What is the integral from-1 to 0 of xe^{-ikx} ?

The integral from-1 to 0 of xe^{-ikx} is (1-cos(k)-ksin(k))/(k^2)+i(-sin(k)+kcos(k))/(k^2)

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integral from-1 to 0 of xe^{-ikx}

\int_{- 1}^{0} x e^{- ik x} d x

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

Solution steps

\int_{- 1}^{0} x e^{- ik x} d x

Apply u-substitution

: \int_{ik}^{0} - \frac{e ^{u} u}{k ^{2}} d u

= \int_{ik}^{0} - \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_{ik}^{0} e^{u} u d u

Apply Integration By Parts

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

= - \frac{1}{k ^{2}} [e^{u} u - \int e^{u} d u]_{ik}^{0}

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

= - \frac{1}{k ^{2}} [e^{u} u - e^{u}]_{ik}^{0}

Simplify

- \frac{1}{k ^{2}} [e^{u} u - e^{u}]_{ik}^{0} : - \frac{[ e ^{u} u - e ^{u} ] _{ik}^{0}}{k ^{2}}

= - \frac{[ e ^{u} u - e ^{u} ] _{ik}^{0}}{k ^{2}}

Compute the boundaries:

(- 1 + k sin (k) + cos (k)) + i (sin (k) - k cos (k))

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

Simplify

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

\int_{1}^{3} x^{2} + \frac{1}{x ^{2}} d x

\int_{\frac{π}{2}}^{x} sin (15 t) d t

\int_{0}^{8} \frac{x}{8} d x

\int_{1}^{2} \frac{5}{t} d t

\int_{- 1}^{2} [x] d x

What is the integral from-1 to 0 of xe^{-ikx} ?
The integral from-1 to 0 of xe^{-ikx} is (1-cos(k)-ksin(k))/(k^2)+i(-sin(k)+kcos(k))/(k^2)