TITLE: "unexpected" residue formula for $\Gamma^3(s)/(\Gamma(3s)(e^{2\pi is}-1)) $
QUESTION [10 upvotes]: There is a related problem in my current work: to find the residue of the following function at any negative integer $s=-n$:
$$f(s)=\frac{\Gamma^3(s)}{\Gamma(3s)(e^{2\pi is}-1)}$$
It seems to be a tedious calculation. As far as I know, the first few terms of Laurent series of $\Gamma(s)$ around $s=-n$ are given by
$$\Gamma(s)=\frac{(-1)^n}{n!}\left(\frac 1{s+n}+\psi(n+1)+\frac 16(3\psi(n+1)^2+\pi^2-3\psi'(n+1))(s+n))+\frac16(\psi(n+1)^3+(\pi^2-3\psi'(n+1))\psi(n+1)+\psi''(n+1))(s+n)^2+O((s+n)^3)\right)$$
where $\psi(s):=\frac{\Gamma'(s)}{\Gamma(s)}$ is the digamma function.
So we might expect that the residue formula of $f(s)$ at $s=-n$ involves the values of digamma function and/or its derivatives. However, numerical values indicate that the formula for $Res (f,-n)$ might only involve $\pi$, $i$, and rational numbers, which is quite unexpected. Here are a few examples:
$$Res (f,-3)=7527 + \frac{4299 i}{4 \pi} + 2100i\pi$$
$$Res (f,-10)=\frac {1065144125784453}{40} - \frac {17136782690536253 i}{11200 \pi} + 6938745989175 i \pi$$
I have verified some other integers as well. Can we find any reason for this pattern or any counterexample? It also seems that $f(s)$ is very special, for instance, if we multiply another $\Gamma(s)$ to the numerator, i.e. $g(s)=\frac{\Gamma^4(s)}{\Gamma(3s)(e^{2\pi is}-1)}$, then the formula for $Res(g,-n)$ does not have such property.

REPLY [9 votes]: $\def\Res{\operatorname*{Res}}
\def\G{\Gamma}
\def\e{\varepsilon}
\def\p{\pi} 
\def\ZZ{\mathbb{Z}}
\def\QQ{\mathbb{Q}}
\def\NN{\mathbb{N}}
\def\j{\psi}
\def\z{\zeta}
\def\To{\rightarrow}
\def\f{\varphi}
\def\g{\gamma} 
\def\a{\alpha}
\def\b{\beta}
\def\i{\mathrm{i}}
\def\bell{\mathcal{B}}
\def\bern{B}
\newcommand\set[1]{\{#1\}}$After a straightforward but tedious calculation one finds that the residue is given by
\begin{align*}
\frac{(3 n)! i}{4 \pi (n!)^3}
    &\Big(8 \pi ^2 +
    18\pi i (H_n-H_{3 n}) -27 (H_n-H_{3 n})^2 \\
    &\quad +9 \big(\psi^{(1)}(n+1)-3 \psi ^{(1)}(3 n+1)\big)\Big)
\end{align*}
Now note that
$H_n, H_{3n}\in\mathbb{Q}$
and
$\psi^{(1)}(m+1) = \pi^2/6 - \underbrace{\sum_{k=1}^m 1/k^2}_{\in\mathbb{Q}}$ for $m\in\mathbb{Z}^+$.
Thus,
$$\Res_{s=-n}f(s) = a + bi/\pi + ci\pi,\quad
\textrm{where }a,b,c\in\mathbb{Q},$$
as claimed.
Addendum
Below we consider a wider class of residues of the form
\begin{align*}
\Res_{s=-n} 
\frac{\G(s)^m}{\G(m s)}\a(s),
\end{align*}
where $m\in\NN$ and $\a$ has an isolated pole of order $p$ at $s=-n$.
That is, we examine
\begin{align*}
\Res_{\e=0} 
\frac{\G(-n+\e)^m}{\G(-m n+m\e)} \frac{\b(\e)}{\e^p},
\end{align*}
where $\b$ is analytic at $\e=0$.
