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.

  • $\begingroup$ You doesn't even need uniform convergence, you trivially have a function in $L^2(-\pi,\pi)$ and the integral of its square is readily given by Parseval's formula. $\endgroup$ – Jack D'Aurizio Dec 7 '17 at 18:36

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 $$\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx = \pi \sum_{n=1}^{\infty} \frac{1}{2^{2n}} = \frac{\pi}{3}$$

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 $b_n=\frac{1}{2^n}$.

  • $\begingroup$ Could you please elaborate on how to use Paserval's theorem. I can't seem to get the right answer. $\endgroup$ – Itay4 Aug 18 '17 at 12:35
  • $\begingroup$ I added some details to the answer. $\endgroup$ – pisco Aug 18 '17 at 12:41
  • 1
    $\begingroup$ Dang it, you covered everything :P $\endgroup$ – Simply Beautiful Art Aug 18 '17 at 14:16

Hint: First, $$\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}= Im\bigg( \sum_{n=1}^\infty\bigg(\frac{e^{ix}}{2}\bigg)^n\bigg) = Im\bigg(\frac{1}{1-\frac{e^{ix}}{2}}\bigg) = Im\bigg(2\frac{2-\cos x+i\sin x}{(2-\cos x)^2+\sin^2 x}\bigg)= \frac{2\sin(x)}{5-4\cos(x)}$$


$$\int_{-\pi}^\pi\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx = 2\int^{\pi}_0\bigg(\sum_{n=1}^\infty\frac{\sin(nx)}{2^n}\bigg)^2dx = 8\int^{\pi}_0\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx$$

From here you can use standard rule for integration of trigonometries functions. See here: What is tha simpliest way to compute :$\int^{\pi}_0\bigg(\frac{\sin(x)}{5-4\cos(x)}\bigg)^2dx$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.