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I'm trying to calculate the two dimensional Fourier integral

$$\iint \mathrm d\vec{R} \; (x^2 + y^2) \; e^{-2 \sqrt{ x^2 + y^2 + z^2}} \; e^{i\vec{K}\vec{R}} \;,$$

with $\vec{R}=(x,y)$. Switching to polar coordinates I arrive at

$$\int_0^{\infty} \mathrm d{R} \int_0^{2\pi} \mathrm d\Phi \; R^3 \; e^{-2 \sqrt{ R^2 + z^2}} \; e^{i K R \cos(\Phi)} \;.$$

The angle integration yields a Bessel function and the remaining integral reads

$$2 \pi \int_0^{\infty} \mathrm d{R} \; R^3 \; e^{-2 \sqrt{ R^2 + z^2}} \; J_0(K R) \;.$$

Using the transfromations $x^2 = R^2 + z^2$ and $y = \frac{x}{|z|}$ this can also be written as

$$2 \pi |z|^4 \int_{1}^{\infty} \mathrm d{y} \; y \; (y^2 - 1) \; e^{-2 |z| y} \; J_0\left(K |z| \sqrt{y^2 - 1}\right) \;.$$

And this is where I'm stuck. I have no idea how to carry out the integration. I also checked various integral tables but without success. Can anyone give me pointers?


Edit 1

Following Andrew's advice I employed the expansion

$$J_0\left(K |z| \sqrt{y^2 - 1}\right) = \sum_{m=0}^\infty \frac{(-1)^m}{m! \; \Gamma(m+1)} \left( \frac{K |z|}{2} \right)^{2m} (y^2 - 1)^m \;.$$

The integral then becomes

$$2 \pi |z|^4 \sum_{m=0}^\infty \frac{(-1)^m}{m! \; \Gamma(m+1)} \left( \frac{K |z|}{2} \right)^{2m} \int_{1}^{\infty} \mathrm d{y} \; y \; (y^2 - 1)^{m+1} \; e^{-2 |z| y} \;.$$

I found the integral tabulated in Gradshteyn/Ryzhik (3.389 4.) and finally arrived at

$$\frac{2 \pi}{\sqrt{\pi}} \sum_{m=0}^\infty \frac{(-1)^m \; (m+1)}{m!} \left( \frac{K}{2} \right)^{2m} |z|^{m+\frac{5}{2}} K_{m+\frac{5}{2}}\left( 2|z| \right) \;,$$

where $K_{m+\frac{5}{2}}$ is the irregular modified cylindrical Bessel function of order $m+\frac{5}{2}$. Unfortunately it doesn't seem like this expression can be simplified any further. Moreover, I'd expect the convergence to be rather slow, since for $m\rightarrow\infty$ with $|z| > 1$ the only limiting term is the factorial $m!$ in the denominator.

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It is possible to expand $J_0$ and integrating termwise to present the result as series. –  Andrew Jul 28 '11 at 15:41
    
@Andrew Thank you, I will try that. –  hennes Jul 28 '11 at 16:15
    
@Andrew I updated the above question with the result of the term by term integration you suggested. Unfortunately this probably can't by simplified any further. Anyway, thanks again for your very valuable input. –  hennes Jul 29 '11 at 8:56

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