Complex Integration with a removable singularity

In summary, the conversation discusses a difficult integral and the possibility of using complex analysis and contour integration to solve it. However, due to a removable singularity, the traditional method of considering the residue may not work. The speaker is open to suggestions and is seeking help in solving the integral.
  • #1
Hi,

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

[tex] \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 [/tex]


where [itex] d,r, \alpha, \gamma ,t \in \mathbb{R}^+[/itex] and [itex]J_o[/itex] 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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  • #2
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.
 

1. What is a removable singularity in complex integration?

A removable singularity is a point in a complex function where the function is undefined, but can be made continuous by assigning a value to that point. This means that the function has a hole or a jump at that point, but it can be filled in to make the function continuous.

2. How do you integrate a function with a removable singularity?

To integrate a function with a removable singularity, you must first identify the location of the singularity. Then, you can use a technique called "analytic continuation" to assign a value to the singularity and make the function continuous. Once the function is continuous, you can integrate it using standard techniques.

3. Can a function have more than one removable singularity?

Yes, a function can have multiple removable singularities. Each singularity must be treated separately by using analytic continuation to assign a value and make the function continuous.

4. What is the significance of removable singularities in complex integration?

Removable singularities are important because they allow us to extend the domain of a function and make it continuous. This allows us to integrate functions that would otherwise be impossible to integrate due to their singularities. It also allows us to study the behavior of functions near these points.

5. Are removable singularities the only type of singularity in complex integration?

No, there are other types of singularities such as poles and essential singularities. These types of singularities cannot be removed and have different properties in terms of integration and the behavior of the function near these points.

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