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I have to work out the integral:

$$ \int_0^{\infty} dq \frac{J_0(q \xi)}{q+1} $$

where $ J_0(z) $ is the Bessel function of the first type and order zero, $\xi \in \mathbb{R}$, $\xi \ge 0 $. So far, I only collected a good amount of trials and errors.

This problem actually comes from physics. I am Fourier transforming the 2-dimensional Coulomb potential $W(q)$ to real space coordinates, when screening effects in the the electron gas are accounted for as in the RPA, in the static and long wavelength limit so that:

$$ W(q) = \frac{2\pi e^2}{A} \frac{1}{q + \Lambda} $$

where $e$ is the electron charge, $A$ is the area, $q$ the modulus of wavevector $\mathbf{q}$ and $\Lambda$ (real positive constant) the screening wavevector. When transforming to real space coordinates $\mathbf{x}$, one arrives at the integral above (unless I made some mistakes), with $\xi = \Lambda x$, $x$ being the modulus of $\mathbf{x}$. So it would be physically interesting to have the large $x$ limit and small $x$ behaviour.

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  • $\begingroup$ Hi Manuel, welcome to MO. (i) What is RPA? (ii) What do you mean by "work out the integral"? what do you expect to get in the end - a bound? an explicit expression (hard to believe)? $\endgroup$ – Amir Sagiv Jul 8 at 18:06
  • $\begingroup$ I mean to evaluate the integral, as it will be a function of $\xi$. It would be nice to find an analytical expression for it, I imagine involving some other special functions, maybe the Bessel of second type, or Struve functions. If there is no expression for that, at least understanding how to have the asymptotic limit for large $\xi$ and for small $\xi$. $\endgroup$ – Manuel Simonato Jul 8 at 18:09
  • $\begingroup$ About the RPA: it is an approximation used in condensed matter physics. Do not worry for it, if it happens the reader is familiar with this approximation, he might give me some other useful insights in what I am doing. Anyway, thank you very much! :) $\endgroup$ – Manuel Simonato Jul 8 at 18:11
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The integral can be expressed in terms of Bessel and Struve functions, $$I(\xi)=\int_0^{\infty} dq\, \frac{J_0(q \xi)}{q+1}=\tfrac{1}{2} \pi \bigl(\pmb{H}_0(\xi)-Y_0(\xi)\bigr).$$ The small-$\xi$ behavior is $$I(\xi)=\xi-\ln \xi-\gamma_{\rm Euler} +\ln 2 +{\cal O}(\xi^2).$$ For large $\xi$ one has $$I(\xi)=1/\xi-\tfrac{1}{16}\sqrt\pi\xi^{-3/2}(\cos\xi+\sin\xi)+{\cal O}(\xi^{-2}).$$

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  • $\begingroup$ According to wikipedia, this is in fact $\frac12\pi \mathbf{K}_0(\xi)$ $\endgroup$ – username Jul 18 at 18:01
  • $\begingroup$ your $\mathbf{K}_0$ is again a Struve function, right, not a Bessel function? $\endgroup$ – Carlo Beenakker Jul 18 at 18:27

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