The Bessel function $J_n(x)$ has the following integral representation:
$$
J_n(x)=\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(x\sin\tau-n\tau)}d\tau
$$
Here $x$ is a complex number and $n$ is an integer.
I am now wondering results of the following integral
$$
\frac{1}{2\pi}\int_{-\pi}^{\pi}e^{i(x\sin m\tau-n\tau)}d\tau
$$
with $m\in\mathbb{Z}$.
I have calculated this integral numerically by mathematica and the results are:
$$
\left(
\begin{array}{c|ccccccc}
&n=1&n=2&n=3&n=4&n=5&n=6&n=7\\
\hline
m=1&1&0&0&0&0&0&0\\
m=2&J_0(x)&J_1(x)&J_2(x)&J_3(x)&J_4(x)&J_5(x)&J_6(x)\\
m=3&J_0(x)&0&J_1(x)&0&J_2(x)&0&J_3(x)\\
m=4&J_0(x)&0&0&J_1(x)&0&0&J_2(x)\\
m=5&J_0(x)&0&0&0&J_1(x)&0&0\\
m=6&J_0(x)&0&0&0&0&J_1(x)&0\\
m=7&J_0(x)&0&0&0&0&0&J_1(x)
\end{array}
\right)
$$
The calculation detail is provided here.
It seems that, for $n>=1$, $\forall$ $m$ with $n\%m\neq0$, this integral is zero.
I want to know the proof of the above observations.
PS: It is obvious that when $n\%m=0$, this integral can be reduced to $m=1$. I am just wondering why other entries are zero.
Best Answer
We have (switching summation/integration is valid because $\sum\limits_{k\in\mathbb{Z}}|J_k(x)|$ converges) $$e^{ix\sin t}=\sum_{k\in\mathbb{Z}}J_k(x)e^{ikt}\implies\int_{-\pi}^{\pi}e^{i(x\sin m\tau)-n\tau}\,d\tau=\sum_{k\in\mathbb{Z}}J_k(x)\int_{-\pi}^{\pi}e^{i(mk-n)\tau}\,d\tau.$$ If $m\not\mid n$ then $mk-n\neq 0$ for all $k$, and thus all integrals on the RHS are zero.