What is the integral of e^{ipiz} ?

The integral of e^{ipiz} is 1/pi sin(piz)-i 1/pi cos(piz)+C

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integral of e^{ipiz}

\int e^{iπ z} d z

\frac{1}{π} sin (π z) - i \frac{1}{π} cos (π z) + C

Solution steps

\int e^{iπ z} d z

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= \int cos (π z) + i sin (π z) d z

Apply the Sum Rule:

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

= \int cos (π z) d z + \int i sin (π z) d z

\int cos (π z) d z = \frac{1}{π} sin (π z)

\int i sin (π z) d z = - i \frac{1}{π} cos (π z)

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

Add a constant to the solution

= \frac{1}{π} sin (π z) - i \frac{1}{π} cos (π z) + C

x \to \infty lim (- 4 + \frac{3}{x})

x (x - 4)

n = 6 \sum \infty e^{5 - 6 n}

\frac{d}{d x} (ln (\frac{4 + x ^{2}}{x}))

f (t) = t - \frac{1}{t}

What is the integral of e^{ipiz} ?
The integral of e^{ipiz} is 1/pi sin(piz)-i 1/pi cos(piz)+C