My question starts from the calculation of the inverse Laplace transform of $F(s)=\frac{1}{s+z_0}$ where I assume $\mathrm{Re}(z_0)>0$ WLOG. Simply I used the Mellin formula and calculated the residue of $F(s)e^{st}$ at $s=-z_0$ so $f(t)=\mathcal{L}^{-1}\{\frac{1}{s+z_0}\}=e^{-z_0t}$. However, the bilateral Laplace transform of $e^{-z_0t}$ seems meaningless, rather I must use $e^{-z_0t}u(t)$ instead where $u(t)$ is the unit step function. Although the unilateral Laplace transform gives the correct result, I want to make sure that $t$ takes value on the whole real line to have connections with Fourier transform, which means we only talk about bilateral Laplace transform.
Here's my first question: What is wrong with the above calculation using bilateral Laplace transform? Or is there any hidden information I missed?
Secondly I also considered $F(s)=\frac{1}{s+z_0}+\frac{1}{s-z_0}$. In this case the inverse Laplace transform varies as I choose different imaginary axis $\mathrm{Re}(s)=\gamma$ in the Mellin formula $f(t)=\frac{1}{2\pi\mathrm{i}}\int_{\gamma-\mathrm{i}\infty}^{\gamma+\mathrm{i}\infty}F(s)e^{st}\mathrm{d}s$. Three possible answers are $f_1(t)=0,\ f_2(t)=e^{-z_0t} $ and $f_3(t)=e^{-z_0t}+e^{z_0t}$.
My Second question: Do $f_1$ and $f_2$ actually mean anything? Does there exist any interpretation of the choice of $\gamma$?