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

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

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

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

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

Solution steps

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

Simplify

e^{- ik x} : cos (k x) - i sin (k x)

= \int_{0}^{1} cos (k x) - i sin (k x) d x

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 (k x) d x - \int_{0}^{1} i sin (k x) d x

\int_{0}^{1} cos (k x) d x = \frac{sin ( k )}{k}

\int_{0}^{1} i sin (k x) d x = i \frac{1 - cos ( k )}{k}

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

Simplify

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

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

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