- Prove that
\[ \frac {d ^ {2} \log \Gamma (s)}{d s ^ {2}} = \sum_ {n = 0} ^ {\infty} \frac {1}{(s + n) ^ {2}} \]
whenever s is a positive number. Show that if the left-hand side is interpreted as \(( \Gamma ^ { \prime } / \Gamma ) ^ { \prime }\) , then the above formula also holds for all complex numbers s with \(s \neq 0 , - 1 , - 2 , . . .\)
Solution
Start from the Weierstrass product \[ \frac1{\Gamma(s)}=s e^{\gamma s}\prod_{n=1}^{\infty}\left(1+\frac{s}{n}\right)e^{-s/n}. \] On compact subsets avoiding \(0,-1,-2,\ldots\), logarithmic differentiation is justified by locally uniform convergence and gives \[ -\frac{\Gamma'(s)}{\Gamma(s)} =\frac1s+\gamma+\sum_{n=1}^{\infty}\left(\frac1{s+n}-\frac1n\right). \] Differentiating once more, again locally uniformly, \[ -\left(\frac{\Gamma'}{\Gamma}\right)'(s) =-\frac1{s^2}-\sum_{n=1}^{\infty}\frac1{(s+n)^2}. \] Hence \[ \left(\frac{\Gamma'}{\Gamma}\right)'(s) =\sum_{n=0}^{\infty}\frac1{(s+n)^2}. \] For positive real \(s\), \((\Gamma'/\Gamma)'=d^2(\log\Gamma)/ds^2\), proving the stated formula. The same derivation holds for every complex \(s\notin\{0,-1,-2,\ldots\}\), and the series is locally uniformly convergent there.