式展開: ディガンマ凾数の相反公式

\begin{array}{rcl} ψ (z) & = & \frac{d}{d z} \log (Γ (z)) \\ = & \frac{Γ^{'} (z)}{Γ (z)} \end{array}

\begin{array}{rcl} ψ (1 - z) & = & \frac{d}{d z} \log (Γ (1 - z)) \\ = & \frac{d}{d u} \log (Γ (u)) \frac{d u}{d z} \dots u = 1 - z, \frac{d u}{d z} = - 1 \\ = & - \frac{d}{d z} \log (Γ (1 - z)) \\ = & - \frac{d}{d z} \log (\frac{Γ (z)}{Γ (z)} Γ (1 - z)) \\ = & - \frac{d}{d z} \log (\frac{1}{Γ (z)} \frac{π}{\sin (π z)}) \dots Γ (z) Γ (1 - z) = \frac{π}{\sin (π z)} \\ = & - {\frac{d}{d z} \log (π) - \frac{d}{d z} \log (\sin (π z)) - \frac{d}{d z} \log (Γ (z))} \\ = & - {0 - π \cot (π z) - ψ (z)} \\ \dots \frac{d}{d z} \log \sin (π z) = π \cot (π z) \\ \dots \frac{d}{d z} \log Γ (z) = ψ (z), (デ ィ ガ ン マ 凾 数 の 定 義) \\ = & π \cot (π z) + ψ (z) \end{array}

\begin{array}{rcl} ψ (1 - z) & = & π \cot (π z) + ψ (z) \\ ψ (1 - z) - ψ (z) & = & π \cot (π z) \end{array}

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