Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Does anybody happen to know what the Fourier transform of $f(x) = \sin(x + \sin(x))$ is?

More generally, what is the Fourier transform of $g(x) = \sin(\phi_1 x + \mu \sin(\phi_2 x))$? ($\phi_1$, $\phi_2$ and $\mu$ are constants.)

I'm pretty sure I saw these written down somewhere, but Wolfram|Alpha insists that it cannot determine the result...

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Since $\sin(x+\sin(x))$ is an odd function with period $2 \pi$, its Fourier series is $$ \sin(x+\sin(x)) = \sum_{n=1}^\infty c_n \sin(n x) $$ where $$ \begin{eqnarray} c_n &=& \frac{1}{\pi} \int_{-\pi}^\pi \sin(n x) \sin(x+\sin(x)) \mathrm{d} x = \frac{2}{\pi} \int_{0}^\pi \sin(n x) \sin(x+\sin(x)) \mathrm{d} x \\ &=& \frac{1}{\pi} \int_{0}^\pi \cos((n-1)x - \sin(x) ) \mathrm{d} x - \frac{1}{\pi} \int_{0}^\pi \cos((n+1)x + \sin(x) ) \mathrm{d} x \\ &=& J_{n-1}(1) - J_{n+1}(-1) = J_{n-1}(1) + (-1)^{n} J_{n+1}(1) \end{eqnarray} $$ where the last line was evaluated using Bessel integrals.

share|improve this answer
    
I thought I remembered Bessel-J being in there somewhere... Is there an easy expression for the more complicated function? –  MathematicalOrchid May 3 '12 at 8:54

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.