Complex contour integral with a second order pole at origin

[Finnjessy]

Homework Statement

Hello all. I'm currently attempting to prove the central limit theorem using a simple case of two uniformly distributed random variables. Aside from being able to solve it using convolutions, I also wish to solve it by using the Dirac Delta function. That aside, the integral I'm trying to solve is

$$\int_{-\infty}^{\infty} \! -\frac{e^{ikx}}{k^2} \, \mathrm{d}k$$

with $x \in [0,2]$

Homework Equations

The Cauchy Residue Theorem

$$Res(f,c) = \frac{1}{2\pi i}\oint_\gamma \! f(z) \, \mathrm{d}z =$$

The Attempt at a Solution

According to Plemelj's theorem if we have a simple pole on the contour of our chosen integral, the result is equal to $\pm i\pi Res(f,c)$, where c is the location of the pole.

I know from both a Mathematica calculation and a Monte Carlo simulation that the solution to this integral is in fact $x\pi$. However, if I calculate the residue of the function and use Plemelj's theorem, I get the exact result that I would want. The problem is that as far as I know, Plemelj's theorem should only work for simple poles and the second order term should change this calculation. My question therefor is, why does the theorem work for my second order pole?

As far as alternate ways of calculating the integral, I've tried several methods but haven't found anything that would work for second order poles. I've also attempted several different substitutions to get rid of the second order pole, but all seem to blow up to infinity or go to zero. So assuming that Plemelj's theory indeed can't be applied to higher order poles, I'm kindly asking for help as far as to what direction I should head with the calculation. Thank you.

 

[Greg Bernhardt]

I'm sorry you are not generating any responses at the moment. Is there any additional information you can share with us? Any new findings?

 

[vela]

The theorem you're citing doesn't apply to a second-order pole, so it's not really correct to say that it works for your problem. The expression just happens to give the right result.

That said, if you apply the Cauchy residue theorem with the appropriate contour, you can derive the same result. Here's a page with different contours. Try the middle one.

http://mathworld.wolfram.com/Contour.html