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Tricky semi-infinite integral

by ab959
Tags: integral, semiinfinite, tricky
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ab959
#1
Feb9-09, 01:59 AM
P: 11
1. The problem statement, all variables and given/known data
I am trying to integrate

[tex]\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp[/tex].

2. Relevant equations
I know that

[tex]\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} dp= \frac{\pi}{2b}e^{tb^2}erfc(\sqrt{a}x)[/tex]


3. The attempt at a solution
I rewrote the problem in terms of the complex exponential and reduced the integrand to a form similar to

[tex]\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp= Re(\int^\infty_0 \frac{e^{-t(p-s)^2}}{{p^2+b^2}} dp)[/tex]

where s is complex but I was stuck here. Any suggestions would be much appreciated.
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Tom Mattson
#2
Feb9-09, 11:27 AM
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The integrand is an even function of p, so I would divide it by 2 and integrate over [itex](-\infty ,\infty )[/itex]. Then I would try to use the Residue Theorem. Are you familiar with that result?
ab959
#3
Feb9-09, 05:09 PM
P: 11
Yes I am familiar with it however not in this form. Are you suggesting forming a curve on the complex plane which encloses the points p=+/-ib. The integral over this curve would equal 2*pi*i* Sum of residues?

Tom Mattson
#4
Feb9-09, 05:52 PM
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Tricky semi-infinite integral

Not quite. If the real axis is part of the contour, then you would close it with a semicircular arc in the upper half plane (or the lower half, your choice). In either case the contour only encloses one of the poles, not both.
ab959
#5
Feb9-09, 05:58 PM
P: 11
Cool. Thanks for the help! It was a very nice step.

Here is the solution for anyone interested:

[tex]
\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp = \frac{\pi e^{tb^2} cos(ibu)}{2|b|}}[/tex]
ab959
#6
Feb9-09, 08:48 PM
P: 11
Hi Tom. As my b is a real number, for the case b=0 this doesn't work. I want to find out

the limit as b-> 0 of b * integral. I know that it is equal to zero but I am having proving this. All I really have to do is show that the integral with b=0 is finite.

Any suggestions?
Tom Mattson
#7
Feb10-09, 10:18 AM
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How about setting b=0 in the original integral and calculating the residue of the pole of order 2 that results? You'll need a different contour of course.
ab959
#8
Feb10-09, 05:28 PM
P: 11
I ended up just using L'Hopital's rule and it came out quite easily. Thanks Tom you were a lot of help! Very much appreciated.


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