Proving that $\int_0^\infty\frac{J_{2a}(2x)~J_{2b}(2x)}{x^{2n+1}}~dx~=~\frac12\cdot\frac{(a+b-n-1)!~(2n)!}{(n+a+b)!~(n+a-b)!~(n-a+b)!}$ 
How could we prove that $$\int_0^\infty\frac{J_{2a}(2x)~J_{2b}(2x)}{x^{2n+1}}~dx~=~\frac12\cdot\frac{(a+b-n-1)!~(2n)!}{(n+a+b)!~(n+a-b)!~(n-a+b)!}$$ for $a+b>n>-\dfrac12$ ?


Inspired by this question, I sought to find $($a justification for$)$ the closed form expressions of 
the following two integrals: $~\displaystyle\int_0^\infty\frac{J_A(x)}{x^N}~dx~$ and $~\displaystyle\int_0^\infty\frac{J_A(x)~J_B(x)}{x^N}~dx.~$ For the former, 
we have $~\displaystyle\int_0^\infty\frac{J_{2k+1}(2x)}{x^{2n}}~dx~=~\frac12\cdot\frac{(k-n)!}{(k+n)!}~,~$ for $k>n>\dfrac14~,~$ which I was ultimately 
able to “justify” $($sort of$)$ in a highly unorthodox manner, using a certain trigonometric integral  expression for the Bessel function, and then carelessly $($and shamelessly$)$ exchanging the order  of integration. Unfortunately, even such underhanded tricks have failed me when attempting  to approach the latter. Can anybody here help me ? Thank you !
 A: Using the product cited above, viz.
\begin{align}
J_{\mu}(x) \, J_{\nu}(x) &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, \Gamma(2n+\mu+\nu+1) \, \left(\frac{x}{2}\right)^{2n+\mu+\nu}}{n! \, \Gamma(\mu+\nu+n+1) \, \Gamma(\mu+n+1) \, \Gamma(\nu+n+1)} \\
\implies  \frac{J_{\mu}(\sqrt x) \, J_{\nu}(\sqrt x)}{x^{\mu/2+\nu/2}} &= \sum_{n=0}^{\infty} \frac{(-1)^{n} \, \Gamma(2n+\mu+\nu+1) \, x^n}{2^{2n+\mu+\nu} n! \, \Gamma(\mu+\nu+n+1) \, \Gamma(\mu+n+1) \, \Gamma(\nu+n+1)} \\
\end{align}
we can invoke Ramanujan's Master Theorem, to get
\begin{align}
\int_0^\infty \frac{J_{\mu}(\sqrt x) \, J_{\nu}(\sqrt x)}{x^{\mu/2+\nu/2}} x^{s-1}\,dx = \Gamma(s) \frac{\Gamma(-2s+\mu+\nu+1)}{2^{-2s+\mu+\nu}\, \Gamma(\mu+\nu-s+1) \, \Gamma(\mu-s+1) \, \Gamma(\nu-s+1)}
\end{align}
If $x=4 u^2 \implies dx = 8u\,du$, the LHS becomes
\begin{align}
2^{2s+1-\mu-\nu}\int_0^\infty J_{\mu}(2u) \, J_{\nu}(2u) u^{2s-\mu-\nu-1}\,du 
\end{align}
Therefore,
\begin{align}
\int_0^\infty J_{\mu}(2u) \, J_{\nu}(2u) u^{2s-\mu-\nu-1}\,du = \frac{\Gamma(s) \Gamma(-2s+\mu+\nu+1)}{2\, \Gamma(\mu+\nu-s+1) \, \Gamma(\mu-s+1) \, \Gamma(\nu-s+1)} \\
\int_0^\infty J_{2\mu}(2u) \, J_{2\nu}(2u) u^{2s-2\mu-2\nu-1}\,du = \frac{\Gamma(s) \Gamma(-2s+2\mu+2\nu+1)}{2\, \Gamma(2\mu+2\nu-s+1) \, \Gamma(2\mu-s+1) \, \Gamma(2\nu-s+1)}
\end{align}
Set $s=\mu+\nu-n$.
\begin{align}
\int_0^\infty \frac{J_{2\mu}(2u) \, J_{2\nu}(2u)}{u^{2n+1}}\,du &= 
\frac{\Gamma(\mu+\nu-n) \Gamma(2n+1)}{2\, \Gamma(\mu+\nu+n+1) \, \Gamma(-\mu+\nu+n+1) \, \Gamma(\mu-\nu+n+1)} \\
&= \frac{(\mu+\nu-n-1)!(2n)!}{2(\mu+\nu+n)!(-\mu+\nu+n)!(\mu-\nu+n)!}
\end{align}
This is your solution.
A: To save from typing a result a highlight will be listed for now. 


*

*The integral in question is a reduction of the more general integral
\begin{align}
\int_{0}^{\infty} \frac{J_{\mu}(at) \, J_{\nu}(bt)}{t^{\lambda}} \, dt = \frac{b^{\nu} \Gamma\left( \frac{\mu + \nu - \lambda +1}{2}\right)}{2^{\lambda} \, a^{\nu - \lambda +1} \, \Gamma\left( \frac{\lambda + \mu - \nu +1}{2} \right) } \, {}_{2}F_{1}\left( \frac{\mu+\nu-\lambda+1}{2}, \frac{\nu-\lambda-\mu+1}{2}; \nu+1; \frac{b^{2}}{a^{2}} \right).
\end{align}
When $\mu \rightarrow 2 \mu$, $\nu \rightarrow 2 \nu$, $a=b=2$, $\lambda = 2n+1$ the result is obtained

*A method to obtain the above listed result is to consider the integral as
\begin{align}
\int_{0}^{\infty} \frac{J_{\mu}(at) \, J_{\nu}(bt)}{t^{\lambda}} \, dt = \lim_{s \rightarrow 0} \, \int_{0}^{\infty} e^{-s t} \, t^{-\lambda} \, J_{\mu}(at) \, J_{\nu}(bt) \, dt
\end{align}

*See G. N. Watson's Bessel function Book section 13.4, p.401

Edit:
The product of two Bessel functions, as required by this problem, is
\begin{align}
J_{\mu}(x) \, J_{\nu}(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, \Gamma(2n+\mu+\nu+1) \, \left(\frac{x}{2}\right)^{2n+\mu+\nu}}{n! \, \Gamma(\mu+\nu+n+1) \, \Gamma(\mu+n+1) \, \Gamma(\nu+n+1)}
\end{align}
When $x = 2t$ it is seen that
\begin{align}
\int_{0}^{\infty} e^{-st} \, t^{- \lambda} \, J_{\mu}(2t) \, J_{\nu}(2t) \, dx = \sum_{n=0}^{\infty} \frac{(-1)^{n} \, \Gamma(2n+\mu+\nu+1) \, \Gamma(2n+\mu+\nu-\lambda+1) \, \left(\frac{1}{s}\right)^{2n+\mu+\nu-\lambda+1}}{n! \, \Gamma(\mu+\nu+n+1) \, \Gamma(\mu+n+1) \, \Gamma(\nu+n+1)}
\end{align}
Reducing this series and a possible transformation of the resulting hypergeometric series along with the limiting value for $s$ will yield the desired result. 
