TITLE: Poisson Integral relation
QUESTION [5 upvotes]: If $$ I_n(r) = \int_0^\pi \frac{\cos nx}{r^2-2r\cos x+1} \, dx $$
How to prove that 
$$ I_{n-1}(r)+I_{n+1}(r)= \left(r+\frac{1}{r}\right)I_n(r)\text{ ?}$$
I only find that $$I_{n-1}(r)+I_{n+1}(r)= \int_0^\pi \frac{2\cos nx\cos x}{r^2-2r\cos x+1} \, dx$$

REPLY [6 votes]: $$\begin{aligned}
I_{n-1}(r)+I_{n+1}(r) &=\int_0^{\pi}\frac{2\cos(nx) \cos x}{r^2-2r\cos x+1}\,dx \\
&=-\frac{1}{r}\int_0^{\pi} \frac{\cos(nx)(r^2-2r\cos x+1-r^2-1)}{r^2-2r\cos x+1}\,dx \\
&=-\frac{1}{r}\int_0^{\pi}\cos(nx)\,dx+\frac{r^2+1}{r}\int_0^{\pi} \frac{\cos nx}{r^2-2r\cos x+1}\,dx\\
&=\left(r+\frac{1}{r}\right)I_n(r) \\
\end{aligned}$$