# Inverse Fourier transform

1. Aug 13, 2009

### 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?

2. Aug 13, 2009

### jpreed

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

It should be:

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

3. Aug 13, 2009

### mathman

A < 0 is necessary.

4. Aug 13, 2009

### Redbelly98

Staff Emeritus
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.

5. Aug 14, 2009

### 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?

6. Aug 14, 2009

### Redbelly98

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

That would become

-A < 0​
or in other words
A > 0​