What is the integral from-1 to 1 of-t*e^{-iwt} ?

The integral from-1 to 1 of-t*e^{-iwt} is i(2sin(w)-2wcos(w))/(w^2)

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integral from-1 to 1 of-t*e^{-iwt}

\int_{- 1}^{1} - t \cdot e^{- i wt} d t

i \frac{2 sin ( w ) - 2 w cos ( w )}{w ^{2}}

Solution steps

\int_{- 1}^{1} - t e^{- i wt} d t

Take the constant out:

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

= - \int_{- 1}^{1} e^{- w i t} t d t

Apply u-substitution

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

= - \int_{i w}^{- i w} - \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_{i w}^{- i w} e^{u} u d u)

Apply Integration By Parts

: [e^{u} u - \int e^{u} d u]_{i w}^{- i w}

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

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

= - (- \frac{1}{w ^{2}} [e^{u} u - e^{u}]_{i w}^{- i w})

Simplify

- (- \frac{1}{w ^{2}} [e^{u} u - e^{u}]_{i w}^{- i w}) : \frac{[ e ^{u} u - e ^{u} ] _{i w}^{- i w}}{w ^{2}}

= \frac{[ e ^{u} u - e ^{u} ] _{i w}^{- i w}}{w ^{2}}

Compute the boundaries:

i (2 sin (w) - 2 w cos (w))

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

Simplify

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

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What is the integral from-1 to 1 of-t*e^{-iwt} ?
The integral from-1 to 1 of-t*e^{-iwt} is i(2sin(w)-2wcos(w))/(w^2)