Calculate limit of $\Gamma$ function for special values.

Question

I would like to calculate the limit without any software but have no idea how to do it.

$$f(n) = \lim_{c \rightarrow 0} \frac{\Gamma(-n + c) + \Gamma(-n - c))}{2}$$ $$n = 0, 1, 2, ...$$

Wolfram in some way claculates it, for example: $$f(0) = - \gamma$$ $$f(1) = \gamma - 1$$ $$f(2) = \frac{3 - 2 \gamma}{4}$$ $$f(3) = \frac{6 \gamma - 11}{36}$$ $$(...)$$

It seems that the solutions will be somethink like that: $$f(n) = \frac{(a - \gamma)(-1)^{n}}{b}$$

Answer 1

Using series expansions for the gamma function we have $$ \begin{aligned} f(n) &=\frac{1}{2}\lim_{\epsilon\to 0}(\Gamma(-n+\epsilon)+\Gamma(-n-\epsilon))\\ &=\frac{(-1)^n}{2 n!}\lim_{\epsilon\to 0}(\tfrac{1}{\epsilon}(1+\psi(n+1)\epsilon)-\tfrac{1}{\epsilon}(1-\psi(n+1)\epsilon)+\mathcal O(\epsilon))\\ &=\frac{(-1)^n}{2 n!}\lim_{\epsilon\to 0}(\tfrac{1}{\epsilon}+\psi(n+1)-\tfrac{1}{\epsilon}+\psi(n+1)+\mathcal O(\epsilon))\\ &=\frac{(-1)^n}{n!}\lim_{\epsilon\to 0}(\psi(n+1)+\mathcal O(\epsilon))\\ &=\frac{(-1)^n}{n!}\psi(n+1), \end{aligned} $$ where $\psi(z)$ is the Digamma function. Then using specialized values for $\psi(n+1)$ we find $$ \begin{aligned} f(n) &=\frac{(-1)^n}{n!}(H_n-\gamma), \end{aligned} $$ where $H_n$ is the harmonic number and $\gamma=0.57721\dots$ is the Euler-Mascheroni constant.

We have for the first several values of $n$: $$ \left( \begin{array}{cc} n & f(n)\\ 0 & -\gamma \\ 1 & \gamma-1 \\ 2 & -\frac{1}{2} \left(\gamma-\frac{3}{2} \right) \\ 3 & \frac{1}{6} \left(\gamma -\frac{11}{6}\right) \\ 4 & -\frac{1}{24} \left(\gamma-\frac{25}{12} \right) \\ 5 & \frac{1}{120} \left(\gamma -\frac{137}{60}\right) \\ \end{array} \right) $$

Answer 2

Using results from gamma function with negative argument

$$\Gamma (-n + c) = \frac{ a_n}{c} + b_n + O(c)$$

with $$b_n~=~-\dfrac{(-1)^n}{n!}\cdot\gamma~-~\dfrac{S_1(n+1,~2)}{n!^2}$$

thus your limit is $$ f(n)=\lim_{c\rightarrow 0}\frac{\Gamma (-n + c) + \Gamma (-n - c)}{2}= b_n $$

For $n=0, 2, \dots, 5$:

$$ \begin{array}{l} f(0) = -\gamma \\ f(1) = -1+\gamma \\ f(2) = \frac{3}{4}-\frac{\gamma }{2} \\ f(3) = -\frac{11}{36}+\frac{\gamma }{6} \\ f(4) = \frac{25}{288}-\frac{\gamma }{24} \\ f(5) = -\frac{137}{7200}+\frac{\gamma }{120} \\ \end{array} $$

Answer 3

Here's another way to do it. Using the reflection formula for the gamma function write $$ \begin{aligned} f(n) &=\frac{1}{2}\lim_{\epsilon\to 0}(\Gamma(-n+\epsilon)+\Gamma(-n-\epsilon))\\ &=\frac{\pi}{2}\lim_{\epsilon\to 0}\left(\frac{1}{\Gamma(1+n-\epsilon)\sin\pi(-n+\epsilon)}+\frac{1}{\Gamma(1+n+\epsilon)\sin\pi(-n-\epsilon)}\right)\\ &=\frac{(-1)^n\pi}{2}\lim_{\epsilon\to 0}\left(\frac{1}{\Gamma(1+n-\epsilon)\sin\pi\epsilon}-\frac{1}{\Gamma(1+n+\epsilon)\sin\pi\epsilon}\right)\\ &=\frac{(-1)^n\pi}{2\Gamma(1+n)^2}\lim_{\epsilon\to 0}\left(\frac{\Gamma(1+n+\epsilon)-\Gamma(1+n-\epsilon)}{\underbrace{\sin\pi\epsilon}_{\sim\pi\epsilon}}\right)\\ &=\frac{(-1)^n}{2(n!)^2}\lim_{\epsilon\to 0}\left(\frac{\Gamma(1+n+\epsilon)-\Gamma(1+n-\epsilon)}{\epsilon}\right),\quad^\ast\text{L'Hôpital}\\ &=\frac{(-1)^n}{2(n!)^2}\lim_{\epsilon\to 0}\left(\psi(1+n+\epsilon)\Gamma(1+n+\epsilon)+\psi(1+n-\epsilon)\Gamma(1+n-\epsilon)\right)\\ &=\frac{(-1)^n}{n!}\psi(n+1). \end{aligned} $$ Then again use the specialized values for $\psi(n+1)$.