Find limit: $\lim_{s\to n}\frac{\Gamma'(1-s)}{\Gamma^2(1-s)}$

Question

When I am playing with Dirichlet beta function, I encountered the limit $$\lim_{s\to n}\frac{\Gamma'(1-s)}{\Gamma^2(1-s)}=(-1)^{n}(n-1)!$$ where $n$ is a positive integer number. Do I need the Laurent series of the Gamma function to compute this limit? Can someone who took special function course help me?

Thanks in advance.

Answer 1

I will prove the correct assertion that... $$ \lim_{s \to n} \frac{\Gamma'(1-s)}{\Gamma^{2}(1-s)} = (-1)^n (n-1)! \ , \ \forall n\in \mathbb{Z}^{+} $$

Start by using the following formula for the Digamma function. $$ \psi(x) = \frac{\Gamma'(x)}{\Gamma(x)} $$ This allows us to write the limit as follows. $$ \lim_{s \to n} \frac{\psi(1-s)}{\Gamma(1-s)} $$ We can then use the reflection formual for both the Gamma function and Digamma function... $$ \Gamma(1-s)=\frac{\pi}{\Gamma(s)\sin(\pi s)} $$ $$ \psi(1-s) = \pi \cot(\pi s) + \psi(s) $$ (Note that these are only valid for non-integer s as they diverge otherwise) we can then use these in our limit. $$ \lim_{s \to n} \Gamma(s)\cos(\pi s) + \frac{\sin(\pi s)}{\pi}\Gamma'(s) $$ Where we have again used the formula for the Digamma function. This expression contains no divergences over the positive integers, and hence we are free to evaluate it. Using the following facts... $$ \sin(\pi n) = 0 \ , \ \forall n \in \mathbb{Z}^{+} $$ $$ \cos(\pi n) = (-1)^n \ , \ \forall n \in \mathbb{Z}^{+} $$ $$ \Gamma(n)=(n-1)! \ , \ \forall n \in \mathbb{Z}^{+} $$ We obtain that.... $$ \lim_{s \to n} \Gamma(s)\cos(\pi s) + \frac{\sin(\pi s)}{\pi}\Gamma'(s) = (-1)^n (n-1)! $$ QED

I hope this helps.

Answer 2

Here is an elementary solution: For each given positive integer $n$, repeatedly applying the functional identity $\Gamma(s+1) = s \Gamma(s)$ gives gives

$$ \frac{1}{\Gamma(1-s)} = \frac{(n-s)(n-1-s) \cdots (1-s)}{\Gamma(n+1-s)} = (n-s)f(s), $$

where $f(s)$ is defined by

$$ f(s) = \frac{(n-1-s) \cdots (1-s)}{\Gamma(n+1-s)}. $$

Note that $f(s)$ is analytic near $s = n$. Then

\begin{align*} \frac{\Gamma'(1-s)}{\Gamma(1-s)^2} &= \frac{\mathrm{d}}{\mathrm{d}s} \frac{1}{\Gamma(1-s)} = \frac{\mathrm{d}}{\mathrm{d}s} (n-s) f(s) \\ &= -f(s) + (n-s)f'(s). \end{align*}

Hence, taking $s \to n$ gives

$$ \lim_{s \to n} \frac{\Gamma'(1-s)}{\Gamma(1-s)^2} = -f(n) = - \frac{(-1)(-2)\cdots(1-n)}{\Gamma(1)} = (-1)^n (n-1)! $$