The Signal
$ (t e^{-4t} \sin{6 \pi t}) u(t) $
The Fourier Transform
$ X(\omega)=\int_{-\infty}^{\infty} x(t) e^{-j\omega t}dt $
$ X(\omega)=\int_{-\infty}^{\infty} (te^{-4t}\sin{6\pi t})u(t) e^{-j\omega t}dt $
$ \sout{X(\omega)=\int_{0}^{\infty} (te^{-4t}\sin{6\pi t}) e^{-j\omega t}dt} $
$ X(\omega)=\int_{0}^{\infty} (te^{-4t})(\frac {e^{j 6 \pi t} - e^{-j 6 \pi t}}{2 j}) e^{-j\omega t}dt $
$ X(\omega)=\int_{0}^{\infty} \frac {t e^{-4t} e^{j 6 \pi t} e^{-j\omega t}}{2 j} - \frac {t e^{-4t} e^{-j 6 \pi t} e^{-j\omega t}}{2 j}dt $
