I would like to compute the following integral:


$$
I\_\ell(\alpha) := \int\_{-1}^1 dx \, |x| J\_0(\alpha \sqrt{1 - x^2}) P\_\ell(x)
\tag{1}
\label{1}
$$


where $\alpha \geq 0$, $J\_0$ is the zeroth-order Bessel function of the first kind, $P\_\ell(x)$ is the Legendre polynomial of order $\ell$, and $\ell$ is an arbitrary positive integer or zero.


Since the integrand is odd if $\ell$ is odd, we have that $I\_\ell(\alpha) = 0, \ell \text{ odd}$, so we just need to care about even $\ell$s.


Mathematica reports remarkably simple results for some actual values of $\ell$:


$$
I\_0(\alpha) = \frac{2 J\_1(\alpha )}{\alpha }\\
I\_2(\alpha) = \frac{6 J\_2(\alpha )-\alpha J\_1(\alpha )}{\alpha ^2}\\
I\_4(\alpha) = \frac{3 \alpha ^2 J\_1(\alpha )-60 \alpha J\_2(\alpha )+280 J\_3(\alpha )}{4 \alpha ^3}
$$


This seems to suggest that we have something along the lines of (purely heuristically, not necessarily true):


$$
I\_\ell(\alpha)
= \sum\limits\_k a\_k \alpha^{b\_k} J\_k(\alpha)
$$


where $b\_k$ seem to be integers.


Now, one idea I had in mind was to use the expansion ([DLMF 10.60](https://dlmf.nist.gov/10.60#E10), written out in a more suitable form):


$$
J\_0\left(\alpha\sqrt{1 - x^2}\right)=\sum\_{n=0}^\infty (4n+1) \frac{(2n)!}{2^{2n}(n!)^2} j\_{2n}(\alpha) P\_{2n} (x)
$$


along with the following identities (see [here](https://dlmf.nist.gov/34.3#E19) and [here](https://mathworld.wolfram.com/LegendrePolynomial.html#eqn51)):


$$
P\_k P\_\ell
= \sum\limits\_{m=|k - \ell|}^{k + \ell}
\begin{pmatrix}k & \ell & m\\ 0 & 0 & 0\end{pmatrix}^2
(2m + 1) P\_m \\
|x| = \begin{cases} -P\_1(x),\quad x \leq 0\\ P\_1(x),\quad x > 0 \end{cases}
\\
\int\_0^1 dx\; P\_m P\_n
= \begin{cases}
\frac{1}{2n + 1}, & m=n\\
0, & m \neq n,m,n \text{ both even or odd}\\
f\_{m,n}, & m \text{ even},n\text{ odd}\\
f\_{n,m} ,& m \text{ odd},n\text{ even}
\end{cases}
$$


where I'll call $g(m,n) \equiv \int\_0^1 dx\; P\_m P\_n$ for brevity, and:


$$
f\_{m,n}
\equiv
\frac{(-1)^{(m+n+1)/2}m!n!}{2^{m+n-1} (m - n) (m + n + 1)
\big[\big(\frac{1}{2}m\big)!\big]^2 \big\{\big[\frac{1}{2}(n - 1)\big]!\big\}^2 }
$$


We can rewrite:


$$
\int\_{-1}^1
d\mu\;
|\mu|
P\_{2n} P\_\ell
=
[(-1)^\ell + 1]
\int\_0^1
d\mu\;
P\_1 P\_{2n} P\_\ell
$$


and likewise:


\begin{align\*}
\int\_0^1
d\mu\;
P\_1 P\_{2n} P\_\ell
&=
\sum\limits\_{m=|2n - \ell|}^{2n + \ell}
\begin{pmatrix}
2n & \ell & m\\
0 & 0 & 0
\end{pmatrix}^2
(2m + 1)
\int\_0^1
d\mu\;
P\_1 P\_m\\
&=
\sum\limits\_{m=|2n - \ell|}^{2n + \ell}
\begin{pmatrix}
2n & \ell & m\\
0 & 0 & 0
\end{pmatrix}^2
(2m + 1)
g(1, m)
\end{align\*}


so that we have:


\begin{align}
I\_\ell(\alpha)
&=
\sum\_{n=0}^\infty
\sum\limits\_{m=|2n - \ell|}^{2n + \ell}
(4n+1) \frac{(2n)!}{2^{2n}(n!)^2} j\_{2n}\left(\alpha\right)
[(-1)^\ell + 1]
\begin{pmatrix}
2n & \ell & m\\
0 & 0 & 0
\end{pmatrix}^2
(2m + 1) g(1, m)
\tag{2}
\label{2}
\end{align}


This is where I kind of got stuck, as I have no idea how to evaluate the double sum.


