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Tricky semiinfinite integral 
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#1
Feb909, 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(ps)^2}}{{p^2+b^2}} dp)[/tex] where s is complex but I was stuck here. Any suggestions would be much appreciated. 


#2
Feb909, 11:27 AM

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PF Gold
P: 5,533

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?



#3
Feb909, 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?



#4
Feb909, 05:52 PM

Emeritus
Sci Advisor
PF Gold
P: 5,533

Tricky semiinfinite 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.



#5
Feb909, 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)}{2b}}[/tex] 


#6
Feb909, 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? 


#7
Feb1009, 10:18 AM

Emeritus
Sci Advisor
PF Gold
P: 5,533

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.



#8
Feb1009, 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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