# Tricky semi-infinite integral

1. Feb 9, 2009

### ab959

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.

2. Feb 9, 2009

### Tom Mattson

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?

3. Feb 9, 2009

### ab959

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. Feb 9, 2009

### Tom Mattson

Staff Emeritus
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. Feb 9, 2009

### ab959

Cool. Thanks for the help! It was a very nice step.

Here is the solution for anyone interested:

$$\int^\infty_0 \frac{e^{-tp^2}}{{p^2+b^2}} cos(pu) dp = \frac{\pi e^{tb^2} cos(ibu)}{2|b|}}$$

Last edited: Feb 9, 2009
6. Feb 9, 2009

### ab959

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. Feb 10, 2009

### Tom Mattson

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.

8. Feb 10, 2009

### ab959

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.