Fourier Transform and Infinity Problem

[frenzal_dude]

Homework Statement

Hi, I need to find the Fourier Transform of: g(t)=\frac{1}{x}e^{\frac{-\pi t^2}{x^2}}

Homework Equations

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

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! :)

 

[gabbagabbahey]

Check your antiderivative; you are essentially claiming that

\frac{d}{dt}\left(\frac{\rm{e}^{h(t)}}{h&#039;(t)}\right)=\rm{e}^{h(t)}

which isn't true.

 

[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}<br />
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}]

 

[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.