I am trying to solve the following integral (which can be seen also as a Hankel Transform):

$\int_{0}^{\infty} \frac{1}{\left( k-1 \right)^2} J_0(k r) \, dk$

where $J_0(k r)$ is the zeroth order Bessel function of the first kind, and $r$ is a positive real number.

My idea was to use integration by parts considering that $\frac{1}{\left( k-1 \right)^2} = \frac{d}{dk} \left(\frac{-1}{\left( k-1 \right)} \right)$ that is to say:

$\int_{0}^{\infty} \frac{1}{\left( k-1 \right)^2} J_0(k r) dk= \left[\frac{-J_0(k r)}{\left( k-1 \right)} \right]_{0}^{\infty}+r \, \int_{0}^{\infty} \frac{1}{\left( k-1 \right)} J_1(k r) dk$

however i cannot find any solution for the integral $\, \int_{0}^{\infty} \frac{1}{\left( k-1 \right)} J_1(k r) dk$

also, as this problem comes from a physical problem, it is important that the solution (which is a function of $r$) has to go to zero at $r \rightarrow \infty$ (however it can diverge at $r=0$), do you know if this can easily checked?

I checked Proudnikov Integrals and Series Vol.2 but I couldn't find anything helpful.

do you have any hint?

  • $\begingroup$ Don't you think that the integral diverges at $k=1$, or are you looking for a principal part integral? $\endgroup$
    – Paul Enta
    Mar 16 '17 at 23:02
  • $\begingroup$ I know that the integrando diverges for $k=1$. But maybe the integral can still have a closed form. For example the integral of $\frac{J_0 (k r)}{k-1}$ has a closed form in terms of Struve H and Bessel Y functions (from proudnikov vol2 pag176) $\endgroup$
    – SSC Napoli
    Mar 16 '17 at 23:07

One can decompose $$I=\int_0^\infty \frac{J_1(kr)}{k-1}\,dk=-\int_0^\infty J_1(kr)\,dk+\int_0^\infty \frac{kJ_1(kr)}{k-1}\,dk$$. Which can be evaluated as$$I=-\frac{1}{r}-\int_0^\infty \frac{d}{dr} \frac{J_0(kr)}{k-1}\,dk$$ Though validity of the inversion of the order integration / derivation is unclear, using the expression you quote: $$\int_0^\infty \frac{J_0(kr)}{k-1}\,dk=\varphi(r) $$ you get$$I=-\frac{1}{r}-\varphi'(r)$$

  • $\begingroup$ Thanks a lot for your answer. Is it possible to exchange the derivation sign in the integration? I should have a look at Leibniz rule. Also from the physical meaning of the problem it is important that the solution goes to zero as r goes to infinity . Whereas from your result it is clear that it will go to 1 $\endgroup$
    – SSC Napoli
    Mar 17 '17 at 9:41

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