After reduction of a problem, I find myself in front of these integrals,$$ \int_0^{2\pi} cn \left( \frac{2K(1/2)}{\pi} \theta, \frac{1}{2} \right) \cos(m \theta) \mathrm{d}\theta, $$with $m$ positive integers. For now,
- I think it vanishes for $m$ even,
- I am not sure how to get the value for $m=1$,
- I am also not sure whether an expansion of $\cos(m\theta)$ then integration by parts would yield something nice (in terms of the case $m=1$ hopefully).
I would like to hear your thoughts on these ones!
$\endgroup$ 31 Answer
$\begingroup$As pointed by @achille hui, the Fourier series of Jacobi elliptic functions are known. In particular here,$$ cn \left( \frac{2K(1/2)}{\pi} \theta, \frac{1}{2} \right) = \frac{2\sqrt{2}\pi}{K(1/2)} \sum_{n=0}^\infty \frac{e^{-(n+1/2)\pi}}{1+e^{-(2n+1)\pi}} \cos ((2n+1) \theta), $$in turn,$$ \int_0^{2\pi} cn \left( \frac{2K}{\pi} \theta, \frac{1}{2} \right) \cos(m \theta) \mathrm{d}\theta = \begin{cases} \frac{2\sqrt{2}}{K(1/2)} \frac{e^{-m\pi/2}}{1+e^{-m\pi}} \pi^2 &\textrm{if $m$ is odd}\\ 0 &\textrm{otherwise} \end{cases} $$
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