calculus - Calculate $int_{-pi}^pibigg(sum_{n=1}^inftyfrac{sin(nx)}{2^n}bigg)^2dx$

Calculate:

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

One can prove $\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}$ converges uniformly by Dirichlet's test, integrate term-by-term, and since $\int_{-\pi}^\pi\frac{\sin(nx)}{2^n}dx=0,$ we get series of $0$'s and the final result would be $0.$

Thing is I'm not sure how to deal with the square.

Any help appreciated.

Answer

Note that

$$\int_{-\pi}^{\pi} \sin(nx)\sin(mx) dx = \begin{cases} 0 \quad \text{ if } n\neq m\\ \pi \quad \text{ if } n=m\end{cases}$$
Hence $$\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx = \pi \sum_{n=1}^{\infty} \frac{1}{2^{2n}} = \frac{\pi}{3}$$

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Alternatively, we can use the Parseval's theorem on the $C^\infty$ function $f(x) = \sum_{n=1}^{\infty} \frac{\sin(nx)}{2^n}$.

If the Fourier series of $g(x)$ is

$$g(x) \sim \frac{a_0}{2} + \sum_{n=1}^{\infty} [a_n \cos(nx) + b_n \sin(nx)]$$
Then $$\frac{1}{\pi}\int_{-\pi}^{\pi} g^2(x) dx = \frac{a_0^2}{2}+\sum_{n=1}^{\infty} (a_n^2+b_n^2)$$

Here $a_n=0$ and $b_n=\frac{1}{2^n}$.