I am getting two different answers for the Fourier transform of the same function $h(t) = {e}^{-2t} cos{\pi}t \space u(t-1).$
I don't find anything wrong in the two methods. What is the right answer and how?
Method 1 (image)
$$ \mathcal{F}[h(t)] = \int_{-\infty}^{\infty} e^{-2t} \cos\pi t \space u(t-1) e^{-j\omega t} dt = \int_{1}^{\infty} {e}^{-2t} \cos{\pi}t \space e^{-j\omega t} dt \\ = \int_{1}^{\infty} {e}^{-2t} \frac {(e^{-j \pi t}+e^{j \pi t})}{2} e^{-j\omega t} dt = \frac{e^{-(2 + j \omega )+ j \pi}}{2(2 +j\omega - j\pi)} + \frac{e^{-(2 + j \omega )- j \pi}}{2(2 +j\omega + j\pi)} $$
Method 2 (image)
$$ h(t) = {e}^{-2t} \cos{\pi}t \space u(t-1) = \frac{-1}{e^{2}}{e}^{-2(t-1)} \cos{\pi}(t-1) \space u(t-1) $$ Consider $ g(t) = {e}^{-2t} \cos{\pi}t \space u(t).$ $$ \mathcal{F}[g(t)] = \int_{-\infty}^{\infty} {e}^{-2t} \space \cos{\pi}t \space u(t) e^{-j\omega t} dt = \int_{0}^{\infty} {e}^{-2t} \frac {(e^{-j \pi t}+e^{j \pi t})}{2} e^{-j\omega t} dt = \frac{(2+j \omega)}{{(2+j \omega)}^{2} + {\pi}^{2}} $$ Using time shift property, i.e. $f (t − a)u(t − a) = e^{−j \omega a} F (\omega),$ $$ \mathcal{F}[h(t)] = \frac{-1}{e^{2}} e^{-j \omega} \frac{(2+j \omega)}{{(2+j \omega)}^{2} + {\pi}^{2}} $$
Method 2 and Method 1 answers don't match after simplification. What is wrong?
They do match: $$ \frac{e^{-(2 + j \omega )+ j \pi}}{2(2 +j\omega - j\pi)} + \frac{e^{-(2 + j \omega )- j \pi}}{2(2 +j\omega + j\pi)} \\ = \frac{e^{-(2+j\omega)}}{2} \left( \frac{e^{j\pi}}{2+j\omega-j\pi} + \frac{e^{-j\pi}}{2+j\omega+j\pi} \right) \\ = \frac{e^{-(2+j\omega)}}{2} \frac{(2+j\omega+j\pi)e^{j\pi} + (2+j\omega-j\pi)e^{-j\pi}}{(2+j\omega-j\pi)(2+j\omega+j\pi)} \\ = \frac{e^{-(2+j\omega)}}{2} \frac{(2+j\omega)(e^{j\pi}+e^{-j\pi})+j\pi(e^{j\pi}-e^{-j\pi})}{(2+j\omega)^2+\pi^2} \\ = \frac{e^{-(2+j\omega)}}{2} \frac{(2+j\omega)2\cos\pi+j\pi\cdot2j\sin\pi}{(2+j\omega)^2+\pi^2} \\ = \frac{e^{-(2+j\omega)}}{2} \frac{(2+j\omega)2(-1)+j\pi\cdot 0}{(2+j\omega)^2+\pi^2} \\ = -\frac{(2+j\omega)e^{-(2+j\omega)}}{(2+j\omega)^2+\pi^2} \\ = -\frac{1}{e^2}e^{-j\omega}\frac{(2+j\omega)}{(2+j\omega)^2+\pi^2} \\ $$