Im given that $J_n(x)$ are the fourier coefficients of $e^{ix\cdot\sin t} = \sum_{-\infty}^\infty J_n(x)e^{int}.$ I want to show that the sum $\sum_{-\infty}^\infty |J_n(x)|^2$ is not dependent on the variable $x$. I was thinking that the since the $J_{n}(x)$ are the coefficients, we can compute them: $J_n(x) = \frac 1 {2\pi} \int_{-\pi}^\pi e^{ix\sin t} e^{-int} \, dt.$ But this integral equals to $0$, so Im not sure what I should do?
Computing fourier coefficients
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fourier-analysis
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0[Parseval's Theorem](http://mathworld.wolfram.com/ParsevalsTheorem.html) deals with the sum of the squares of the fourier coefficients for a function. See equation number 5 on that page specifically. – 2017-02-11
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0Ok, but why doesnt my method work? Shouldnt this integral compute the $J_{n}(x)$ ? – 2017-02-11
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0And also if I compute the integral I will end up with the sum of the series, but will this answer my question whether the sum is dependent on x or not? – 2017-02-11
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0Your function $f_x(t) = e^{ix\sin t}$. You want to compute $\int_{-\pi}^\pi |f_x(t)|^2 dt = \int_{-\pi}^\pi f_x(t)\overline{f_x(t)}dt$. Do you see any reason why this shouldn't be that bad? – 2017-02-11
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0Not really.. and I didnt follow on the step $\int_{-\pi}^{\pi} |f_{x}(t)|^{2} dt = \int_{-\pi}^{\pi} f_{x}(t)\overline{f_{x}(t)} dt $ – 2017-02-11
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0Or I can see that the right hand side will make us integrate a constant.. But I dont know why the equality holds.. – 2017-02-11
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0For a complex number $z$ (which is all that $f_x(t)$ is in disguise), we have that $|z|^2 = z\overline{z}$. This is all that was used. – 2017-02-11
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0Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53439/discussion-between-mark-and-fejz1234). – 2017-02-11