Calculate limit of $\Gamma$ function for special values. 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:
$$f(n) = \frac{(a - \gamma)(-1)^{n}}{b}$$
 A: 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 $n$:
$$
\left(
\begin{array}{cc}
 n & f(n)\\
 0 & -\gamma  \\
 1 & \gamma-1 \\
 2 & -\frac{1}{2} \left(\gamma-\frac{3}{2} \right) \\
 3 & \frac{1}{6} \left(\gamma -\frac{11}{6}\right) \\
 4 & -\frac{1}{24} \left(\gamma-\frac{25}{12} \right) \\
 5 & \frac{1}{120} \left(\gamma -\frac{137}{60}\right) \\
\end{array}
\right)
$$
A: 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 $\psi(n+1)$.
A: 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 $n=0, 2, \dots, 5$:
$$
\begin{array}{l}
 f(0) =  -\gamma  \\
f(1) = -1+\gamma  \\
f(2) = \frac{3}{4}-\frac{\gamma }{2} \\
f(3) = -\frac{11}{36}+\frac{\gamma }{6} \\
f(4) = \frac{25}{288}-\frac{\gamma }{24} \\
f(5) = -\frac{137}{7200}+\frac{\gamma }{120} \\
\end{array}
$$
