How to compute $\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\cos(nx)}{2^n}\bigg)^2dx$

Question

I am trying to compute: $$\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\cos(nx)}{2^n}\bigg)^2dx$$

I tired to expand the integrand but it is too messy and lengthy.The Thing is I'm not sure how to deal with the square. If anyone has an idea on how to compute I will appreciate.

The Changing integral and summation and Compute $\int 0 ^{2\pi} (\sum{n=1}^\infty \frac {\cos(nx)}{2^n})^2 dx$ could give you some idea. — Sil, Nov 09 '17 at 22:28
That's the same as asking how to compute $\sum_{n\geq 1}\frac{\pi}{4^n}$. Do you realize why? — Jack D'Aurizio, Nov 09 '17 at 22:48
@zhw.: maybe true, but we should mention that the OP asked almost the same thing (in disguise) in his previous question and got nice answers. — Jack D'Aurizio, Nov 10 '17 at 00:37

score 1 · Answer 1 · answered Nov 09 '17 at 22:18

1

Hint: The functions $\cos (nx)$ are orthogonal in $L^2[-\pi,\pi].$

answered Nov 09 '17 at 22:18

zhw.

107,943

How to compute $\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\cos(nx)}{2^n}\bigg)^2dx$

1 Answers1