$J_n(p)=\int^{2\pi}_0 \cos(n\theta-p\sin\theta)\, d\theta$ with $n \in \mathbb{N} $ and $p \in \mathbb{R} $
Problem: Prove $J_n(p)= \displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k}{(k!)(n+k)!}(\frac{p}{2})^{n+2k} $
So the hint is to use contour integration on the function $f(z)=z^{n-1}e^{\frac{pz}{2}}e^{\frac{p}{2z}}$.
I worked out to get rid of the nasty exponent of $\frac{1}{z}$ you should integrate over the unit circle- then $\frac{1}{z}=\bar{z}$ when you multiply the two exponentials it works out nicely and you get the real part of $z$ disappears and get $\int f(z) \,dz$ over the unit circle is equal to
(I dont know how to do sign for line integrals over a path- I'd appreciate if someone could please say in the comments)
$i\int^{2\pi}_0 \cos(n\theta-p\sin\theta)\, d\theta + \int^{2\pi}_0 \sin(n\theta-p\sin\theta)\, d\theta$
So we're interested in the imaginary part. I would like to now calculate the actual value of the integral using Cauchy's Residue Theorem. But the only singularity there seems to be of $f(z)$ is when $z=0$ with the problem being caused at $e^{\frac{p}{2z}}$. But I couldn't tell what kind of singularity this was so I wikipediaed it and found it's an essential singularity. What do I do? How do I solve this problem? Is the residue theorem even the way to go about it anymore?
EDIT: AHH! I think I'm getting there- find the series expansion of $f(z)$ by multiplying out taylor expansion of $e^{\frac{pz}{2}}$ and the laurent expansion of $e^{\frac{p}{2z}}$ and the other term of course and get the coefficient of $\frac{1}{z}$ In both brackets there are infinitely many pairs (one from each bracket) which multiply to give a $\frac{1}{z}$ term- sum up all of these to get the coefficient of $\frac{1}{z}$ in the expansion of $f(z)$. Am I on the right track? I will try it now.
Yes I am correct. The series expansion of $f(z)$ is $z^{n-1}(1+\frac{\left(\frac{-pz}{2}\right)}{1!}+\frac{\left(\frac{-pz}{2}\right)^2}{2!}...)(1+\frac{\left(\frac{p}{2z}\right)}{1!}+\frac{\left(\frac{p}{2z}\right)^2}{2!}...)$ We want $c_{n-1}$ the coefficient of $\frac{1}{z}$ for the residue. If we pick a $k^{th}$ power in the first bracket we want the $(n+k)^{th}$ power from the second bracket because $n-1+(k)+(-(n+k))= -1$
We have$-(n+k)$ because of the negative powers in the second bracket. So multiplying the two respective terms we get $\frac{\left(\frac{-p}{2}\right)^k}{k!}.\frac{\left(\frac{p}{2}\right)^{n+k}}{n+k!}=\frac{(-1)^k}{(k!)(n+k)!}(\frac{p}{2})^{n+2k} $. We sum over k to get $\displaystyle\sum_{k=0}^{\infty} \frac{(-1)^k}{(k!)(n+k)!}(\frac{p}{2})^{n+2k}$
Problem: I havent taken into account the factor of $2i\pi$ in the Residue Theorem- and I've completely forgot about equating the real and imaginary parts of the integral. Can someone please finish this off?
Ah but of course!- $p$ was real and the factor of $2i\pi$ makes the sum now from pure real to pure imaginary which makes the extra real integral zero and we can indeed equate to the imaginary integral which is $J_n(p)=\int^{2\pi}_0 cos(n\theta-psin\theta)\ d\theta$ and so the proof is complete :)