Evaluating a definite integral of a Bessel-type function

Question

I have an expression as follows:

$\int_{0}^{2\pi} \sin{(x\sin{(\theta}) - n\theta)}\mathrm{d}\theta$

For real $x$ and $\theta$ and positive integer $n$. From plugging it into Mathematica with arbitrary $n$ and $x$, I know that the expression equals $0$, but since it doesn't have an analytic form for $n \gt 1$ I have no idea how to rigorously show that it equals $0$ for any $n$ and $x$.

Answer 1

$$\int_{0}^{2\pi}\sin(x\sin(\theta)-n\theta)\,d\theta = \text{Im}\int_{0}^{2\pi} e^{ix\sin(\theta)}e^{-ni\theta}\,d\theta \tag{1}$$ but the exponential function is an entire function, and: $$ \int_{0}^{2\pi}\sin(\theta)^m e^{-ni\theta}\,d\theta = \frac{1}{(2i)^m}\int_{0}^{2\pi}\left(e^{i\theta}-e^{-i\theta}\right)^m e^{-ni\theta}\,d\theta \tag{2}$$ differs from zero only if $m$ and $n$ have the same parity. However, that implies that $$ \int_{0}^{2\pi} e^{ix\sin(\theta)}e^{-ni\theta}\,d\theta =2\pi\cdot J_n(x)\in\mathbb{R},\tag{3}$$ hence our original integral (the LHS of $(1)$) is just zero.

Answer 2

Let $t=\theta-\pi$. Then your integral transforms to $$ \int_{-\pi}^{\pi}\sin\bigl(x\sin(t+\pi)-n(t+\pi)\bigr)\,dx=\pm\int_{-\pi}^{\pi}\sin(x\sin t+nt)\,dt $$ where the $\pm$ depends on if $n$ is even or odd. In any case, $$ t\mapsto \sin(x\sin t+nt) $$ is odd, so the integral is zero.