Challenge Problem:

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- Jun 10th 2011, 03:06 PMRandom Variableimproper integral #3
Challenge Problem:

Show that - Jun 10th 2011, 04:03 PMDeMath
Can we use the residue theorem?

- Jun 10th 2011, 04:10 PMRandom Variable
- Jun 10th 2011, 10:48 PMTheCoffeeMachine
I wonder whether the following fact is of any use here

I'm not familiar with double integrals, unfortunately! (Doh) - Jun 11th 2011, 11:52 AMRandom Variable
If you use that fact, then we have

But then changing the order of integration (which is justified because the original integral converges absolutely) doesn't really simply things.

The approach I had in mind was to use contour integration using a slightly unusual contour. - Jun 11th 2011, 01:27 PMRandom Variable
__Spoiler__: - Jun 11th 2011, 09:35 PMchisigma
- Jun 12th 2011, 07:52 AMchisigma
Performing the partial fraction expansion od the result (2) of my previous post we obtain...

(3)

The integrals in x can be solved in standard way with an approriate path in the complex plane obtaining...

(4)

(5)

(6)

(7)

Now is we substitute (4),(5), (6) and (7) in (3) we obtain the [incredible] result...

(8)

... so that is...

(9)

... and finally...

(10)

Kind regards

- Jun 12th 2011, 10:10 AMRandom Variable
Let

Let be the contour consisting of the line segment from the origin to (0,R); the arc from (R,0) to (0, IR); and the line segment from (0,IR) back to the origin.

has 4 simple poles at

The only pole inside of the contour is

So we have

(using L'Hospital's rule)

(ML inequality and triangle inequality)

(since )

therefore

Now we have

Now take the limit as R goes to infinity and equate the real portions on both sides to get