In solving a particular physical problem I have had to perform inverse Laplace transforms of sum and products of Gamma functions. Since my actual problem is complicated, I will state a simple example. Suppose: $$F(s)=\Gamma(s+a)-\Gamma(s,b)\\ \Gamma(z)\equiv\int_0^\infty dt~ t^{z-1}e^{-t}\\ \Gamma(z,b)\equiv\int_b^\infty dt~ t^{z-1}e^{-t}$$ where $a,b>0$ are real.
The inverse Laplace transform is then: $$f(t)=\frac{1}{2\pi i}\int_Cds~e^{st}F(s)=\frac{1}{2\pi i}\int_Cds~e^{st}(\Gamma(s+a)-\Gamma(s,b))$$ where $C$ is a vertical line in the complex $s$-plane such that all the singularities of $F(s)$ lie on the left of it.
$\Gamma(s+a)$ has simple poles at $-(a+n),~n\in\{0,1,2,\ldots\}$ but no branch points. $\Gamma(s,b)$ has a branch point at $0$ but has no poles. So I take my branch cut to be the positive real axis including origin. The rationale for this choice of the branch cut is that I want to evaluate the integral by summing the residues inside the closed contour shown in the figure above in the limit $R\to\infty$. Following the logic in this post sum of residues of the integrand is simply: $\sum_{n=0}^\infty e^{-(a+n)t}(-1)^n/n!$. Therefore:
$$\int_C ds~e^{st}F(s)+\lim_{R\to\infty}\int_{\tilde{C}}ds~e^{st}F(s)=\sum_{n=0}^\infty e^{-(a+n)t}\frac{(-1)^n}{n!}$$
Assuming that everything I have done so far is correct then: How do I evaluate the second term on the LHS? Is it simply zero (how do I justify that)? Does the conclusion change if $F(s)=\Gamma(s+a)\Gamma(s,b)$?
P.S. I have read post1, post2, post3, but I still can't figure the answer to my question.
Using the relation between the inverse Laplace transform and the inverse Mellin transform: $$\mathcal L^{-1}[F](-\ln t) = \mathcal M^{-1}[F](t),$$ we get $$\mathcal M^{-1}[\Gamma(s + a) - \Gamma(s, b)] = e^{-t} (t^a - H(t - b)), \\ \mathcal L^{-1}[\Gamma(s + a) - \Gamma(s, b)] = e^{-e^{-t}} (e^{-a t} - H(e^{-t} - b)).$$ Since $\mathcal L^{-1}[F](t)$ takes non-zero values for negative $t$ as well, we need to take the two-sided Laplace transform to recover $F$ (with the region of convergence $\operatorname{Re} s > -a$).