An alternate method would be to use the expansion from [here](https://mathworld.wolfram.com/LegendrePolynomial.html#eqn42):


$$
J\_0(\alpha \sqrt{1 - x^2})
= e^{-\alpha x} \sum\limits\_{n=0}^\infty
\frac{P\_n(x)}{n!}\alpha^n
$$


but then I end up with integrals of the form:


$$
\int\_{-1}^1 dx\; |x| P\_\ell (x) P\_n(x) e^{-\alpha x}
$$


which seems even more challenging to evaluate.
One idea for this one would be to expand $e^{-\alpha x} = \sum\_k \frac{1}{k!} (-1)^k \alpha^k x^k$, and then rewrite $x^k$ as a linear combination of Legendre polynomials, but this again yields an integral over three Legendre polynomials, so I'd probably just obtain eq. \ref{2} in a more roundabout way.


Any hints would be appreciated!



Computing the integral $\int_{-1}^1 dx \, |x| J_0(\alpha \sqrt{1 - x^2}) P_\ell(x)$
Thanks to the [comment](https://mathoverflow.net/questions/365861/computing-the-integral-int-11-dx-x-j-0-alpha-sqrt1-x2-p-ell#comment924242_365861) by Johannes, the solution can indeed be obtained by using the following identities:


\begin{equation}
P\_\ell(z)
=
\frac{1}{2^\ell}
\sum\limits\_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor}
(-1)^k
\begin{pmatrix} \ell \\ k \end{pmatrix}
\begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix}
z^{\ell - 2k}
\tag{3}
\label{3}
\end{equation}


and:


\begin{equation}
\int\_{0}^{\frac{1}{2}\pi}J\_{\mu}\left(z\sin\theta\right)(\sin\theta)^{\mu+1}(%
\cos\theta)^{2\nu+1}\mathrm{d}\theta=2^{\nu}\Gamma\left(\nu+1\right)z^{-\nu-1}%
J\_{\mu+\nu+1}\left(z\right)
\tag{4}
\label{4}
\end{equation}


Transforming the integral yields:


\begin{align}
I\_\ell(\alpha)
&=
\int\_{-1}^1 dx\, |x|\, J\_0(\alpha\sqrt{1 - x^2}) P\_\ell(x)\\
&=
[(-1)^\ell + 1]
\int\_0^1 dx\, x\, J\_0(\alpha \sqrt{1 - x^2}) P\_\ell(x)\\
&=
|\mathrm{substitution}\;x = \cos \phi|\\
&=
[(-1)^\ell + 1]
\int\_0^\frac{\pi}{2} d\phi\, \sin \phi\, \cos \phi\, P\_\ell(\cos \phi)\, J\_0 (\alpha \sin \phi)\\
&=
|\mathrm{expansion\;of}\;P\_\ell|\\
&=
[(-1)^\ell + 1]
\sum\limits\_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor}
(-1)^k
\begin{pmatrix} \ell \\ k \end{pmatrix}
\begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix}
\int\_0^\frac{\pi}{2} d\phi\, \sin \phi\, \cos \phi\, (\cos \phi)^{\ell - 2k} J\_0 (\alpha \sin \phi)\\
&=
[(-1)^\ell + 1]
\sum\limits\_{k=0}^{\left\lfloor \frac{\ell}{2}\right\rfloor}
(-1)^k
\begin{pmatrix} \ell \\ k \end{pmatrix}
\begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix}
\int\_0^\frac{\pi}{2} d\phi\, \sin \phi\, (\cos \phi)^{\ell - 2k + 1} J\_0 (\alpha \sin \phi)
\end{align}


The integral in the above sum has the same form as the Bessel identity \ref{4}, with $\mu = 0$ and $\nu = \ell / 2 - k$, so that the final result is:


\begin{equation}
\boxed{
I\_\ell(\alpha)
=
\frac{
[(-1)^\ell + 1]
}
{
2^\frac{\ell}{2}
}
\sum\limits\_{k = 0}^{\left \lfloor \frac{\ell}{2} \right \rfloor}
\frac{(-1)^k}{2^k}
\begin{pmatrix} \ell \\ k \end{pmatrix}
\begin{pmatrix} 2\ell - 2k \\ \ell \end{pmatrix}
\Gamma\left[\frac{\ell}{2} - k + 1\right]
\frac{
J\_{\frac{\ell}{2} - k + 1} (\alpha)
}
{
\alpha^{\frac{\ell}{2} - k + 1}
}
}
\tag{5}
\label{5}
\end{equation}


