2
$\begingroup$

I'm trying to show that

$$\int_{0}^{2\pi}\cos(n\theta-2\sin\theta)d\theta = 2\pi\sum_{r=0}^{\infty}\frac{(-1)^n}{r!(n+r)!}$$

The question hints that I should consider

$$e^{z-z^{-1}}$$

but I don't see how to get to the answer. Considering the nature of the integral, the unit circle seems like a sensible contour. The integrand has only one pole in the unit circle (at zero), which has residue $-1$. So:

$$\int e^{z-z^{-1}}dz=-2\pi i$$

Switching to $\theta$:

$$\int_0^{2\pi}ie^{i\theta}e^{e^{i\theta}-e^{-i\theta}}d\theta=-2\pi i$$

$$\int_0^{2\pi}e^{2\sin\theta+i\theta}d\theta=-2\pi$$

That integrand looks similar, but doesn't match up with the required integral, and I don't see how to get them to match. More troublesome, the required infinite sum on the right hand side is missing.

I've considered switching to a contour of radius $n$, but that just gives me $(2n+1)\pi$ on the right-hand side, and I still can't get it to match on the left. How do I solve this?

EDIT: Infinite sum corrected.

$\endgroup$
  • 1
    $\begingroup$ It seems you have a slight notational problem, as '$n$' is a dummy index on the RHS, while it is a specific number (integer?) on the LHS... $\endgroup$ – Pierpaolo Vivo Dec 23 '16 at 17:24
  • $\begingroup$ This integral is related to an integral representation of Bessel function $J_n$. See formula (10.9.2) in (dlmf.nist.gov/10.9) with $z=2$ $\endgroup$ – Jean Marie Dec 23 '16 at 17:56
  • $\begingroup$ Infinite sum corrected. @JeanMarie's hint about the Bessel function brought me to: math.stackexchange.com/questions/1051558/…, which solves a more general case of this. $\endgroup$ – Number 34 Dec 23 '16 at 23:00
  • $\begingroup$ Rewriting the whole thing in terms of complex exponentials gives me $\int_0^{2\pi}\cos(n\theta-2\sin\theta)d\theta &= \Re\int_0^{2\pi}e^{in\theta-2i\sin\theta}d\theta = \Re\int_0^{2\pi}\left(e^{i\theta}\right)^n\exp\left(-e^{i\theta}+e^{-i\theta}\right)d\theta=\Re\oint_{|z|=1}z^ne^{-\left(z-\frac{1}{z}\right)}dz$, which contains the quantity you were given as a hint and also a dependancy on $n$. Can you take it from there? $\endgroup$ – Tom Dec 24 '16 at 0:03
  • $\begingroup$ Sorry for that. Now I can't even click on the edit button, can anybody help with that? $\endgroup$ – Tom Dec 24 '16 at 0:08
3
$\begingroup$

The integral is the real part of

$$\int_0^{2 \pi} d\theta \, e^{i n \theta} \, e^{-i 2 \sin{\theta}} $$

To convert this into a complex integral over the unit circle, sub $z=e^{i \theta}$ and the integral is

$$-i \oint_{|z|=1} dz \, z^{n-1} \, e^{-\left ( z-z^{-1} \right )} $$

This integrand has an essential singlarity at $z=0$. To evaluate the integral using the residue theorem, we replace the exponential by its series representation and reverse order of sum and integral. The integral is then equal to

$$-i \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \oint_{|z|=1} dz \, z^{n-1} \, \left ( z-z^{-1} \right )^k$$

We can then expand the binomial inside the integral:

$$-i \sum_{k=0}^{\infty} \frac{(-1)^k}{k!} \oint_{|z|=1} dz \, z^{n-1} \, \sum_{j=0}^k (-1)^{k-j} \binom{k}{j} z^{2 j-k}$$

Note that the integral is going to be zero unless $k \ge n$. To help us see this a bit better, let's shift the sum to begin there:

$$-i \sum_{k=0}^{\infty} \frac{(-1)^{n+k}}{(n+k)!} \oint_{|z|=1} dz \, z^{-1} \sum_{j=0}^{n+k} (-1)^{k-j} \binom{n+k}{j} z^{2 j-k}$$

The residue is then the coefficient of $z^0$ inside the sum. The exponent of $z$ is zero when $j=k/2$, so only even values of $k$. Let's clean this up a bit by only summing over even value of $k$:

$$-i \sum_{k=0}^{\infty} \frac{(-1)^{n+2 k}}{(n+2 k)!} \oint_{|z|=1} dz \, z^{-1} \sum_{j=0}^{n+2 k} (-1)^{2 k-j} \binom{n+2 k}{j} z^{2 j-2 k}$$

Now we see that the exponent of $z$ is zero when $j=k$. Thus, we may now apply the residue theorem and the integral is equal to

$$2 \pi \sum_{k=0}^{\infty} \frac{(-1)^{n}}{(n+2 k)!} (-1)^k \binom{n+2 k}{k}$$

Simplifying, and noting the the integral is entirely real, we get that

$$\int_0^{2 \pi} d\theta \, \cos{\left ( n \theta - 2 \sin{\theta} \right )} = 2 \pi \sum_{k=0}^{\infty} \frac{(-1)^{n+k}}{k! (n+k)!}= 2 \pi (-1)^n J_n(2)$$

I believe my answer differs from that of the OP by a factor of $(-1)^k$ in the sum.

$\endgroup$
  • $\begingroup$ really a pity that such a nice answer recieves absolutly no attention (not even from the OP)..(+1) from me anyways $\endgroup$ – tired Jan 7 '17 at 13:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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