# Inverse Fourier Transform involving a Bessel Function, Exponential, and Power

I'm interested in this integral as a function of $r$ for various spectral densities $S(s)$: $\frac{2 \pi}{r^{p/2}-1} \int_{0}^{\infty} S(s) J_{p/2-1}(2 \pi r s) s^{p/2} ds$, where $J_{p/2-1}$ is a Bessel function of order $(p/2-1)$.

For example, if $S(s)$ is a Gaussian centred at the origin, we can use the integral identity $\int_{0}^{\infty} x^{\nu+1} e^{-\alpha x^2} J_{\nu}(\beta x) dx = \frac{\beta^\nu}{(2\alpha)^{\nu+1}} \exp(-\frac{\beta^2}{4 \alpha}) \qquad \text{Re}(\alpha) > 0 \,, \quad \text{Re}(\nu) > -1$
to recover a zero mean Gaussian in $r$ centred at 0.

However, suppose $S(s)$ is a Gaussian with mean $\mu$, in which case I want to analytically solve $\frac{2 \pi}{r^{p/2}-1} \int_{0}^{\infty} \exp[-\frac{(s-\mu)^2}{2 \sigma^2}] J_{p/2-1}(2 \pi r s) s^{p/2} ds$. I am not sure how I should best do this integral. I found the possibly helpful identities $\int_{0}^{\infty} x^{\nu+1} e^{-\alpha x} J_{\nu}(\beta x) dx = \frac{2 \alpha (2 \beta)^{\nu} \Gamma(\nu+\frac{3}{2})}{\sqrt{\pi}(\alpha^2 + \beta^2)^{\nu+\frac{3}{2}}} \qquad \text{Re}(\nu) > -1, \text{Re}(\alpha) > |\text{Im}(\beta)|$

$\int_{0}^{\infty} x^{m+1} e^{-\alpha x} J_{\nu}(\beta x) dx = (-1)^{m+1} \beta^{-\nu} \frac{d^{m+1}}{d\alpha^{m+1}} [\frac{(\sqrt{\alpha^2 + \beta^2} - \alpha)^\nu}{\sqrt{\alpha^2+\beta^2}}] \,, \beta > 0, \text{Re}(\nu) > -m -2 \,.$

It doesn't seem I can simply do a change of variables $z = s-\mu$. I feel there should be a nice clean solution to this problem.

I would greatly appreciate your help.

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I am still really hoping to answer this question. I am happy to clarify, or expand on the question, so that it will be of more general interest. – Gordon Jul 26 '13 at 17:02