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Sorry to ask yet another "how do I do this integral" question! But I've really been having a hard time with this one.

$$\int_0^\infty \frac{J_0(x)}{|C^2 - x^2|}\mathrm{d}x$$

I've been through a lot of identities giving results for expressions of the form $\int Z_\nu(x) x^p \mathrm{d}x$, where $Z_\nu$ is a Bessel function or one of its variants. But I haven't found a way to apply them to the integral I'm trying to do; I don't see any way to change variables such that the denominator becomes a simple power of $x$ without shifting the argument of $J_0$ becomes shifted.

For what it's worth, I've tried expanding $J_0$ in a power series (for which every term after the first diverged), partial fraction decomposition on the denominator, integration by parts, etc. etc. I considered contour integration, but there is the slight problem that the integrand isn't an analytic function, and if I break it into domains $x < C$ and $x > C$ I'm not really sure how to set up an appropriate contour. I've also tried various other manipulations but nothing looks promising so far, which suggests that some heavy mathematical machinery that I'm not familiar with is called for. So can anyone here suggest a method or point me to a helpful reference?

If the integral is divergent, that'll be nice to know, but I'd also like to have a way of characterizing the divergence, since it will be added to a couple other terms with a similar structure and any divergences may cancel out.

In case it's useful, the particular expression that prompted me to ask this question is

$$\int_0^\infty\mathrm{d}k\ k J_0(kr)\int_0^\infty \frac{\mathrm{d}k'^2}{k'^2}\int_0^\infty \frac{\mathrm{d}r'}{r'}\left[\frac{k'^2 J_0(k'r')}{|k^2 - k'^2|} - \Biggl(\frac{k^2}{|k^2 - k'^2|} - \frac{k^2}{\sqrt{4 k'^{4} + k^{4}}}\Biggr)J_0(kr')\right]N(r')$$

More generally, I'm trying to invert the Fourier transform of the BFKL equation in the paper, basically going from equation (9) to equation (5), and in various methods I've encountered several of these integrals involving "shifted" Bessel functions.

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I tried $C=1$ and wolfram alpha claims it is divergent. Would $C$ be a root of $J_0(x)$? – Aryabhata Jan 20 '11 at 2:26
@Moron: unfortunately no, $C$ is an arbitrary non-negative real number. (As a matter of fact this whole thing is also inside an integral $\int_0^\infty \mathrm{d}C$.) – David Z Jan 20 '11 at 6:00
Using the product representation (Weierstrass factorization) for $J_0(x)$, I believe we can show that the integral converges only when $C$ is one of the roots. See this for the representation. – Aryabhata Jan 20 '11 at 6:31
If C is not a root of $J_0$ then the integral is divergent. It would be interesting to see the double integral you are trying to solve. – Michael Ulm Jan 20 '11 at 6:42
@Michael: I've edited that in. – David Z Jan 20 '11 at 19:14

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