Where are the Bessel functions in $\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha[\sin(\theta)-\sin(\varphi)]}d\theta d\varphi$ ?
Due to a misread in
this question
- square brackets instead of absolute value (modulus) signs - I've produced an answer that was subsequently deleted.
There is a question remaining, though, due to a comment by
Chris where he says:
It is already known that the integral can be done when there is no modulus sign (one obtains Bessel functions).
That is exactly my question: I have not found any of the kind in the deleted answer below.
Building on top of an
answer
by Smriti Sivakumar [featuring square brackets instead of an absolute value in the exponent]:
$$
\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha[\sin(\theta)-\sin(\varphi)]}d\theta d\varphi = \\
\int_{0}^{2\pi}e^{-i\alpha \sin(\varphi)}d\varphi\int_{0}^{2\pi}e^{i\alpha \sin(\theta)}d\theta = \\
\int_{0}^{2\pi}\Bigl(\cos({\alpha \sin\varphi})-i\sin({\alpha \sin\varphi})\Bigr)d\varphi
\times \int_{0}^{2\pi}\Bigl(\cos({\alpha \sin\theta})+i\sin({\alpha \sin\theta})\Bigr)d\theta \\
= \left[\int_0^{2\pi}\cos(\alpha\sin t)dt \right]^2+\left[\int_0^{2\pi}\sin(\alpha\sin t)dt\right]^2
= A^2 + B^2
$$
This leaves us with two real valued integrals:
$$
A = \int_0^{2\pi}\cos(\alpha\sin t)dt \\
B = \int_0^{2\pi}\sin(\alpha\sin t)dt
$$
Developing into Taylor series:
$$
\cos(\alpha\sin t) = 1 - \frac{1}{2}(\alpha\sin t)^2 + \frac{1}{4!}(\alpha\sin t)^4 - \cdots + \frac{(-1)^k}{(2k)!}(\alpha\sin t)^{2k}
\\ \Longrightarrow \quad A = \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}\alpha^{2k}\int_0^{2\pi}\sin(t)^{2k}dt =
\sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}\alpha^{2k} I_{2k}\\
\sin(\alpha\sin t) = \alpha\sin t - \frac{1}{3!}(\alpha\sin t)^3 + \frac{1}{5!}(\alpha\sin t)^5 - \cdots + \frac{(-1)^k}{(2k+1)!}(\alpha \sin t)^{2k+1}
\\ \Longrightarrow \quad B = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\alpha^{2k+1}\int_0^{2\pi}\sin(t)^{2k+1}dt =
\sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!}\alpha^{2k+1} I_{2k+1}
$$
So we have to evaluate integrals of the form:
$$
I_n = \int_0^{2\pi}\sin(t)^n dt = - \int_0^{2\pi}\sin(t)^{n-1} d\cos(t) = \\
- \left[\sin(t)^{n-1}\cos(t)\right]_0^{2\pi}+\int_0^{2\pi}\cos(t)d\sin(t)^{n-1} = \\
\int_0^{2\pi}(n-1)\sin(t)^{n-2}\cos(t)^2dt = (n-1)\left[\int_0^{2\pi}\sin(t)^{n-2}dt - \int_0^{2\pi} \sin(t)^n dt \right]
\\ \Longrightarrow \quad I_n = (n-1)I_{n-2} - (n-1)I_n \quad \Longrightarrow \quad I_n = \frac{n-1}{n}I_{n-2}
$$
Two cases have to be distinguished: $n$ is odd or $n$ is even.
For $n$ is odd, the reduction ends in a zero integral - perhaps we could have seen that immediately:
$$
I_1 = \int_0^{2\pi}\sin(t) dt = 0 \quad \Longrightarrow \quad I_{2k+1}=\int_0^{2\pi}\sin(t)^{2k+1}dt=0 \quad \Longrightarrow \quad B=0
$$
For $n$ is even, the reduction ends as follows:
$$
I_0 = \int_0^{2\pi} dt = 2\pi \quad \Longrightarrow \quad I_{2k} = \frac{2k-1}{2k}\frac{2k-3}{2k-2} \cdots \frac{3}{4}\frac{1}{2}2\pi
$$
where:
$$
\frac{2k-1}{2k} = \frac{2k(2k-1)}{2^2k^2} \quad ; \quad \frac{2k-3}{2k-2} = \frac{(2k-2)(2k-3)}{2^2(k-1)^2} \quad \cdots
$$
It follows that
$$
I_{2k} = \int_0^{2\pi}\sin(t)^{2k}dt = \frac{(2k)!}{2^{2k}(k!)^2}2\pi
$$
And so:
$$
\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha[\sin(\theta)-\sin(\varphi)]}d\theta d\varphi = A^2 \quad \mbox{with}\\
A = \sum_{k=0}^\infty \frac{(-1)^k}{(2k)!}\alpha^{2k}\frac{(2k)!}{2^{2k}(k!)^2}2\pi =
2\pi \sum_{k=0}^\infty \frac{(-1)^k}{2^{2k}(k!)^2}\alpha^{2k}
$$
The comment by J.G. contains the information that enables me to formulate the answer myself.
From
Wikipedia we quote:
$$
{\displaystyle J_{\alpha }(x)=\sum _{m=0}^{\infty }{\frac {(-1)^{m}}{m!\Gamma (m+\alpha +1)}}{\left({\frac {x}{2}}\right)}^{2m+\alpha },}
$$
In our case $\alpha \rightarrow 0$ , $x \rightarrow \alpha$ and $\Gamma(m+1)=m!$ , consequently:
$$
\sum_{k=0}^\infty \frac{(-1)^k}{2^{2k}(k!)^2}\alpha^{2k} =
\sum _{m=0}^\infty \frac{(-1)^m}{m!\Gamma(m+1)}\left(\frac{\alpha}{2}\right)^{2m} = J_0(\alpha)
$$
And so, finally:
$$
\int_{0}^{2\pi}\int_{0}^{2\pi}e^{i\alpha[\sin(\theta)-\sin(\varphi)]}d\theta d\varphi = 4\pi^2J_0^2(\alpha)
$$