Fourier transform ##f(t) = te^{-at}##

[DragonBlight]

Homework Statement

Finding the Fourier transform of $f(t) = te^{-at}$ if t > 0 and $f(t) = 0$ otherwise.

Relevant Equations

$F(w) = \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} te^{-at}e^{-iwt} dt$

Doing the Fourier transform for the function above I'm getting a result, but since I can't get the function f(t) with the inverse Fourier transform, I'm wondering where I made a mistake.

$F(w) = \frac{1}{\sqrt{2 \pi}} \int_{0}^{\infty} te^{-t(a + iw)} dt$
By integrating by part, where G = -a - iw
$F(w) = \frac{te^{Gt}}{G}|_{0}^{\infty} - \int_{0}^{\infty} \frac{e^{Gt}}{G} dt$

$= \frac{1}{G^2}$

Thus,
$F(w) = \frac{1}{\sqrt{2 \pi}} \frac{1}{(-a -iw)^2}$

 

[Orodruin]

Is your function $t^{-at}$ as you write in the statement and title or $t e^{-at}$ as you have used in your computation?

 

[DragonBlight]

I made I mistake. It's $te^{-at}$

 

[Orodruin]

So, here is an alternative way: Fourier's trick (differentiating with respect to a parameter inside the integral)

Let
$$
G(a,\omega) = \frac{1}{\sqrt{2\pi}} \int_0^\infty e^{-at} e^{-i\omega t} dt = -\frac{1}{\sqrt{2\pi}} \left[\frac{e^{-at-i\omega t}}{a+i\omega}\right]_0^\infty = \frac{1}{\sqrt{2\pi}}\frac{1}{a+i\omega}.
$$
Now,
$$
\frac{\partial G}{\partial a} = - \frac{1}{\sqrt{2\pi}}\int_0^\infty t e^{-at} e^{-i\omega t} dt = -F(\omega).
$$
It follows that
$$
F(\omega) = -\frac{1}{\sqrt{2\pi}} \frac{d}{da}\frac{1}{a+i\omega} =\frac{1}{\sqrt{2\pi}} \frac{1}{(a+i\omega)^2}.
$$

Why do you think there is a mistake?

 

[vela]

since I can't get the function f(t) with the inverse Fourier transform, I'm wondering where I made a mistake.

As @Orodruin has noted, you calculated the Fourier transform correctly, so it sounds like the problem is in your calculation of the reverse Fourier transform. Can you show us your attempt at that?

 

[Mark44]

I made I mistake. It's $te^{-at}$

Edited title and Homework Statement to correct this typo.

 

[DragonBlight]

As @Orodruin has noted, you calculated the Fourier transform correctly, so it sounds like the problem is in your calculation of the reverse Fourier transform. Can you show us your attempt at that?

$f(t) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}} \frac{1}{(-\alpha -i\omega)^2} e^{i\omega t} d\omega$

I don't see how to find the residue since the imaginary part is $\omega$. There's a singularity $\omega = i \alpha$
I mean, if I had $(-i\alpha - \omega)$ I could find it.

I think I found. if I factor i at the denominator then I have a pole of order 2 at $\omega = \frac{- \alpha}{i}$