C

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integral of xe^{-ikx}

\int x e^{- ik x} d x

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

Solution steps

\int x e^{- ik x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

Apply Integration By Parts

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

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

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

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

Substitute back

u = - ki x

= - \frac{1}{k ^{2}} (e^{- ki x} (- ki x) - e^{- ki x})

Simplify

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

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

Add a constant to the solution

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

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What is the integral of xe^{-ikx} ?
The integral of xe^{-ikx} is 1/(k^2)(cos(kx)+kxsin(kx))+i 1/(k^2)(-sin(kx)+kxcos(kx))+C