# Improper integral

1. Jul 29, 2010

### techmologist

How do you get this:

$$\int_{-\infty}^{\infty}e^{-(ax^2+b/x^2)}dx = \sqrt{\frac{\pi}{a}}e^{-2\sqrt{ab}}$$

I've been trying all the tricks I know, like differentiating under the integral sign and whatnot, but I can't get it. Thanks.

2. Jul 29, 2010

### Coriolis314

In which course did you run into this monster?

3. Jul 29, 2010

### techmologist

Not in any course. I came across it in a paper where they calculate the probability that a Brownian motion X(t) crosses the line at+b at some point during the interval (0, T).

4. Jul 29, 2010

### Coriolis314

Send a link of the paper?

5. Jul 30, 2010

### lurflurf

2)Find I' by differentiating under the integral sign with respect to sqrt(b)
3)Effect the change of variable u=sqrt(b/a)/x (break up the interval at 0)
4)Notice I=-2sqrt(a)I'
5)Conclude I=I0 exp(-2sqrt(ab))
6)Compute I0

sqrt() denotes the square root
I' is the derivative of I with respect to sqrt(b)
I0 is your integral when b=0
Feynman's Calculus

6. Jul 30, 2010

### techmologist

Thanks, lurflurf! That's exactly what I needed. Well done. That's an excellent thread you linked to, also. Surely, You're Joking is where I first learned about differentiating under the integral sign. I apparently need more practice at applying it, though.

Sure. Here is the paper where the probability is calculated (near the end) using Laplace transforms:

http://projecteuclid.org/DPubS?service=UI&version=1.0&verb=Display&handle=euclid.pjm/1102911625

Here is a paper that makes use of the result. Scroll down to where it says "Kelly Criterion"--that's it. In the section on probability of reaching a goal within a certain time.

http://edwardothorp.com/id10.html

Last edited: Jul 30, 2010