Example of Computation of Fourier transform of a CT SIGNAL
A practice problem on Fourier transform
$ X(t)=e^{-5t}cos{(2t)}u(t)dt $
$ \,\mathcal{X}(\omega)=\int_{-\infty}^{+\infty}x(t)e^{-j\omega t}\,dt\, $
$ \,\mathcal{X}(\omega)=\int_{0}^{+\infty}e^{-5t}cos{(2t)}e^{-j\omega t}\,dt, $
$ \,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-5t}e^{2jt}e^{-j\omega t}\,dt + 1/2\int_{0}^{+\infty}e^{-5t}e^{-2jt}e^{-j\omega t}\,dt\, $
$ \,\mathcal{X}(\omega)=1/2\int_{0}^{+\infty}e^{-t(5-2j+j\omega)}\,dt + 1/2\int_{0}^{+\infty}e^{-t(5+2j+j\omega)}\,dt\, $
$ \,\mathcal{X}(\omega)=e^{-15}\int_{1}^{+\infty}e^{-(j\omega +5)t}\,dt + e^{-j(\omega +\pi)\frac{\pi}{2}}\, $
$ \,\mathcal{X}(\omega)=\left. 1/2\frac{e^{-t(5-2j+j\omega)}}{-(5-2j+j\omega)}\right]_{0}^{+\infty} + \left. 1/2\frac{e^{-t(5+2j+j\omega)}}{-(5+2j+j\omega)}\right]_{0}^{+\infty} $
$ \,\mathcal{X}(\omega)=\frac{1}{2(5-2j+j\omega)} + \frac{1}{2(5+2j+j\omega)} $