We find the general form of such residues.
This is an extension of the types of residues suggested by @Wolfgang in the comments.
In the posted question, $m=3$, $p=1$, and $\b(\e) = \e/(e^{2\p \i\e}-1)$.
A routine calculation results in
\begin{align*}
\frac{\G(-n+\e)^m}{\G(-m n+m\e)}\frac{\b(\e)}{\e^p}
    &= 
    \frac{1}{\e^{m+p-1}}
    m\b(\e) 
    \left(\frac{\p\e}{\sin\p\e}\right)^m
    \frac{\sin\p m\e}{\p m\e} 
    \frac{\G(mn+1-m\e)}{\G(n+1-\e)^m} 
\end{align*}
where $s = -n+\e$.
Thus,
\begin{align*}
\Res_{s=-n}
\frac{\G(s)^m}{\G(m s)}\a(s) 
    &= 
    \Res_{\e=0}
    \frac{1}{\e^{m+p-1}} m\b(\e)f(\e)g(\e),
\end{align*}
where
\begin{align*}
f(\e) &= 
    \left(\frac{\p\e}{\sin\p\e}\right)^m
    \frac{\sin\p m\e}{\p m\e} \\
g(\e) &= \frac{\G(m n+1-m\e)}{\G(n+1-\e)^m}.
\end{align*}
We find
\begin{align*}
\Res_{s=-n} 
\frac{\G(s)^m}{\G(m s)}\a(s) 
&= 
    m
    \sum_{i+j+k=m+p-2}
    \frac{1}{i!j!k!}
    \b^{(i)}(0)
    f^{(j)}(0)
    g^{(k)}(0). 
        \tag{1}
\end{align*}
For the original problem $\b^{(i)}(0) = a_i (\i\p)^{i-1}$, where $a_i\in\QQ$.
In fact, since $\b(\e)$ in this case is essentially the generating function for the Bernoulli numbers $B_i^-$ we find
\begin{align*}
\b^{(i)}(0) = (2\p\i)^{i-1}B_i^-.
    \tag{2}
\end{align*}
The first few derivatives of $f$ are given as
\begin{align*}
f(0) &\equiv\lim_{\e\To0}f(\e) = 1 \\ 
f'(0) &= 0 \\
f''(0) &= -\frac{1}{3} (m-1) m \pi^2 \\
f'''(0) &= 0.
\end{align*}
Note that $f^{(2i+1)}(0) = 0$ generally.
One can show that, as suggested above,
\begin{align*}
f^{(2i)}(0) &= b_{2i}\p^{2i},
\end{align*}
where $b_{2i}\in\QQ$.
In fact, an explicit form may be found for these coefficients,
\begin{align*}
f^{(2i)}(0) &= (-1)^i \p^{2i} \bell_{2i}(x_1,\ldots,x_{2i}),
    \tag{3}
\end{align*}
where $\bell_i(\cdots)$ is the $i$th complete Bell polynomial,
\begin{align*}
x_j &= \begin{cases}
\frac{m}{j} 2^j (m^{j-1}-1) B_j, & \textrm{$j$ even} \\
0, & \textrm{else},
\end{cases} \tag{4}
\end{align*}
and where $\bern_j$ is the $j$th Bernoulli number.
(Since $j$ is even, there is no distinction between $\bern^-_j$ and $\bern^+_j$.)
The first few complete Bell polynomials are
\begin{align*}
\bell_0 &= 1 \\
\bell_1(t_1) &= t_1 \\
\bell_2(t_1,t_2) &= t_1^2+t_2 \\
\bell_3(t_1,t_2,t_3) &= t_1^3 + 3t_1 t_2+t_3.
