In a physics textbook I'm working through I found an interesting integral identity which I want to prove:
\begin{equation}
\int_0^\infty t^{\nu +1} J_\nu(\beta t) e^{-\alpha t} \, dt = \frac{2\alpha (2\beta)^\nu \Gamma(\nu + \tfrac{3}{2})}{\sqrt{\pi} (\alpha^2 + \beta^2)^{\nu + 3/2}}
\end{equation}
where $J_\nu$ is the Bessel function and $\Re(\nu)>-1, \Re(\alpha) > |\Im(\beta)|$. In my case for the application of this identity these conditions are always fulfilled since it will be $\alpha > 0, \beta \geq 0, \nu \geq 1/2$.
One could write this integral as the Hankel transform of the function $f(t) = t^\nu e^{-\alpha t}$ such that
\begin{equation}
\mathcal{H}_\nu [f(t)](\beta) = \int_0^\infty f(t) J_\nu (\beta t) t \, dt \,.
\end{equation}
However this is more of an interesting fact about this integral and not helpful for the proof I assume. The result can perhaps be found in a table of Hankel transforms but I doubt it will be given alongside a proof.
My idea and calculations so far are the following. Using the series definition of the Bessel function and then switching summation and integration we have
\begin{align}
\int_0^\infty t^{\nu +1} J_\nu(\beta t) e^{-\alpha t} \, dt &= \sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(\nu+m+1)} \int_0^\infty t^{\nu+1} \left( \frac{\beta t}{2} \right)^{2m+\nu} e^{-\alpha t} \, dt \\
&= \sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(\nu+m+1)} \left( \frac{\beta}{2} \right)^{2m+\nu} \int_0^\infty t^{2(m+\nu)+1} e^{-\alpha t} \, dt \,.
\end{align}
Using the integral identity
\begin{equation}
\int_0^\infty x^n e^{-ax} \, dx = \frac{\Gamma(n+1)}{a^{n+1}}
\end{equation}
we can write this as
\begin{equation}
\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(\nu+m+1)} \left( \frac{\beta}{2} \right)^{2m+\nu} \frac{\Gamma(2(m+\nu+1))}{\alpha^{2(m+\nu+1)}} \,.
\end{equation}
Now my idea was to use the duplication formula for the Gamma function
\begin{equation}
\Gamma(z)\Gamma(z+\tfrac{1}{2}) = 2^{1-2z} \sqrt{\pi} \, \Gamma(2z)
\end{equation}
which leads to
\begin{gather}
\sum_{m=0}^\infty \frac{(-1)^m}{m! \Gamma(\nu+m+1)} \left( \frac{\beta}{2} \right)^{2m+\nu} \frac{\Gamma(\nu+m+1) \Gamma(\nu+m+\tfrac{3}{2})}{2^{1-2(m+\nu+1)} \sqrt{\pi} \, \alpha^{2(m+\nu+1)}} \\
= \frac{2^{\nu+1} \beta^\nu}{\sqrt{\pi} \alpha^{2(\nu+1)}} \sum_{m=0}^\infty \frac{(-1)^m}{m!} \left( \frac{\beta^2}{\alpha^2} \right)^m \Gamma(\nu+m+\tfrac{3}{2}) \,.
\end{gather}
I hope that I made no mistake while simplifying the expression. Now I'm stuck at the evaluation of the sum. So far I see not much resemblance to the desired result.
Using this approach I was able to show that
\begin{equation}
\int_0^\infty t^{\nu +1} J_\nu(\beta t) e^{-\alpha t^2} \, dt = \frac{\beta^\nu}{(2\alpha)^{\nu+1}} \exp{(\tfrac{-\beta^2}{4\alpha})}
\end{equation}
quite easily, so I think this way would work here too. I would be thankful for any input on how to proceed.
Best Answer
You've done most of the work! Using the fact that $$\Gamma(z+1)=z\Gamma(z)$$ we get: $$\Gamma\left(\nu+m+\frac 3 2\right)=\left(\nu+\frac 3 2\right)\left(\nu+\frac 5 2\right)\dots\left(\nu+m+\frac 1 2\right)\Gamma\left(\nu+\frac 3 2\right)$$ Replacing in your sum: $$\begin{split} \sum_{m=0}^\infty \frac{(-1)^m}{m!} \left( \frac{\beta^2}{\alpha^2} \right)^m \Gamma(\nu+m+\tfrac{3}{2}) &= \Gamma(\nu+\tfrac{3}{2})\sum_{m=0}^\infty \frac{(-1)^m\left(\nu+\frac 3 2\right)\dots\left(\nu+m+\frac 1 2\right)}{m!} \left( \frac{\beta^2}{\alpha^2} \right)^m \\ &= \Gamma(\nu+\tfrac{3}{2})\left( 1+\frac {\beta^2}{\alpha^2}\right)^{-\nu-\frac 3 2} \end{split}$$ where we have used the well-known binomial series.
This should give you the result you wanted.