# 双伽玛函数

（重定向自双Γ函数

${\displaystyle \psi (x)={\frac {d}{dx}}\ln {\Gamma (x)}={\frac {\Gamma '(x)}{\Gamma (x)}}.}$

## 与调和数的关系

${\displaystyle \psi (n)=H_{n-1}-\gamma \!}$

${\displaystyle \psi \left(n+{\frac {1}{2}}\right)=-\gamma -2\ln 2+\sum _{k=1}^{n}{\frac {2}{2k-1}}}$

## 积分表示法

${\displaystyle \psi (x)=\int _{0}^{\infty }\left({\frac {e^{-t}}{t}}-{\frac {e^{-xt}}{1-e^{-t}}}\right)\,dt}$

${\displaystyle \psi (s+1)=-\gamma +\int _{0}^{1}{\frac {1-x^{s}}{1-x}}dx}$

## 泰勒级数

${\displaystyle \psi (z+1)=-\gamma -\sum _{k=1}^{\infty }\zeta (k+1)\;(-z)^{k}}$ ,

## 牛顿级数

${\displaystyle \psi (s+1)=-\gamma -\sum _{k=1}^{\infty }{\frac {(-1)^{k}}{k}}{s \choose k}}$

## 反射公式

${\displaystyle \psi (1-x)-\psi (x)=\pi \,\!\cot {\left(\pi x\right)}}$

## 递推关系

${\displaystyle \psi (x+1)=\psi (x)+{\frac {1}{x}}}$

## 高斯和

${\displaystyle {\frac {-1}{\pi k}}\sum _{n=1}^{k}\sin \left({\frac {2\pi nm}{k}}\right)\psi \left({\frac {n}{k}}\right)=\zeta \left(0,{\frac {m}{k}}\right)=-B_{1}\left({\frac {m}{k}}\right)={\frac {1}{2}}-{\frac {m}{k}}}$

${\displaystyle \sum _{n=1}^{k}\psi \left({\frac {n}{k}}\right)=-k(\gamma +\log k),}$

${\displaystyle \sum _{p=0}^{q-1}\psi \left(a+{\frac {p}{q}}\right)=q[\psi (qa)-\ln(q)],}$

## 高斯双伽玛定理

${\displaystyle \psi \left({\frac {m}{k}}\right)=-\gamma -\ln(2k)-{\frac {\pi }{2}}\cot \left({\frac {m\pi }{k}}\right)+2\sum _{n=1}^{\lfloor {\frac {k-1}{2}}\rfloor }\cos \left({\frac {2\pi nm}{k}}\right)\ln \sin \left({\frac {n\pi }{k}}\right)}$

## 特殊值

${\displaystyle \psi (1)=-\gamma \,\!}$
${\displaystyle \psi \left({\frac {1}{2}}\right)=-2\ln {2}-\gamma }$
${\displaystyle \psi \left({\frac {1}{3}}\right)=-{\frac {\pi }{2{\sqrt {3}}}}-{\frac {3}{2}}\ln {3}-\gamma }$
${\displaystyle \psi \left({\frac {1}{4}}\right)=-{\frac {\pi }{2}}-3\ln 2-\gamma }$
${\displaystyle \psi \left({\frac {1}{6}}\right)=-{\frac {\pi }{2}}{\sqrt {3}}-2\ln {2}-{\frac {3}{2}}\ln 3-\gamma }$
${\displaystyle \psi \left({\frac {1}{8}}\right)=-{\frac {\pi }{2}}-4\ln {2}-{\frac {\sqrt {2}}{2}}\left[\pi +\ln(3+2{\sqrt {2}})\right]-\gamma }$
${\displaystyle \psi \left({\frac {3}{4}}\right)={\frac {\pi }{2}}-3\ln {2}-\gamma }$