Specify a signal x(t) and compute its Fourier transform using the integral formula. (Make sure your signal is not trivial to transform; it should be hard enough to be on a test).
Defining x(t):
$ x(t) = te^{-4t}u(t-3) $
By the integral formula:
$ \mathcal{X}(\omega) = \int_{-\infty}^{\infty}x(t)e^{-j\omega t}\,dt\, $
Therefore:
$ \mathcal{X}(\omega) = \int_{-\infty}^{\infty}te^{-4t}u(t-3)e^{-j\omega t}\,dt\, $
$ \mathcal{X}(\omega) = \int_{3}^{\infty}te^{-4t}e^{-j\omega t}\,dt\, $ (maybe remove)
$ \mathcal{X}(\omega) = \int_{3}^{\infty}te^{-t(4+ j\omega)}\,dt\, $
Integrating by parts
$ \int UdV=UV - \int VdU $
Where $ U=t; dU=1; V= \frac{-1}{4+j \omega} e^{-t(4+j \omega)}; dV= e^{-t(4+j \omega)} $
