Integral over a product of polynomial, exponential and Bessel function

Question

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.

Answer 1

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.

Answer 2

To use that approach we need the additional restriction $|\beta| < |\alpha|$ so that the series converges.

The hypergeometric representation of $(1-z)^{-a}$ is $_{1}F_{0}(a;;z)$. See here for an explanation.

Therefore, $$ \begin{align} \sum_{m=0}^\infty \frac{(-1)^m}{m!} \left( \frac{\beta^2}{\alpha^2} \right)^m \Gamma\left(\nu+m+\frac{3}{2}\right) &= \Gamma \left(\nu + \frac{3}{2} \right) \sum_{m=0}^\infty \frac{(-1)^m}{m!} \left( \frac{\beta^2}{\alpha^2} \right)^m \frac{\Gamma(\nu+m+\tfrac{3}{2})}{\Gamma\left(\nu+\tfrac{3}{2}\right)} \\ &= \Gamma \left(\nu + \frac{3}{2} \right) \, _{1}F_{0} \left(\nu + \frac{3}{2}; ; - \frac{\beta^{2}}{\alpha^{2}} \right) \\ &= \Gamma \left(\nu + \frac{3}{2} \right) \left(1+ \frac{\beta^{2}}{\alpha^{2}} \right)^{-\nu-3/2}, \end{align}$$

and $$\begin{align} \frac{2^{\nu+1} \beta^\nu}{\sqrt{\pi} \alpha^{2(\nu+1)}} \sum_{m=0}^\infty \frac{(-1)^m}{m!} \left( \frac{\beta^2}{\alpha^2} \right)^m \Gamma\left(\nu+m+\frac{3}{2}\right) &= \frac{2^{\nu+1} \beta^\nu}{\sqrt{\pi} \alpha^{2(\nu+1)}}\Gamma \left(\nu + \frac{3}{2} \right) \left(1+ \frac{\beta^{2}}{\alpha^{2}} \right)^{-\nu-3/2} \\ &= \frac{2^{\nu+1} \beta^\nu}{\sqrt{\pi} \alpha^{2(\nu+1)}} \frac{\alpha^{2\nu+3} \, \Gamma \left(\nu + \tfrac{3}{2} \right)}{\left(\alpha^{2}+\beta^{2} \right)^{\nu+3/2}}\\ &= \frac{2\alpha (2\beta)^\nu \, \Gamma(\nu + \tfrac{3}{2})}{\sqrt{\pi} (\alpha^2 + \beta^2)^{\nu + 3/2}}. \end{align} $$

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The integral $$\int_0^\infty t^{\nu} J_{\nu}(\beta t) e^{-\alpha t} \, \mathrm dt $$ has a simple closed-form expression as well, which we can find using the formula for the Laplace transform of $\frac{f(t)}{t}$.

Assume that $\Re(\nu) >- \frac{1}{2} $.

Then

$$ \begin{align} \int_0^\infty t^{\nu} J_\nu(\beta t) e^{-\alpha t} \, \mathrm dt &= \int_0^\infty \frac{t^{\nu+1} J_\nu(\beta t)}{t} e^{-\alpha t} \, \mathrm dt \\ &= \frac{(2 \beta)^{\mu} \, \Gamma \left(\nu + \tfrac{3}{2} \right)}{\sqrt{\pi}} \int_{\alpha}^{\infty} \frac{2 p}{(p^2 + \beta^2)^{\nu + 3/2}} \, \mathrm dp \\ &= \frac{(2 \beta)^{\nu} \, \Gamma \left(\nu + \tfrac{3}{2} \right)}{\sqrt{\pi}} \frac{1}{\nu + \tfrac{1}{2}} \frac{1}{(\alpha^{2}+\beta^{2})^{\nu+1/2}} \\ &= \frac{(2 \beta)^{\nu} \, \Gamma \left(\nu + \frac{1}{2} \right)}{\sqrt{\pi} (\alpha^{2}+\beta^{2})^{\nu+ 1/2}}. \end{align}$$