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Calculate the integral: \begin{equation} \int_{-\pi}^{\pi}\left|\sum_{n=1}^{\infty}\frac{1}{2^{n}}e^{inx}\right|^{2}dx\end{equation}

I'm familiar with Parseval's identity which states that for each piecewise continuous complex function $f$ we have the equality \begin{equation} \int_{-\pi}^{\pi}\left|f(x) \right|^{2}dx=\frac{|a_{0}|^{2}}{2}+\sum_{n=1}^{\infty}\left(|a_{n}|^{2}+|b_{n}|^{2} \right) \end{equation} where $a_{n}$ and $b_{n}$ are the Fourier coefficients of $f$.

I'm naturally drawn to wanting to use this result or its complex variant:

\begin{equation} \int_{-\pi}^{\pi}\left|f(x) \right|^{2}dx=\sum_{n \in \mathbb{Z}}|c_{n}|^{2} \end{equation}

where $c_{n}$ is the complex Fourier coefficient of $f$.

Is this a trick-question where all I need to do is to take $c_{n}=\frac{1}{2^{n}}$, or is there something else in the play?

user113898
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1 Answers1

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Here is how,

$$ \int_{-\pi}^{\pi}\left|\sum_{n=1}^{\infty}\frac{1}{2^{n}}e^{inx}\right|^{2}dx = \sum_{n=1}^{\infty}\frac{1}{2^{n}}\sum_{m=1}^{\infty}\frac{1}{2^{m}} \int_{-\pi}^{\pi} e^{i(n-m)x}dx .$$

Now, see here for details and how to finish the problem. Another related technique.

Mhenni Benghorbal
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