Complex Integration with a removable singularity

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SUMMARY

The discussion centers on evaluating the integral \(\int_0^{\infty} x^{\frac{3}{2}}e^{-xd} J_o(rx) \frac{\sin (\gamma \sqrt{x}\sqrt{x^2+\alpha^2}t)}{\sqrt{x^2+\alpha^2}}\ dx\), where \(d, r, \alpha, \gamma, t \in \mathbb{R}^+\) and \(J_o\) is the zeroth order Bessel function of the first kind. The integral presents a removable singularity, prompting the use of complex analysis techniques, particularly contour integration. The initial approach involved a branch cut along the positive real axis and a modified keyhole contour, but this did not yield productive results. It is noted that while contour integration is advisable, the presence of a removable singularity means that the residue is zero, complicating the straightforward application of residue theory.

PREREQUISITES
  • Understanding of complex analysis and contour integration
  • Familiarity with Bessel functions, specifically the zeroth order Bessel function \(J_o\)
  • Knowledge of singularities in complex functions
  • Experience with integral transforms, such as the Bromwich integral
NEXT STEPS
  • Research advanced techniques in contour integration for integrals with removable singularities
  • Study the properties and applications of Bessel functions, focusing on \(J_o\)
  • Explore the use of branch cuts in complex analysis
  • Investigate the Bromwich integral and its relation to Laplace transforms
USEFUL FOR

Mathematicians, physicists, and engineers engaged in advanced calculus, particularly those working with integrals involving Bessel functions and complex analysis.

nickthequick
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Hi,

I'm trying to make headway on the following ghastly integral:

\int_0^{\infty} x^{\frac{3}{2}}e^{-xd} J_o(rx) \frac{\sin (\gamma \sqrt{x}\sqrt{x^2+\alpha^2}t)}{\sqrt{x^2+\alpha^2}}\ dx


where d,r, \alpha, \gamma ,t \in \mathbb{R}^+ and J_o is the zeroth order Bessel function of the first kind. Normally I wouldn't think I'd have a shot at finding this in closed form, but because there is a (removable) singularity, maybe there's a chance to exploit some complex analysis.

My attempts so far have all focused on making a branch cut along the positive real axis and integrating around a modified keyhole, which goes along the real axis, then follows an arc to the imaginary axis, going around the singularities, before arcing back to the real axis. This has not led to anything productive. It reminds of the contour used in a Bromwich integral.

Any suggestions? They'd be greatly appreciated.

Cheers,

Nick
 
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In principal, I would say it's a good idea to exploit contour integration, but keep in mind that when you have a removable singularity, the residue is 0, so the straight-forward way where you just consider the residue and argue that the arc doesn't contribute won't work.
 

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