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

The integral from-1 to 1 of e^{-iwx} is 2/w sin(w)

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integral from-1 to 1 of e^{-iwx}

\int_{- 1}^{1} e^{- i w x} d x

\frac{2}{w} sin (w)

Solution steps

\int_{- 1}^{1} e^{- i w y} d y

Simplify

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

= \int_{- 1}^{1} cos (w y) - i sin (w y) d y

Apply the Sum Rule:

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

= \int_{- 1}^{1} cos (w y) d y - \int_{- 1}^{1} i sin (w y) d y

\int_{- 1}^{1} cos (w y) d y = \frac{2 sin ( w )}{w}

\int_{- 1}^{1} i sin (w y) d y = 0

= \frac{2 sin ( w )}{w} - 0

Simplify

\frac{2 sin ( w )}{w} - 0 : \frac{2}{w} sin (w)

= \frac{2}{w} sin (w)

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

\int_{0}^{π} 7 + sin (x) d x

\int_{0}^{\frac{π}{3}} tan (x) + sec (x) d x

\int_{2}^{6} ln (x + 3) d x

\int_{2}^{3} (\frac{1}{t - 1} + t) d t

What is the integral from-1 to 1 of e^{-iwx} ?
The integral from-1 to 1 of e^{-iwx} is 2/w sin(w)