In this paper, page 252, there is an equality (D.69), the numerical results integrals on the left is of interest:

$$\int_{-\infty}^{\infty}\dfrac{x\cdot J_0(x\cdot r)}{x^2-k^2}dx=\left\{ \begin{array}{cc} \dfrac{i \cdot\pi \cdot H_0^{(1)}(k\cdot r)}{2}&\text{Im}(k)>0 \\ \\ \dfrac{i \cdot\pi \cdot H_0^{(2)}(k\cdot r)}{2} &\text{Im}(k)<0\\ \end{array} \right. \tag{D.69}$$

Where $J_0$ is the BesselJ, $H_0$'s are the Hankel functions respectively.

I don't know how to prove the integral equality or whether it holds only for the non-real $k$'s .

But I am interested in the numerical (or analytical when possible) result when $k$ is a real positive number.

But how to obtain a numerical solution of such integrals? e.g., I tried to let $k=10$ and $r=6$, but found most of the available integral algorithms in NIntegrate didnot lead to easy convergence probably due to the existence of a discontinuity point $x=k$ of the integrand and the improper upper limit.


NIntegrate[(x BesselJ[0, 6 x])/(x^2 - 100), {x, 0, Infinity}, 
 Method -> {"OscillatorySelection", "TermwiseOscillatory" -> True, 
   Method -> {"GlobalAdaptive", "MaxErrorIncreases" -> 10000}}, 
 MaxRecursion -> 100, PrecisionGoal -> 8, AccuracyGoal -> 5]



and warning messages:

NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.

NIntegrate::ncvb: NIntegrate failed to converge to prescribed accuracy after 9 recursive bisections in x near {x} = {10.0009}. NIntegrate obtained 0.16591870616051807and 0.2705956661691797 for the integral and error estimates.

  • $\begingroup$ The integration doesn't converge: Integrate[(x BesselJ[0, 6 x])/(x^2 - 100), {x, -Infinity, Infinity}] $\endgroup$
    – xzczd
    Jul 30, 2017 at 5:22

1 Answer 1

Integrate[(x BesselJ[0, r x])/(x^2 - k^2), {x, 0, Infinity}, 
    PrincipalValue -> True, Assumptions -> k > 0 && r > 0]
(* 1/2 π BesselY[0, k r] *)

N[% /. k -> 10 /. r -> 6]
(* -0.0743913 *)
  • 1
    $\begingroup$ You only took half of the domain in the OP's integral. $\endgroup$ Jul 30, 2017 at 6:35
  • 1
    $\begingroup$ @J.M. OP's actual code shows the integral from zero to Infinity, so I presume that is what he meant. Furthermore, the integrand is antisymmetric, so the integral from -Infinity to Infinity is zero. $\endgroup$
    – bbgodfrey
    Jul 30, 2017 at 6:37
  • 1
    $\begingroup$ Indeed, so we'll have to wait for OP to clarify. The only problem now is $k=0$, where the integral is genuinely divergent. $\endgroup$ Jul 30, 2017 at 6:40
  • $\begingroup$ Thank you. This already answers my question $\endgroup$ Jul 30, 2017 at 8:09

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