# Tricky semi-infinite integral

by ab959
Tags: integral, semiinfinite, tricky
 P: 11 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.
 Emeritus Sci Advisor PF Gold P: 5,540 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?
 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?
Emeritus
PF Gold
P: 5,540

## 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.
 P: 11 Cool. Thanks for the help! It was a very nice step. Here is the solution for anyone interested: $$\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp = \frac{\pi e^{tb^2} cos(ibu)}{2|b|}}$$
 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?
 Emeritus Sci Advisor PF Gold P: 5,540 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.
 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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