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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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

= - \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

\frac{d}{d θ} ((1 + 2 sin (θ)) cos (θ))

d er i v a t i v e - 7 x^{5} - 1

\int \frac{1}{x ^{2} - 4 x + 13} d x

\frac{\partial}{\partial x} (4 x^{8 y})

\int x 2 - 1 - x^{2} d x

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