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Is there a closed form solution to \begin{equation} \int_0^\infty \frac{J_0(kr)}{k}dk, \end{equation} where $J_0$ is Bessel function and $r$ is a constant?

Amey Joshi
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  • Your integral is not convergent. You have, essentially a $1/k$ as $k\to0^+$. – mickep Mar 05 '16 at 12:05
  • I got this integral while I was trying to find inverse Fourier transform of $1/k^2$ in two dimensions. Was expecting to get a logarithm, something like $\ln r$. – Amey Joshi Mar 05 '16 at 12:22
  • The general solution can be found here. – Lucian Mar 05 '16 at 12:39
  • The integral is indeed divergent. If that helps, though $$\lim_{\epsilon\to 0^+}\left[\ln\epsilon+\int_{\epsilon}^\infty dk\frac{J_0(kr)}{k}\right]=\ln \left(\frac{2}{r}\right)-\gamma\ ,$$ where $\gamma$ is the Euler's constant. – Pierpaolo Vivo Mar 05 '16 at 12:41
  • Thanks everyone for the comments. The problem of finding inverse Fourier transform of $1/k^2$ is briefly discussed at https://math.stackexchange.com/questions/847706/does-the-integral-in-the-formal-2d-fourier-transform-of-the-logarithm-converge. – Amey Joshi Mar 08 '16 at 05:28

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