Is $\int_0^{\pi/2}e^{inx}dx=0$ for $n \neq 0$?

Question

How can we prove this using Cauchy's Integral? We know that $\int_{-\pi}^{\pi}e^{inx} dx = 0$ for $n$ not equal to $0$, and is equal to $2\pi$ for $n=0$, how can we prove that $\int_0^{\pi/2} e^{inx} dx=0$ for $n \neq 0$

Answer 1

$\int_0^{\pi/2} e^{inx} dx= [\frac{1}{in}e^{inx}]_0^{\pi/2}.$

Can you proceed ?

Answer 2

The integral is$$\left[\frac{1}{in}e^{inx}\right]_0^{\pi/2}=\frac{e^{in\pi/2}-1}{in}=\frac2n\sin\frac{n\pi}{4}e^{in\pi/4}.$$This is nonzero if $4\nmid n$.