## Tricky semi-infinite integral

1. The problem statement, all variables and given/known data
I am trying to integrate

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

2. Relevant equations
I know that

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

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

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

where s is complex but I was stuck here. Any suggestions would be much appreciated.
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus The integrand is an even function of p, so I would divide it by 2 and integrate over $(-\infty ,\infty )$. Then I would try to use the Residue Theorem. Are you familiar with that result?
 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?

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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.
 Cool. Thanks for the help! It was a very nice step. Here is the solution for anyone interested: $$\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp = \frac{\pi e^{tb^2} cos(ibu)}{2|b|}}$$
 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?
 Recognitions: Gold Member Homework Help Science Advisor Staff Emeritus 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.
 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.