How to prove this integration value is one ????

Question

For $0 \leq r<1$ and $\theta \in [-\pi,\pi]$ how to prove that the integral value is one. $$ \frac{1}{2 \pi} \int_{-\pi}^\pi P_r(\theta) d \theta=1 $$

The function here defined by the function $$ P_r(\theta)=\sum_{n=-\infty}^{\infty} r^{|n|} e^{i n \theta} $$

Answer 1

Since the series converges uniformly, we can interchange the series and the integral to get that

$$\frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta)~\mathrm{d}\theta=\sum_{n=-\infty}^\infty \frac{r^{\lvert n\rvert}}{2\pi}\int_{-\pi}^\pi e^{in\theta}~\mathrm{d}\theta.$$

Now recall that

$$\int_{-\pi}^\pi e^{in\theta}~\mathrm{d}\theta=\begin{cases} 2\pi,&n=0,\\ 0,&n\neq 0, \end{cases}$$

and so the above reduces down to

$$\frac{1}{2\pi}\int_{-\pi}^\pi P_r(\theta)~\mathrm{d}\theta=\frac{r^{\lvert 0\rvert}}{2\pi}2\pi=1.$$