Fourier Transform and Infinity Problem

1. May 2, 2010

frenzal_dude

1. The problem statement, all variables and given/known data
Hi, I need to find the Fourier Transform of: $$g(t)=\frac{1}{x}e^{\frac{-\pi t^2}{x^2}}$$

2. Relevant equations

$$G(f)=\int_{-\infty}^{\infty }g(t)e^{-j2\pi ft}dt \therefore G(f)=\int_{-\infty}^{\infty }\frac{1}{x}e^{\frac{-\pi t^2}{x^2}-j2\pi ft}dt$$

3. The attempt at a solution
$$G(f)=\frac{1}{x}[\frac{x^2e^{-t(\frac{\pi t}{x^2}+j2\pi f)}}{-2\pi t-j2\pi fx^2}] (limits:t=\infty,t=-\infty)$$

If you sub in t=infinity or t=-infinity, both will give you 0 because t is on the denominator.

Thanks for the help in advance! :)

Last edited: May 3, 2010
2. May 2, 2010

gabbagabbahey

Check your antiderivative; you are essentially claiming that

$$\frac{d}{dt}\left(\frac{\rm{e}^{h(t)}}{h'(t)}\right)=\rm{e}^{h(t)}$$

which isn't true.

3. May 3, 2010

frenzal_dude

Hi, I had a - insead of a + in the last exponential.

Basically to integrate the exponential, I took the derivative of the power and divided the exponential by that derivative. Isn't this the right method?

So the derivative of $$\frac{-\pi t^2}{x^2}-j2\pi ft=\frac{-2\pi t}{x^2}-j2\pi f=\frac{-2\pi t-j2\pi fx^2}{x^2}$$
$$G(f)=\int_{-\infty}^{\infty }\frac{1}{x}e^{\frac{-\pi t^2}{x^2}-j2\pi ft}dt=\frac{1}{x}[\frac{e^{\frac{-\pi t^2}{x^2}-j2\pi ft}}{\frac{-2\pi t-j2\pi fx^2}{x^2}}=\frac{1}{x}[\frac{x^2e^{-t(\frac{\pi t}{x^2}+j2\pi f)}}{-2\pi t-j2\pi fx^2}]$$

Last edited: May 3, 2010
4. May 3, 2010

gabbagabbahey

No, for example, $\int e^{-t^2}dt\neq-\frac{1}{2t}e^{-t^2}+C$. Take the derivative of it to convince yourself:

$$\frac{d}{dt}\left(-\frac{1}{2t}e^{-t^2}+C\right)=\left(\frac{1}{2t^2}+1\right)e^{-t^2}$$

via the product rule.

In fact, there is no nice antiderivative for $e^{-t^2}$, or your integrand (Unlees you consider the error function to be nice). You can however integrate it over the interval $(-\infty,\infty)$ easily by completing the square on the exponent, and using the fact that $\int_{-\infty}^{\infty}e^{-x^2}dx=\int_{-\infty}^{\infty}e^{-y^2}dy$, and hence

$$\int_{-\infty}^{\infty}e^{-x^2}dx=\left(\int_{-\infty}^{\infty}e^{-x^2}dx\int_{-\infty}^{\infty}e^{-y^2}dy\right)^{1/2}=\left(\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-(x^2+y^2)}dxdy\right)^{1/2}$$

Which you can integrate by switching to polar coordinates.