Using Parseval's Identity to Calculate an Integral

Question

The exercise is to use Parseval's identity to solve the following integral: \begin{equation} \int_{-\pi}^{\pi}|\sum_{n=1}^{\infty}\frac{1}{2^n}e^{inx}|^2 dx \end{equation}

Now, I know that the Parseval's identity tells us that: \begin{equation} \frac{1}{\pi} \int_{-\pi}^{\pi}|f(x)|^2 dx = \frac{|a_0|^2}{2}+ \sum_{n=1}^{\infty}(|a_n|^2 + |b_n|^2) \end{equation}

I have seen a lot of examples here evaluating integrals but my problem is that the integral and the sum is on the same side in my equation. How do you solve this? Thanks!

Answer 1

let $$ f(x) = \sum_{n = 1}^{+\infty} \frac{1}{2^n}e^{inx} = \sum_{n = 1}^{+\infty} \frac{1}{2^n}(\cos(nx) + i\sin(nx)). $$ Hence $$ a_0 = 0,\quad a_n = \frac{1}{2^n},\quad b_n = \frac{i}{2^n}\quad (n = 1,2,\cdots). $$ From the Parseval's identity, we got $$ \int_{-\pi}^{\pi}\left|f(x)\right| = \pi \sum_{n = 1}^{+\infty} \frac{2}{2^{2n}} = \frac{2\pi}{3}. $$

Answer 2

Let me give you a hint:

$$\sum_{n=1}^{\infty}\frac{1}{2^n}e^{inx}$$ $$=\sum_{n=1}^{\infty}\left(\frac{e^{ix}}{2}\right)^n$$ $$=\dfrac{\dfrac{e^{ix}}{2}}{1-\dfrac{e^{ix}}{2}}$$

which means $$|\sum_{n=1}^{\infty}\frac{1}{2^n}e^{inx}|^2 = \dfrac{1}{2-2\cos x}=\dfrac{1}{4\sin^2(\dfrac{x}{2})}=\left(\frac{\csc(\dfrac{x}{2})}{2}\right)^2$$

So your $f(x)$ is $\dfrac{1}{2}\csc(\dfrac{x}{2})$.

The Fourier expansion can now be done and you can use Parseval's identity. However, you can also directly compute the integral using substitution, but that might be a bit complicated.