Inverse fourier transform of gaussian

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Homework Help Overview

The discussion revolves around proving that the inverse Fourier transform of a Gaussian function, specifically \( e^{-\frac{k^2}{2}} \), results in another Gaussian function. Participants are exploring concepts from complex analysis and Fourier transforms.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to integrate over a circular contour in the complex plane but struggles with the resulting integral. They question the application of Cauchy's theorem regarding the integrand's analyticity. Other participants inquire about the necessity of the circular contour and suggest clarifying the limits of integration for the inverse Fourier transform.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about integration methods and dimensionality. Some guidance has been offered regarding the limits of integration and the nature of the integral involved, but no consensus has been reached on the best approach.

Contextual Notes

There is a focus on understanding the dimensionality of the integral and the implications of integrating over the complex plane. Participants express uncertainty about the correct method to approach the problem, indicating a need for further clarification on complex integration.

Dazedandconfu
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well, i have to prove that the inv. Fourier transform of a gaussian (e^(-(k^2/2)) is a gaussian, i know some elementary complex analysis(never actually taken a class in it), not well enough, it seems, to find the solution to this. I tried to integrate over a circular contour, and let the radius of the circle go to infinity, i couldn't solve the integral that i obtained (it was pretty complicated). Also, i don't quite understand this, the integrand is complex analytic everywhere, so if i integrate it over a circular contour, wouldn't i get 0, by cauchy's theorem?
Any help much appreciated
 
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Dazedandconfu said:
well, i have to prove that the inv. Fourier transform of a gaussian (e^(-(k^2/2)) is a gaussian, i know some elementary complex analysis(never actually taken a class in it), not well enough, it seems, to find the solution to this. I tried to integrate over a circular contour, and let the radius of the circle go to infinity, i couldn't solve the integral that i obtained (it was pretty complicated). Also, i don't quite understand this, the integrand is complex analytic everywhere, so if i integrate it over a circular contour, wouldn't i get 0, by cauchy's theorem?
Sure, but why are you integrating over a circular contour in the first place?

Start by writing out the integral you have to do to find the inverse Fourier transform.
 
well its the integral (e(-k^2)e^ikx)dk over the entire complex plane right?, so unless I am wrong there are two ways to integrate over the complex plane, a circle of infinite radius, or k=x+iy, and x,y go to infinity
riight??
 
For one thing, you can't integrate over the entire complex plane. Well, you can, but it has to be a double integral, which is not what you have here. There's only one differential (dk), so you only get to integrate in one dimension.

Check your references if you need to, in order to find the correct limits of integration for the integral
\int_?^? e^{-k^2/2}e^{ikx}\mathrm{d}k
which is involved in the Fourier transform.
 
im fairly certain that the limits are -inf. to inf . well yes, i get that you only integrate in one variable, but isn't it true that a complex number a+bi, with arbitrary a and b can span the entire complex plane? so you would have a line integral in da and idb, with both a and b (-∞,∞) or equivalently circle of radius of r, a=rcosθ ; b= rsinθ with r going to ∞.
sorry if I am being slow btw
 
Dazedandconfu said:
im fairly certain that the limits are -inf. to inf.
Yes, that's correct.
Dazedandconfu said:
well yes, i get that you only integrate in one variable,
Not just one variable, but one dimension. A single integral with a single differential dk is a one-dimensional integral. A dimension corresponds to one real variable.
Dazedandconfu said:
but isn't it true that a complex number a+bi, with arbitrary a and b can span the entire complex plane? so you would have a line integral in da and idb, with both a and b (-∞,∞) or equivalently circle of radius of r, a=rcosθ ; b= rsinθ with r going to ∞.
Yes, but if you have both da and db, then it's not a line integral, it's a (two-dimensional) surface integral.
 

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