Fourier Transform of $\frac{1}{2\pi} \int\limits_{0}^{2\pi}e^{ir\sin\theta}e^{2i\theta} d\theta$? The integral relationship for a function $J(r)$ is-
$$J(r)=\frac{1}{2\pi} \int\limits_{0}^{2\pi}e^{ir\sin\theta}e^{2i\theta} d\theta$$
How can I determine the Fourier transform of the above function $J(r)$.
I am trying to solve this by using the property of $\delta$-function-
$$X(\omega) = \int\limits_{\infty}^{\infty}e^{i\omega_0 t}e^{i\omega t}dt = 2\pi \delta(\omega-\omega_0)$$ Is this approach correct or is there any other elegant way to solve this problem? Thanks in advance for the help.
 A: I don't know if this is a more elegant approach, but herein we will adopt the general procedure I used in THIS ANSWER to find the Fourier Transform of $J_0(x)$.  Proceeding, we see that
$$\begin{align}
J_2(r)&=\frac1{2\pi}\int_{-\pi}^\pi e^{i(r\sin(\phi)-2\phi)}\,d\phi\\\\
&=\frac1{\pi}\int_0^\pi \cos(r\sin(\phi)-2\phi)\,d\phi\\\\
&=\frac1\pi \int_0^\pi \left(\cos(2\phi)\cos(r\sin(\phi))+\sin(r\sin(\phi))\sin(2\phi)\right)\,d\phi\\\\
&=\frac1\pi \int_0^\pi \cos(2\phi)\cos(r\sin(\phi))\,d\phi \tag1
\end{align}$$
where we exploited symmetries to arrive at $(1)$.
Now, enforcing the substitution
$$\phi=
\begin{cases}
\arcsin (k),&\text{for}\,\,0\le\phi\le\pi/2\\\\
\pi-\arcsin (k),&\text{for}\,\,\pi/2\le\phi\le\pi
\end{cases}$$
into $(1)$ yields
$$\begin{align}
J_2(r)&=\frac{2}{\pi}\int_0^1 \frac{1-2k^2}{\sqrt{1-k^2}}\cos(kr)\,dk\\\\
&=\frac1\pi \int_{-1}^1 \frac{1-2k^2}{\sqrt{1-k^2}}\cos(kr)\,dk\\\\
&=\frac1\pi \int_{-1}^1 \frac{1-2k^2}{\sqrt{1-k^2}}e^{ikr}\,dk\\\\
&=\frac1{2\pi} \int_{-1}^1 \frac{2(1-2k^2)}{\sqrt{1-k^2}}e^{ikr}\,dk\\\\
&=\frac1{2\pi} \int_{-\infty }^\infty \text{rect}\left(\frac k2\right)\frac{2(1-2k^2)}{\sqrt{1-k^2}}e^{ikr}\,dk\tag2
\end{align}$$
Recognizing $(2)$ as the inverse Fourier Transform, we see that the Fourier Transform of $J_2(r)$ is given by
$$\bbox[5px,border:2px solid #C0A000]{\mathscr{F}\{J_2\}(k)=\text{rect}\left(\frac k2\right)\frac{2(1-2k^2)}{\sqrt{1-k^2}}}$$
A: Since the integrand in your problem is $2\pi$-periodic,
\begin{align}
2\pi J(r)&= \int\limits^{3\pi/2}_{-\pi/2}e^{ir\sin\theta}e^{2i\theta} d\theta=
\int^{\pi/2}_{-\pi/2}e^{r\sin\theta} e^{2i\theta}\,d\theta +\int^{3\pi/2}_{\pi/2}e^{r\sin\theta} e^{2i\theta}\\
&=\int^{\pi/2}_{-\pi/2}e^{ir\sin\theta}(2\cos^2(\theta)-1)\,d\theta+2i\int^{\pi/2}_{-\pi/2}e^{ir\sin\theta}\sin\theta\cos\theta\,d\theta\quad-\\
&\quad\Big(\int^{\pi}_{0}e^{ir\cos\theta}(1-\sin^2(\theta))\,d\theta+2i\int^\pi_0 e^{ir\cos\theta}\cos\theta\sin\theta\,d\theta\Big)\\
&=\int^{\pi/2}_{-\pi/2}e^{ir\sin\theta}(2\cos^2(\theta)-1)\,d\theta + \int^{\pi}_{0}e^{ir\cos\theta}(\sin^2(\theta) - 1)\,d\theta\tag{1}\label{one}
\end{align}
The imaginary part vanishes using the substitution by $u=\sin\theta$ and $v=\cos\theta$ respectively.
