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Homework Help: Need help integrating a function involving e^(-t^2)

  1. Aug 10, 2011 #1
    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. jcsd
  3. Aug 10, 2011 #2
    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.
    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.
  4. Aug 10, 2011 #3


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    Note: i = sqrt (-1) is a constant
  5. Aug 10, 2011 #4

    Ray Vickson

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    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.)

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