I need to show that $$\int_{-\pi}^{\pi}\left|\frac{a_0}{2}+\sum_{n=1}^{\infty}\alpha_n\cos(nx-\theta_n)\right|^2dx=2\pi\left(\frac{a_0^2}{4}+\frac12\sum_{n=1}^{\infty}\alpha_n^2\right)\tag{1}$$

Just for reference the trigonometric Fourier series is $$f(x)= \frac{a_0}{2} + \sum_{n=1}^{\infty}\left(a_n\cos(nx)+b_n\sin(nx)\right)$$ and the connection between the trigonometric Fourier series and the power spectrum is given by $$a_n=\alpha_n\cos\theta_n$$ $$b_n=\alpha_n\sin\theta_n$$ $$\alpha_n^2=a_n^2+b_n^2$$ $$\tan\theta_n=\frac{b_n}{a_n}$$

So I start by expanding the LHS of $(\mathrm{1})$

$$\int_{-\pi}^{\pi}\left\{\frac{a_0^2}{4}+a_0\sum_{n=1}^{\infty}\alpha_n\cos(nx-\theta_n)+\left[\sum_{n=1}^{\infty}\alpha_n\cos(nx-\theta_n)\right]^2\right\}dx$$ $$= \int_{-\pi}^{\pi}\frac{a_0^2}{4}dx+a_0\int_{-\pi}^{\pi}\sum_{n=1}^{\infty}\alpha_n\cos(nx-\theta_n)dx+\int_{-\pi}^{\pi}\left[\sum_{n=1}^{\infty}\alpha_n\cos(nx-\theta_n)\right]^2dx$$ $$= 2\pi\frac{a_0^2}{4}+a_0\int_{-\pi}^{\pi}\sum_{n=1}^{\infty}\alpha_n\cos(nx-\theta_n)dx$$ $$+\int_{-\pi}^{\pi}\sum_{n=1}^{\infty}\alpha_n\cos(nx-\theta_n)\sum_{m=1}^{\infty}\alpha_m\cos(mx-\theta_m)dx\tag{2}$$

I don't know how to proceed any further with this but do I know that for integer $n\ne m$ $$\langle\cos(nx)|\cos(mx)\rangle=0$$ but I am struggling to apply the same logic to $(\mathrm{2})$ as the cosines have different phase offsets, I am also confused about how to deal with the 2 sums in the second integral.

The answer just states that:


Does anyone have any advice on how I can complete this proof?

  • 1
    $\begingroup$ You have to argue why the infinite sum can be exchanged with the integral, and then, as $\cos$ is periodic, we get the same integrals for each $n$ if we drop $\theta_n$.. $\endgroup$ – Berci Nov 26 '18 at 17:40
  • $\begingroup$ @Berci Thanks for your reply, I'm not sure why the infinite sum can be exchanged with the integral. In fact, I don't even understand what you mean by 'exchange'. Could you please elaborate on this in an answer? $\endgroup$ – BLAZE Nov 26 '18 at 19:37


Therefore $$\int_{-\pi}^\pi \cos(nx-\theta_n)\cos(mx-\theta_m)dx={\cos(\theta_n)\cos(\theta_m)\int_{-\pi}^\pi \left(\cos(nx)\cos(mx)\right)dx \quad\text{etc.}}$$

For $n\ne m$, $$\int_{-\pi}^\pi \cos(nx)\cos(mx)dx=0$$ and in general $$\int_{-\pi}^\pi \cos(nx)\sin(mx)dx=0$$ Meanwhile $$\int_{-\pi}^\pi \cos^2(nx)dx=\pi$$ This will allow you to complete $(2)$.

  • $\begingroup$ I don't know who edited my answer. Moreover I don't see the point of it. $\endgroup$ – herb steinberg Nov 26 '18 at 20:48
  • $\begingroup$ It was me that edited the format of your answer. Do you not agree that it is easier to read now? If not you can revert back to the original. $\endgroup$ – BLAZE Nov 27 '18 at 15:01
  • $\begingroup$ It is just more spread out. I have found from past experience that reverting to the original is too troublesome, so leave it. $\endgroup$ – herb steinberg Nov 27 '18 at 16:54

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