Consider the integral $\varphi_p(t):=\int^1_{-1}(1-x^2)^pe^{-ixt}\,dx$, with $p>-3/2$ (for your problem, you only need $p\in\{-\frac12,\frac12\}$).
The trigonometric substitution $x=\sin\theta$ in the integral defining $\varphi_p$ gives
\begin{align}
    \varphi_p(r)&=\int^{\pi/2}_{-\pi/2}e^{-ir\sin\theta}\cos^{2p+1}\theta\,d\theta=
    \int^{\pi/2}_{-\pi/2}e^{ir\sin\theta}\cos^{2p+1}\theta\,d\theta=\int^\pi_0 e^{-ir\cos\theta}\sin^{2p+1}\theta\,d\theta
  \end{align}
On the other hand, expanding the exponential in $\varphi_p$ as a power series leads to
\begin{align*}
    \phi_p(t)&=\int^1_{-1}(1-x^2)^p\sum_{n\geq0}\frac{(-ixt)^n}{n!}\,dx =\sum_{n\geq0}\frac{(-it)^n}{n!}\int^1_{-1}(1-x^2)^px^n\,dx\\
    &=2\sum_{k\geq0}\frac{(-it)^{2k}}{(2k)!}\int^1_0(1-x^2)^px^{2k}\,dx\\
    &=\sum_{k\geq0}\frac{(-it)^{2k}}{(2k)!}\int^1_0(1-u)^pu^ku^{-1/2}\,du \\
    &=\sum_{k\geq0}\frac{(-1)^kt^{2k}}{(2k)!}B(p+1,k+\tfrac12)\tag{2}\label{two}
  \end{align*}
Notice that $\phi_p(-t)=\phi_p(t)$.
From \eqref{one} we get
\begin{align}
J(r)&=\frac{1}{2\pi}\big(4\phi_{1/2}(r)-2\phi_{-1/2}(r)\big)\\
&=\frac{1}{2\pi}\int^1_{-1}\Big(4(1-x^2)^{1/2} -2(1-x^2)^{-1/2}\Big)\,e^{-ixr}\,dx\\
&=\frac{1}{\pi}\int e^{-ixr} \mathbb{1}_{[-1,1]}(x)\frac{1-2x^2}{\sqrt{1-x^2}}\,dx\\
&\stackrel{x=2\pi u}{=}2\int e^{-2\pi iur}\mathbb{1}_{[-1,1]}(2\pi u)\frac{1-8\pi^2u^2}{\sqrt{1-4\pi^2 u^2}}\,du
\end{align}
Hence
$$\big(\mathcal{F}^{-1}J\big)(x)=2\mathbb{1}_{[-1,1]}(2\pi x)\frac{1-8\pi^2x^2}{\sqrt{1-4\pi^2x^2}}$$

Edit: Although this is not directly relevant to the OP, I would like to mention that there is a relation between $\phi_p$ and the  Bessel function
$$
  J_p(z)=\sum_{n\geq0}\frac{(-1)^n}{n!\Gamma(n+p+1)}\Big(\frac{z}{2}\Big)^{p+2n}
$$
Using the identities
\begin{align*}
    B(p+1,k+\tfrac12) &= \frac{\Gamma(p+1)\Gamma(k+\tfrac12)}{\Gamma(p+k+\tfrac32)} = \frac{\Gamma(p+1)}{\Gamma(p+k+\tfrac32)}\frac{(2k)!\sqrt{\pi}}{2^{2k}\,k!}
  \end{align*}
we obtain from\eqref{two}
\begin{align}
    \varphi_p(r)&= \int^1_{-1}(1-x^2)^pe^{-ixr}\,dx=\sum_{k\geq0}\frac{(-1)^k\Gamma(p+1)\sqrt{\pi}}{\Gamma(k+ p+ \tfrac32)k!}\Big(\frac{r}{2}\Big)^{2k}\\
    &= \frac{\Gamma(p+1)\sqrt{\pi}}{(r/2)^{p+\tfrac12}}J_{p+\tfrac12}(r)
  \end{align}
