# Inverse Laplace transform of Gamma with branch cut

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.

• Some additional remarks. $\Gamma(s, b)$ is an entire function of $s$. If a branch cut existed, it would have to be placed outside of the region of convergence, otherwise the integral would depend on the choice of the contour. The integral over the left semicircle is negligible when $e^{−t} < b$ but not when $e^{−t} > b$. Moving the contour to the left would give a function whose two-sided Laplace transform is also $\Gamma(s + a) - \Gamma(s, b)$ but with a different region of convergence. – Maxim Apr 17 at 13:37

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 $$\operatorname{Re} s > -a$$).