I got this integral that I have been asked to calculate: $\int_{0}^{2\pi} |3+4e^{10ix}+5e^{100ix}|^{2}dx$

I tried using Parseval's identity and tried to convert it to Fourier series. I think there is an easy way to solve it that I am missing.



There is a much easier way: just multiply the integrand out.

$$\begin{align}|3+4e^{10ix}+5e^{100ix}|^{2} &= (3+4e^{10ix}+5e^{100ix})(3+4e^{-10ix}+5e^{-100ix})\\ &= 9 + 16 + 25 + \text{cosine terms} \end{align}$$

The integral over the cosine terms is zero (why?) Therefore, your answer is $50 \cdot 2 \pi = 100 \pi$.

  • $\begingroup$ $\int_{0}^{2\pi} e^{inx}dx = [-{ie^{inx}\over n}]_{0}^{2\pi} = [{\sin(nx)-icos(nx)\over n}]_{0}^{2\pi} = 0$ Because $\\sin(2\pi n)=\sin(0)=0$ and $\\cos(2\pi n)=\cos(0)=1$ $\endgroup$ – Gal Silberman Feb 15 '13 at 11:33
  • $\begingroup$ @Lag: You have to be super careful with the $n=0$ case, because there, $\sin{(2 \pi n)}/n = 2 \pi$. $\endgroup$ – Ron Gordon Feb 15 '13 at 14:21

Using Parseval's identity is a good idea.

Let $f(x) = \sum\limits_{k=-\infty}^{\infty}c_ke^{ikx}$ --- Fourier series, then the Parseval's identity is $$ \frac{1}{2\pi}\int_{0}^{2\pi}|f(x)|^2\,dx = \sum_{k=-\infty}^{\infty}|c_k|^2 $$

Your function $f(x) = 3+4e^{10ix}+5e^{100ix}$, (this means, that $c_0$=3, $c_10$=4 and so on). In this case Parseval's identity is $$ \frac{1}{2\pi}\int_{0}^{2\pi}|3+4e^{10ix}+5e^{100ix}|^2\,dx = 3^2 + 4^2 + 5^2. $$


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