Fourier Transfer of sawtooth function

[hotvette]

Homework Statement

Find the magnitude and phase of the Fourier transform of $h(t)=t$ over the interval 0,1

Homework Equations

$H(s) = \int^{1}_{0} h(t) e^{-i 2 \pi f t} dt$

The Attempt at a Solution

I found this thread: https://www.physicsforums.com/threads/fourier-transform-of-the-sawtooth-function.502113/ but can't seem to come close to the same result. Letting $a = -i 2 \pi f = -i \omega$ we get:
$H(s) = \int^{1}_{0} t e^{a t} dt \\ = \frac{e^{at}}{a^2}(at-1)|^{1}_{0} \\ = \frac{1}{a^2} \left( a t e^{at} - e^{at}\right)|^{1}_{0} \\ = \frac{1}{a^2}(ae^{a} - e^{a} + 1) \\ = \frac{1}{a^2}\left[(a-1) e^{a} + 1 \right] \\ = \frac{1}{a^2}(ae^{a} - e^{a} + 1) \\ = \frac{1}{(-i \omega)^2}\left[(-i \omega-1) e^{-i \omega} + 1 \right] \\ = \frac{-1}{\omega^2}\left[(-i \omega-1) e^{-i \omega} + 1 \right] \\ = \frac{1}{\omega^2}\left[(i \omega+1) e^{-i \omega} - 1 \right] \\ = \frac{1}{\omega^2}\left[(i \omega+1) (\cos \omega - i \sin \omega) - 1 \right] \\ \vdots \\ = \frac{1}{\omega^2} \left[ i (\omega\cos\omega-\sin \omega) + (\omega \sin \omega + \cos \omega -1)\right]$
I just don't see how this is converging to the solution in the referenced thread. Have I gone wrong somewhere?

 

[Ray Vickson]

Homework Statement

Find the magnitude and phase of the Fourier transform of $h(t)=t$ over the interval 0,1

Homework Equations

$H(s) = \int^{1}_{0} h(t) e^{-i 2 \pi f t} dt$

The Attempt at a Solution

I found this thread: https://www.physicsforums.com/threads/fourier-transform-of-the-sawtooth-function.502113/ but can't seem to come close to the same result. Letting $a = -i 2 \pi f = -i \omega$ we get:
$H(s) = \int^{1}_{0} t e^{a t} dt \\ = \frac{e^{at}}{a^2}(at-1)|^{1}_{0} \\ = \frac{1}{a^2} \left( a t e^{at} - e^{at}\right)|^{1}_{0} \\ = \frac{1}{a^2}(ae^{a} - e^{a} + 1) \\ = \frac{1}{a^2}\left[(a-1) e^{a} + 1 \right] \\ = \frac{1}{a^2}(ae^{a} - e^{a} + 1) \\ = \frac{1}{(-i \omega)^2}\left[(-i \omega-1) e^{-i \omega} + 1 \right] \\ = \frac{-1}{\omega^2}\left[(-i \omega-1) e^{-i \omega} + 1 \right] \\ = \frac{1}{\omega^2}\left[(i \omega+1) e^{-i \omega} - 1 \right] \\ = \frac{1}{\omega^2}\left[(i \omega+1) (\cos \omega - i \sin \omega) - 1 \right] \\ \vdots \\ = \frac{1}{\omega^2} \left[ i (\omega\cos\omega-\sin \omega) + (\omega \sin \omega + \cos \omega -1)\right]$
I just don't see how this is converging to the solution in the referenced thread. Have I gone wrong somewhere?

Your answer is correct; you have not gone wrong anywhere that I can see.