Integral of product of Bessel functions of the first kind I would like to solve the integral:
$$\int_0^{+\infty}\quad rJ_n(ar)J_n(br)\quad dr$$
Is there any closed form for it?
Thanks!
What happens if I have instead:
$$\int_0^{c}\quad rJ_n(ar)J_n(br)\quad dr$$
with $c>0$.
(Since the first part has already a proper answer, the bounty is offered actually for the second part!)
 A: According to Wolfram functions (at the bottom) this is simply (for any $n$ in $\mathbb{R}$) :
$$\int_0^{+\infty}r\,J_n(ar)J_n(br)\ dr=\frac {\delta(a-b)}a$$
The same formula appears in DLMF where this closure equation appears with the constraints $\ \Re(n)>-1,\;a>0,\;b>0$ and additional references (A & W $11.59$ for example).
It may be proved using the Hankel transform : see Wikipedia's 'Hankel transform'.

If we replace the upper (infinite) bound by a finite bound $c$ we get the result kindly indicated by J.M. (changing merely the sign since we want the integral over $(0,c)$ and in conformity with Alpha's result $\not =$ and result $=$ for some specific value of $n$) :
\begin{align}
\tag{1}\int_0^c r\,J_n(a r) J_n(b r) \,dr&=\frac c{a^2-b^2}\left[b \;J_n(a\,c)\,J_{n-1}(b\,c)-a\;J_{n-1}(a\,c)\,J_n(b\,c)\right]\quad\text{if}\ a\not =b\\
\tag{2}\int_0^c r\,J_n(a r)^2 \,dr&=\frac {c^2}2\left[J_n(a\,c)^2-J_{n-1}(a\,c)\,J_{n+1}(a c)\right]\\
\end{align}
The first result appears as entry $03.01.21.0064.01$ at Wolfram functions while the second is entry $03.01.21.0055.01$. They are usually called 'first and second Lommel integrals'.
For a really venerable reference you may consult Whittaker & Watson p.$381$ with this interesting exercise (using our previous notations) :
"Shew that, if $a\not =b$ and $n>-1$",
$$(a^2-b^2)\int_0^c r\,J_n(a r) J_n(b r) \,dr=c\left[J_n(a\,c)\frac d{dc}J_n(b\,c)-J_n(b\,c)\frac d{dc}J_n(a\,c)\,\right],$$
$$2a^2\int_0^c r\,J_n(a r)^2 \,dr=(a^2c^2-n^2)J_n(a\,c)^2-\left[c\frac d{dc}J_n(a\,c)\right]^2$$
For a proof see page $33$ and $34$ of Culham's lecture or Bowman's book page $101$.
The recurrence relations (Wikipedia) :
$$\frac d{dc}J_n(a\,c)=\frac a2\left[J_{n-1}(a\,c)-J_{n+1}(a\,c)\right]= a\,J_{n-1}(a\,c)-\frac ncJ_n(a\,c)$$
should allow you to obtain the $(1)$ and $(2)$ formulas.
(the exercise $19$ from W & W with $a$ and $b$ two different zeros of $J_n$ seems of interest too)


