How to Compute Inverse Fourier Transform for a Specific Function

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The discussion focuses on computing the Inverse Fourier Transform of the function \(\frac{\alpha^2}{2\pi}\exp\left(-\frac{1}{2} \alpha^2 C^2(K) \tau \omega^2\right)\). The key point is that for the transform to be valid, the parameter \(A\) must be negative, which translates to \(A > 0\) when adjusted for the new function. Participants confirm that the resulting expression for the inverse transform is \(\frac{\alpha^2}{2\pi}\frac{1}{\sqrt{2A}}\exp\left(\frac{t^2}{4A}\right)\). A variable transformation can simplify the integral to a standard form that can be referenced in integral tables. Overall, the adjustments in the function's parameters do not significantly alter the computation process.
mathy_girl
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Hi all,

I'm having a bit trouble computing the Inverse Fourier Transform of the following:

\frac{\alpha}{2\pi}\exp\left(\frac{1}{2} \alpha^2 C^2(K) \tau \omega^2\right)

Here, C^2(K), \alpha and \tau can be assumed to be constant. Hence, we have an integral with respect to \omega.

Who can help me out?
 
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So you want to find the inverse Fourier transform of
\frac{\alpha}{2\pi}\exp(A \omega^2)?

It should be:

\frac{\alpha}{2\pi}\frac{1}{\sqrt{-2 A}}\exp\left(\frac{t^2}{4A}\right)
 
jpreed said:
So you want to find the inverse Fourier transform of
\frac{\alpha}{2\pi}\exp(A \omega^2)?

It should be:

\frac{\alpha}{2\pi}\frac{1}{\sqrt{-2 A}}\exp\left(\frac{t^2}{4A}\right)

A < 0 is necessary.
 
One can do a suitable variable transform to get the integral in the form


∫e-x2 dx
with limits -∞ to +∞


which can be looked up in a standard table of integrals. I suspect the answer is what jpreed gave in post #2.
 
Whoops.. I just figured that there are two small mistakes in my first post, I would like to have the Inverse Fourier Transform of:
\frac{\alpha^2}{2\pi}\exp\left(-\frac{1}{2} \alpha^2 C^2(K) \tau \omega^2\right)

Here, note that \alpha is squared, and a minus sign is added in the argument of exp.

Don't know if that makes a lot of difference?
 
Not really. Just replace A with -A in all the responses.

mathman said:
A < 0 is necessary.

That would become

-A < 0​
or in other words
A > 0​
 
Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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