Evaluating this complicated integral using complex analysis

Question

I am trying to evaluate this integral:

$$\boxed{\int_{-\pi}^\pi\sin(2\cos\theta)\cos((2m+1)\theta)\,d\theta}$$ where $m\in\mathbb{N}$, using complex analysis methods.

What I have done is to find the residue of $\sin(z+\frac 1z)$:

Since $$\sin(z+\frac 1z)=\sum_{k=0}^\infty\frac{(-1)^k(z+z^{-1})^{2k+1}}{(2k+1)!}$$, and using the Binomial theorem, we have $$(z+z^{-1})^{2k+1}=\sum_{r=0}^{2k+1}{{2k+1}\choose r}z^{2k+1-2r}$$.

Combining these two facts we have that the residue (coefficient of $z^{-1}$) is: $$\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}{{2k+1}\choose{k+1}}$$.

After this I am kind of stuck.

I am aware of the standard technique of substituting $\cos\theta=\frac{z+z^{-1}}{2}$, $d\theta=\frac{dz}{iz}$, etc, but I am not sure what to do with the $\cos((2m+1)\theta)$.

Thanks for any help.

By the way, if it helps, the answer I am supposed to get is $$\boxed{2\pi\sum_{k=m}^\infty\frac{(-1)^k}{(2k+1)!}{{2k+1}\choose {k-m}}}$$.

Update: Using David Holden's excellent hint, I proceded to find the residue of the integrand.

So the integral becomes $\frac{1}{2i}\int_C\sin(z+z^{-1})(z^{2m}+z^{-2m-2})\,dz$.

I find the residue of the integrand to be

$$\sum_{k=m}^\infty \frac{(-1)^k}{(2k+1)!}{{2k+1}\choose{k+m+1}}+\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)!}{{2k+1}\choose{k-m}}$$

I may have made a mistake along the way, I can't get to the final answer.

We have that $\cos(n\theta)=T_n\cos(\theta)$ where $T_n$ is the Chebyshev Polynomial of the First Kind. Once you get the polynomial in $\cos\theta$ (which is what $T_n\cos\theta$ is), you should be able to substitute in $\cos\theta = \frac{z+z^{-1}}{2}$. While this seems like it might work, it also seems like it'd suck a lot. — Mark Schultz-Wu, Aug 30 '16 at 03:59
One approach (which, alas, doesn't fit your desire for a complex-analysis proof) is to appeal to the Jacobi-Anger expansion. Specifically, you can use the second real-valued expression listed there to get the result as a Bessel function evaluated at $z=2$. — Semiclassical, Aug 30 '16 at 14:02
when you have those $Res(e^{a z+b/z},0)$ it becomes the (complicated) value of Bessel functions — reuns, Sep 02 '16 at 04:31

score 1 · Accepted Answer · answered Sep 02 '16 at 05:27

Consider that $\cos{\left [ (2 m+1) \theta \right ]} = \operatorname{Re}{\left [e^{i (2 m+1) \theta}\right ]} $. Then we consider the integral

$$-i \oint_{|z|=1} dz \, \sin{\left ( z+\frac1{z} \right )} z^{2 m}$$

which is equal to $i 2 \pi$ times the sum of the residues of the integrand at $z=0$. To evaluate the residues and their sum, we Taylor expand the sine term to get, as the integrand,

$$-i \sum_{k=0}^{\infty} \frac{(-1)^k}{(2 k+1)!} \left ( z+\frac1{z} \right )^{2 k+1} z^{2 m} $$

The coefficient of $z^{-1}$ in the $k$th term of the expansion is the coefficient of the $k+m+1$th term. (You can work this out for successive values of $m$.) Thus, the integral is equal to

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

To evaluate the sum, note that the terms for $k \lt m$ are zero. Then we can shift the index of the sum to get

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

The latter sum is recognizable as a Bessel function evaluated at $z=2$. Thus,

$$\int_{-\pi}^{\pi} d\theta \, \sin{(2 \cos{\theta})} \cos{\left [ (2 m+1) \theta \right ]} = 2 \pi (-1)^m J_{2 m+1}(2)$$

Thanks. But how about the other term in $cos(2m+1)\theta=\frac 12(z^{2m+1}+z^{−(2m+1)})$? — yoyostein, Sep 02 '16 at 08:49
Note that I used the real part from which the cosine is derived. No need to complicate things with two terms like that. — Ron Gordon, Sep 02 '16 at 10:19
Can you elaborately briefly how to use the real part? This is the first time I am seeing it. — yoyostein, Sep 02 '16 at 10:50
@yoyostein: you have never seen the trick $$\int dx Re[f(x)] = Re \int dx f(x)$$ ? — Ron Gordon, Sep 02 '16 at 10:51

Evaluating this complicated integral using complex analysis

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