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

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Homework Help Overview

The problem involves integrating a function of the form g(t) = e^(-(a^2)(t^2)) * e^(iwt) with respect to t from -infinity to +infinity. The context is mathematical, although it is related to physics, and the original poster expresses concern about the use of online integrators and the reliance on the error function in the solution.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster simplifies the function and seeks methods to deduce the integral without direct computation. Some participants suggest using polar coordinates and squaring the integral as a technique for solving Gaussian integrals. Others discuss the implications of the imaginary unit in the context of integration.

Discussion Status

Participants are exploring various methods to approach the integral, with some providing insights into techniques that could be useful. There is no explicit consensus on a single method, but several lines of reasoning are being examined.

Contextual Notes

The discussion includes references to the error function and its properties, as well as the challenges of integrating complex functions. The original poster indicates a lack of familiarity with deriving the integral without external resources.

JFuld
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Homework Statement



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.

Homework Equations



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

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


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 won't 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.
 
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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.
 
Note: i = sqrt (-1) is a constant
 
JFuld said:

Homework Statement



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.

Homework Equations



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

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


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 won't 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.

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
 

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