# 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: 5,532 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 Cool. Thanks for the help! It was a very nice step. Here is the solution for anyone interested: $$\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp = \frac{\pi e^{tb^2} cos(ibu)}{2|b|}}$$