$\int^{2\pi}_0 i\sin(x\sin \theta -n\theta) d\theta$

I've read this paper and I don't quite understand the reasoning at the end of page 3.

It seems like that $\int^{2\pi}_0 i\sin(x\sin \theta -n\theta) d\theta=0$, but why?

Solution 1:

Note that inasmuch as $\sin(x\sin(\theta)-n\theta)$ is $2\pi$-periodic, we can write

$$\int_0^{2\pi} \sin(x\sin(\theta)-n\theta)\,d\theta=\int_{-\pi}^\pi \sin(x\sin(\theta)-n\theta)\,d\theta\tag1$$

Then, exploiting the fact that $\sin(x\sin(\theta)-n\theta)$ is an odd function of $\theta$, and the integration limits of the integral on the right-hand side of $(1)$ are symmetrical around the origin, we conclude immediately that

$$\int_0^{2\pi} \sin(x\sin(\theta)-n\theta)\,d\theta=0$$

And we are done!

Solution 2:

Because $\sin(x\sin\theta-n\theta)$ is odd about the center of the interval: \begin{align} \sin(x\sin(2\pi-\theta)-n(2\pi-\theta)) &= \sin(x\sin(-\theta)+n\theta -n2\pi) \\& = \sin(-(x\sin\theta-n\theta) -n2\pi)\,, \end{align} and, since $n$ is an integer, $\sin(\phi+2 n\pi)=\sin(\phi)$, so, \begin{align} \sin(x\sin(2\pi-\theta)-n(2\pi-\theta)) = -\sin(x\sin\theta-n\theta)\,. \end{align}