# Need help integrating a function involving e^(-t^2)

1. Aug 10, 2011

### JFuld

1. The problem statement, all variables and given/known data

The context of this question is physics related but the problem I am having is purely mathematical.

g(t)= e^(-(a^2)(t^2))*e^(iwt) (a and w are just constants, i is sqrt(-1), not a constant)

I need to integrate this function with respect to t from -infinity to +infinity.

2. Relevant equations

error function: erf(x) = 2/sqrt(pi) * e^-t^2dt evaluated from t=0 to t=x

erf(infinity)=1, erf(-infinity) = -1

3. The attempt at a solution

i simplified g(t) to = e^(iwt-(a^2)(t^2)) and used an online integrator to get the answer.

My problem is that the integrated function involves the error function. From the math classes I have taken, the only way for me to have gotten that answer is if I had looked it up.

Clearly I wont be able to use an online integrator on an actual test so I was wondering If there are methods to deduce the answer without actually taking the integral.

Any insight is much appreciated.

2. Aug 10, 2011

### cragar

if you want to integrate e^(-t^2) , and the integral is over all space you can square the integral and then change one of t's to another letter and then do it in polar coordinates.
http://en.wikipedia.org/wiki/Gaussian_integral
look at this article it tells you how to compute using the trick i explained.
As one of my math professors would say its black magic.

3. Aug 10, 2011

### SteamKing

Staff Emeritus
Note: i = sqrt (-1) is a constant

4. Aug 10, 2011

### Ray Vickson

Write a^2*t*2 - i*w*t = a^2*(t^2 - 2*(i*w/ (2*a^2))*t) = a^2*(t - b)^2 - a^2*b^2, where b = i*w/(2*a^2). Thus, int_{-inf..inf} g(t) dt = exp(-a^2*b^2)*int_{u=-inf..inf} exp(-a^2*u^2) du, by changing variables to u = t-b. (Note: the t-integration goes along the real axis, while the u-integration goes along a line parallel to the real (u) axis (because b is imaginary); however, because the function exp(-a^2*u^2) is analytic and goes quickly to zero as |Re(u)| --> infinity, we can apply some standard theorems to show that the same integral is obtained by shifting the contour down to the real u-axis, so we get back again to a real function on the real line.)

RGV