How to Compute Inverse Fourier Transform for a Specific Function

[mathy_girl]

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?

 

[jpreed]

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

It should be:

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

 

[mathman]

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.

 

[Redbelly98]

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.

 

[mathy_girl]

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?

 

[Redbelly98]

Not really. Just replace A with -A in all the responses.

A < 0 is necessary.

That would become

-A < 0​

or in other words

A > 0​