\end{align*}
Note that
$$g(0) = \frac{(m n)!}{n!^m} .$$
The first few derivatives of $g$ are given as
\begin{align*}
\frac{g'(0)}{g(0)} &= -m(H_{m n} - H_n)   \\
\frac{g''(0)}{g(0)} &= 
    m^2 (H_{m n} - H_n)^2 
    + m^2 \j^{(1)}(m n+1)- m\j^{(1)}(n+1) \\
\frac{g'''(0)}{g(0)} &= - \Bigg(m^3 (H_{m n} - H_n)^3 \\
    & \hspace{7ex} + 3 m (H_{m n} - H_n) 
    \left(m^2 \j^{(1)}(m n+1)-m\j^{(1)}(n+1)\right) \\
    & \hspace{7ex} + m^3 \j^{(2)}(m n+1) - m\j^{(2)}(n+1)\Bigg), 
\end{align*}
where
$H_n$ is the $n$th harmonic number
and where
$\j^{(i)}$ is the polygamma function.
Generally we find
\begin{align*}
\frac{g^{(j)}(0)}{g(0)} 
    &= (-1)^j \bell(y_1,\ldots,y_j), 
    \tag{5}
\end{align*}
where
\begin{align*}
y_i &= m^{i}\j^{(i-1)}(m n+1)-m\j^{(i-1)}(n+1).
\end{align*}
Note that
\begin{align*}
y_i &= \begin{cases}
m(H_{m n} - H_{n}), & i=1 \\
(-1)^i (i-1)!\Big(
(m^i-m)\z(i) - m^i H^{(i)}_{m n} + m H^{(i)}_n
\Big) & \textrm{else.}
\end{cases} 
    \tag{6} 
\end{align*}
Thus, residues involving $y_3$ or higher will contain terms involving $\z(3)$.
We find
\begin{align*}
\Res_{s=-n} 
\frac{\G(s)^m}{\G(m s)}\a(s) 
&= 
    m
    \frac{(m n)!}{n!^m}
    \sum_{{0\le i,2j,k\le m+p-2}\atop{i+2j+k=m+p-2}}
    \frac{1}{i!(2j)!k!}
    (-1)^{j+k}\p^{2j} 
    \b^{(i)}(0) \\
    &\hspace{25ex}\times    
    \bell_{2j}(x_1,\ldots,x_{2j})
    \bell_k(y_1,\ldots,y_k) \\
&= \sum_{{0\le i,2j,k\le m+p-2}\atop{i+2j+k=m+p-2}}
    q(i,2j,k)
    \b^{(i)}(0) 
    \p^{2j} \\
    & \hspace{16ex}\times 
    \bell_k\Big(q_1,q_2\z(2)+q'_2,
    \ldots,q_k\z(k)+q_k'\Big),
    \tag{7}
\end{align*}
where $q(i,2j,k),q_i,q_i'\in\QQ$ are explicitly calculable.
Thus, the residues involve sums of rational multiples of $\b^{(i)}(0)$ and products of powers of $\z(2),\z(3),\ldots$.
The residues do not involve the Euler-Mascheroni constant unless $\b^{(i)}(0)$ does.
For $\a$ in the question statement we find
\begin{align*}
\Res_{s=-n} 
\frac{\G(s)^m}{\G(m s)(e^{2\p\i\e}-1)}
&= \sum_{{0\le i,2j,k\le m-1}\atop{i+2j+k=m-1}}
    q'(i,2j,k)
    \i^{i-1}
    \p^{2j+i-1} \\
    & \hspace{14ex}\times
    \bell_k\Big(q_1,q_2\z(2)+q'_2,
    \ldots,q_k\z(k)+q_k'\Big),
    \tag{8}
\end{align*}
where $q'(i,2j,k)\in\QQ$.
For $m=3$, one can quickly verify that this sum results in a residue of the form $a + b\i/\p + c\i\p$, where $a,b,c\in\QQ$.
(Schematically, the sum will be of the form
$ \{2,0,0\}+\{0,2,0\}+\{1,0,1\}+\{0,0,2\}$.)