Let $x>0$, $y>0$, $\alpha>0$ and $m$ be an integer bigger or equal to one.
Consider a following integral:
\begin{equation}
{\mathcal J}_{\alpha,m}(x,y):= \int\limits_0^\pi J_\alpha( x \cdot \sin(\theta) )\cdot e^{\imath y \cdot \cos(\theta)} \cdot \left[ \sin(\theta) \right]^m d\theta
\end{equation}
where $J_\alpha()$ is the Bessel function of the first kind. Now by using the power series expansion of the Bessel function in the integrand then by integrating the the series term by term and then by resuming the resulting series we have obtained the following result:
\begin{equation}
{\mathcal J}_{0,1}(x,y):= 2 \cdot j_0 \left( \sqrt{x^2+y^2}\right)
\end{equation}
In here $j_0()$ is the spherical Bessel function. The code below verifies the result numerically. We have:
{x, y} = RandomReal[{0, 2}, 2, WorkingPrecision -> 50]; M = 50; t =.;
NIntegrate[
BesselJ[0, x Sin[th]] Exp[I y Cos[th]] Abs[Sin[th]], {th, 0, Pi},
WorkingPrecision -> 15]
1/2 Take[Accumulate[
2^(4/2) Table[ ((-(x^2/(2 y)))^m)/m! Sqrt[\[Pi]/2] y^(-(1/2))
BesselJ[1/2 + m, y], {m, 0, M}]], -5] // MatrixForm
(*Here we identify the terms in the sum as spherical Bessel functions \
and we used the generating function identity from \
https://en.wikipedia.org/wiki/Bessel_function .*)
2 D[1/y Cos[Sqrt[y^2 - 2 y (t - x^2/(2 y))]], t] /. t :> 0
(2 Sin[Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]
Now my question would be what is the result like for other values of $n,m$.
Best Answer
Not a general answer, but the derivation of explicit representations for special cases.
We can modify the expression by changing $\theta\to\pi-\theta$ \begin{align} {\mathcal J}_{\alpha,m}(x,y)&= \int_0^\pi J_\alpha( x \sin(\theta) ) e^{\imath y\cos(\theta)} \sin^m(\theta) \,d\theta\\ &= \left( \int_0^{\pi/2}+\int_{\pi/2}^\pi\right) J_\alpha( x \sin(\theta) ) e^{\imath y\cos(\theta)} \sin^m(\theta) \,d\theta\\ &= \int_0^{\pi/2} J_\alpha( x \sin(\theta) ) e^{\imath y\cos(\theta)} \sin^m(\theta) \,d\theta+\int_0^{\pi/2}J_\alpha( x \sin(\theta) ) e^{-\imath y\cos(\theta)} \sin^m(\theta) \,d\theta\\ &=2\int_0^{\pi/2} J_\alpha( x \sin(\theta) )\cos( y\cos(\theta)) \sin^m(\theta) \,d\theta \end{align} Using the Bessel representation \begin{equation} J_{-1/2}\left( y\cos\theta \right)=\sqrt{\frac{2}{\pi}}\frac{\cos( y\cos(\theta))}{\sqrt{y\cos(\theta)}} \end{equation} we can express \begin{equation} {\mathcal J}_{\alpha,m}(x,y)=\sqrt{2\pi y}\int_0^{\pi/2} J_\alpha( x \sin(\theta) )J_{-1/2}\left( y\cos\theta \right) \sin^m(\theta) \cos^{1/2}\theta \,d\theta \end{equation} A similar integral is tabulated (G&R 6.683.2): \begin{equation} \int_0^{\pi/2} J_\nu( z_1 \sin\theta )J_{\mu}\left( z_2\cos\theta \right) \sin^{\nu+1}(\theta) \cos^{\mu+1}\theta \,d\theta=\frac{z_1^\nu z_2^\mu J_{\nu+\mu+1}\left( \sqrt{z_1^2+z_2^2} \right)}{\sqrt{\left( z_1^2+z_2^2 \right)^{\nu+\mu+1}}} \end{equation} when $\Re\nu>-1,\Re\mu>-1$. By choosing $\nu=\alpha,\mu=-1/2,z_1=x,z_2=y$, if $m=\nu+1$, we obtain \begin{equation} {\mathcal J}_{\alpha,\alpha+1}(x,y)=\sqrt{2\pi}\frac{x^\alpha J_{\alpha+1/2}\left( \sqrt{x^2+y^2} \right)}{\left( x^2+y^2 \right)^{\alpha/2+1/4}} \end{equation} When $\alpha=0$, we find ${\mathcal J}_{0,1}(x,y)=2j_0\left( \sqrt{x^2+y^2} \right)$ as expected.
Other results may be obtained from the recurrences relations for the Bessel function. For example, using \begin{equation} J_{\alpha}(z)=\frac{2(\alpha+1)}{z}J_{\alpha+1}(z)-J_{\alpha+2}(z) \end{equation} by taking $z=x\sin\theta$, it comes \begin{equation} {\mathcal J}_{\alpha,\alpha+3}(x,y)=\frac{2(\alpha+1)}{x}{\mathcal J}_{\alpha+1,\alpha+2}(x,y)-{\mathcal J}_{\alpha+2,\alpha+3}(x,y) \end{equation} Both terms of the rhs have an explicit representation from the above expression.