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 $$\Gamma(s)=\frac{(1)^n}{n!}\left(\frac 1{s+n}+\psi(n+1)+\frac 16(3\psi(n+1)^2+\pi^23\psi'(n+1))(s+n))+\frac16(\psi(n+1)^3+(\pi^23\psi'(n+1))\psi(n+1)+\psi''(n+1))(s+n)^2+O((s+n)^3)\right)$$ 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.

2$\begingroup$ A naiveimmediate reaction is that the triplication formula for $\Gamma(s)$ may play a role... But you probably already thought about that? $\endgroup$– paul garrettCommented Mar 9, 2017 at 23:17

$\begingroup$ @paulgarrett Yes, I've tried that but I couldn't get a slick answer. I was wondering if there is something deep going on... $\endgroup$– MickeyCommented Mar 10, 2017 at 1:45

2$\begingroup$ Is it so surprising? Won't the digamma function near negative integers simply look like a rational number + the singularity. And, the derivative should look like a rational number + zeta at 2 + singular part. $\endgroup$– LuciaCommented Mar 10, 2017 at 1:45

$\begingroup$ @Lucia I guess f(s) may have certain speciality that makes the answer so simple. However, g(s) looks very similar, yet its residues are complicated, which also involves Euler–Mascheroni constant and also higher derivatives of digamma function. $\endgroup$– MickeyCommented Mar 10, 2017 at 2:00

2$\begingroup$ Have you tried numerically others than triple, like $ f(s)=\frac{\Gamma^4(s)}{\Gamma(4s)(e^{2\pi is}1)}$? Wouldn't be surprised if the results are similar. $\endgroup$– WolfgangCommented Mar 12, 2017 at 17:12
1 Answer
$\def\Res{\operatorname*{Res}} \def\G{\Gamma} \def\e{\varepsilon} \def\p{\pi} \def\ZZ{\mathbb{Z}} \def\QQ{\mathbb{Q}} \def\NN{\mathbb{N}} \def\j{\psi} \def\z{\zeta} \def\To{\rightarrow} \def\f{\varphi} \def\g{\gamma} \def\a{\alpha} \def\b{\beta} \def\i{\mathrm{i}} \def\bell{\mathcal{B}} \def\bern{B} \newcommand\set[1]{\{#1\}}$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_nH_{3 n}) 27 (H_nH_{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+p1}} m\b(\e) \left(\frac{\p\e}{\sin\p\e}\right)^m \frac{\sin\p m\e}{\p m\e} \frac{\G(mn+1m\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+p1}} 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+1m\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+p2} \frac{1}{i!j!k!} \b^{(i)}(0) f^{(j)}(0) g^{(k)}(0). \tag{1} \end{align*} For the original problem $\b^{(i)}(0) = a_i (\i\p)^{i1}$, 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)^{i1}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} (m1) 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^{j1}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^{(i1)}(m n+1)m\j^{(i1)}(n+1). \end{align*} Note that \begin{align*} y_i &= \begin{cases} m(H_{m n}  H_{n}), & i=1 \\ (1)^i (i1)!\Big( (m^im)\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+p2}\atop{i+2j+k=m+p2}} \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+p2}\atop{i+2j+k=m+p2}} 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 EulerMascheroni 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 m1}\atop{i+2j+k=m1}} q'(i,2j,k) \i^{i1} \p^{2j+i1} \\ & \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\}$.)

3$\begingroup$ The Mathematica 11.3 command Residue[Gamma[s]^3/(Gamma[3*s]*(Exp[2*PiIs]  1)), {s, n}, Assumptions > n [Element] Integers && n > 0] performs $$\frac{i (1)^{2 n} (3 n)! \left(27 \psi ^{(0)}(n+1)^2+18 (3 \psi ^{(0)}(3 n+1)+i \pi ) \psi ^{(0)}(n+1)27 \psi ^{(0)}(3 n+1)^218 i \pi \psi ^{(0)}(3 n+1)+9 \psi ^{(1)}(n+1)27 \psi ^{(1)}(3 n+1)+8 \pi ^2\right)}{4 \pi (n!)^3}, $$ confiming your statement. $\endgroup$ Commented May 24, 2018 at 19:43

1$\begingroup$ Did you try starting from an arbitrary meromorphic function such that $f(s1)=\frac{u(s)}{v(s)} f(s)$ with $u,v\in \Bbb{Q}[x],u(0)v(0)\ne 0$ ? $\endgroup$– reunsCommented Dec 12, 2020 at 7:54

$\begingroup$ @reuns: That is an interesting idea that I did not consider. $\endgroup$ Commented Dec 12, 2020 at 18:11