How to Compute Inverse Fourier Transform for a Specific Function

AI Thread Summary
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​
 
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