What is the integral from 0 to 1 of e^{-iwt} ?

The integral from 0 to 1 of e^{-iwt} is 1/w sin(w)+i(-1+cos(w))/w

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integral from 0 to 1 of e^{-iwt}

\int_{0}^{1} e^{- i wt} d t

\frac{1}{w} sin (w) + i \frac{- 1 + cos ( w )}{w}

Solution steps

\int_{0}^{1} e^{- i wt} d t

Simplify

e^{- w i t} : cos (wt) - i sin (wt)

= \int_{0}^{1} cos (wt) - i sin (wt) d t

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int_{0}^{1} cos (wt) d t - \int_{0}^{1} i sin (wt) d t

\int_{0}^{1} cos (wt) d t = \frac{sin ( w )}{w}

\int_{0}^{1} i sin (wt) d t = i \frac{1 - cos ( w )}{w}

= \frac{sin ( w )}{w} - i \frac{1 - cos ( w )}{w}

Simplify

\frac{sin ( w )}{w} - i \frac{1 - cos ( w )}{w} : \frac{1}{w} sin (w) + i \frac{- 1 + cos ( w )}{w}

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

\int_{0}^{2} \frac{1}{( 2 - x ) ^{\frac{4}{5}}} d x

\int_{0}^{1} 1 + x^{2} x d x

\int_{0}^{2 x} sin (t^{2}) d t

\int_{- 5}^{0} 1 d x

\int_{0}^{x} y e^{- y x} d x

What is the integral from 0 to 1 of e^{-iwt} ?
The integral from 0 to 1 of e^{-iwt} is 1/w sin(w)+i(-1+cos(w))/w