Currently I try to find the Fourier transformed of this function: $ f(t) = -2e^{-3|t+3|} + e^{j4t}\cdot\frac{1}{1+t^2} $
I tried the following:
$$ \mathcal{F}\{f(t)\} = -2\cdot \mathcal{F} \{e^{-3|t+3|}\} + \mathcal{F}\{ \frac{e^{j4t}}{1+t^2}\} $$
My approach for $ \mathcal{F} \{e^{-3|t+3|}\} $ was to split the absolute function.
$$ e^{-3|t+3|} = e^{-3t-9} = e^{-3t}\cdot e^{-9}, t \ge -3 $$ $$ e^{-3|t+3|} = e^{3t+9} = e^{3t}\cdot e^{9}, t \le -3 $$
For the first one: $$ \mathcal{F} \{e^{-3t-9}\} = \mathcal{F}\{e^{-3t}\cdot e^{-9}\} = e^{-9} \cdot \mathcal{F} \{e^{-3t}\} = e^{-9} \cdot \frac{6}{9 + \omega^2} $$
The second one could be: $$ \mathcal{F} \{e^{3t+9}\} = \mathcal{F}\{e^{3t}\cdot e^{9}\} = e^{9} \cdot \mathcal{F} \{e^{3t}\} $$
I can't tell if the first one is correct and if my approach to split up the absolute function was the right way. Also I'm completely failing with $ \mathcal{F}\{ \frac{e^{j4t}}{1+t^2}\} $
Could someone give me some advise?
Start with $e^{-a|t|}$ and $\frac {2a}{a^2+w^2}$ are a Fourier transform pair. In the following all integrals are $-\infty,\infty$.
$\int e^{-3|t+3|}e^{-itw}dt=\int e^{-3|u|}e^{-i(u-3)w}du+\frac{6e^{3iw}}{9+w^2}$. Here $u=t+3$ and the direct transform was used.
For the second term, $\int \frac{e^{4it}}{1+t^2}e^{-itw}dt=\int \frac{e^{is(w-4)}}{1+s^2}ds=\pi e^{-|w-4|}$. Here $s=-t$ and the inverse transform was used.
Net result: transform=$\frac{-12e^{3iw}}{9+w^2}+\pi e^{-|w-4|}$