Evaluate $\lim_{s \rightarrow 1^{-}} \frac{\Gamma(b-as)}{\Gamma(s)\Gamma(1-s)}$, for $a,b>0$

Question

I encountered the following as a result of a Mellin-transform:

$$ \frac{\Gamma(b-as)}{\Gamma(s)\Gamma(1-s)},\quad b>0,\ 0 < a,\text{Re}(s) < 1 $$ where $\Gamma(\cdot)$ is the gamma function.

Because of the restriction that the real part of $s$ should be within $0$ and $1$, I am curious as to whether the following limit exist and if it exists, what is its value in terms of $a$ and $b$.

$$ \lim_{s\rightarrow 1^{-}} \frac{\Gamma(b-as)}{\Gamma(s)\Gamma(1-s)} $$

Direct substitution might lead to something like: $$ \lim_{s\rightarrow 1^{-}} \frac{\Gamma(b-as)}{\Gamma(s)\Gamma(1-s)} = \Gamma(b-a) $$ knowing $\Gamma(0) = \Gamma(1) = 1$. Still I am not sure if I'm doing this right.

EDIT: Additional context.

From Lemma 1.11 (page 44) of Kibas etal's Theory and Applications of Fractional Differential Equations it states:

For $\alpha > 0$ and $z\in\mathbb{C}\ (|\arg(-z)| < \pi)$ the following relation holds: $$ E_{a,b}(z) = \frac{1}{2\pi i} \int_{\gamma - i\infty}^{\gamma + i\infty} \frac{\Gamma(s)\Gamma(1-s)}{\Gamma(b-as)}(-z)^{-s}\,\text{d}s$$ where the path of integration separates all the poles at $s = -k$ ($k \in \mathbb{N}\cup\{0\}$) to the left and all the poles at $s = n + 1$ ($n \in \mathbb{N}\cup\{0\}$) to the right.

The author then followed-up with a remark that the Mellin-transform of the function $E_{a,b}(-z)$is given by $$ \mathcal{M}\left[E_{a,b}(-z)\right](s) = \frac{\Gamma(s)\Gamma(1-s)}{\Gamma(b-as)}\quad (0 < \text{Re}(s) < 1) $$ immediately follows from this lemma. The function $E_{a,b}(z)$ above is the well-known Mittag-Leffler type function defined as $$ E_{a,b}(z) = \sum_{k=0}^{\infty} \frac{z^{n}}{\Gamma(an + b)}\quad (\text{Re}(a) > 0, b\in\mathbb{C}) $$

Thus, follows my question of curiosity.

Answer 1

Near $s = 1$, $\Gamma(s)$ is bounded and $\Gamma(1 - s) = \frac{1}{1 - s} + O(1) ,$ so, if $\Gamma(b - a s)$ does not have a pole at $s = 1$, (equivalently, $\Gamma$ does not have a pole at $b - a$) we have $$\lim_{s \to 1} \frac{\Gamma(b - a s)}{\Gamma(s) \Gamma(1 - s)} = 0 .$$ Now, the poles of $\Gamma$ are all simple and occur exactly at the nonpositive integers $-n$, and their respective residues are $\frac{(-1)^n}{n!}$, so if $a - b \in \Bbb Z_{\geq 0}$, the limit is $$\lim_{s \to 1} \frac{\Gamma(b - a s)}{\Gamma(s) \Gamma(1 - s)} = \frac{(-1)^{a - b}}{a \cdot (a - b)!} ,$$ and for other values of $a, b$, the limit is zero.

Answer 2

$$\frac{\Gamma(b - a s)}{\Gamma(s)\, \Gamma(1 - s)}= \Gamma (b-a)(1-s)\left(1+a \psi ^{(0)}(b-a)(1-s)+O\left((s-1)^2\right) \right)$$

For approximations, the $[n+1,n]$ Padé approximants $P_n$ are quite good. Form example, using $t=(1-s)$ $$P_1=\Gamma (b-a)\, t \,\,\frac {1+\frac{1}{6} \left(3 a \psi ^{(0)}(b-a)+\frac{\frac{\pi ^2}{a}-3 a \psi ^{(1)}(b-a)}{\psi ^{(0)}(b-a)}\right)t }{1+\frac{\pi ^2-3 a^2 \left(\psi ^{(0)}(b-a)^2+\psi ^{(1)}(b-a)\right)}{6 a \psi ^{(0)}(b-a)}t }$$ is $O(t^5)$.