# Balanced polygamma function

(Redirected from Generalized polygamma function)

In mathematics, the generalized polygamma function or balanced negapolygamma function is a function introduced by Olivier Espinosa Aldunate and Victor H. Moll.[1]

It generalizes the polygamma function to negative and fractional order, but remains equal to it for integer positive orders.

## Definition

The generalized polygamma function is defined as follows:

$\psi(z,q)=\frac{\zeta'(z+1,q)+(\psi(-z)+\gamma ) \zeta (z+1,q)}{\Gamma (-z)} \,$

or alternatively,

$\psi(z,q)=e^{- \gamma z}\frac{\partial}{\partial z}\left(e^{\gamma z}\frac{\zeta(z+1,q)}{\Gamma(-z)}\right),$

where $\psi(z)$ is the Polygamma function and $\zeta(z,q),$ is the Hurwitz zeta function.

The function is balanced, in that it satisfies the conditions $f(0)=f(1)$ and $\int_0^1 f(x) dx = 0$.

## Relations

Several special functions can be expressed in terms of generalized polygamma function.

• $\psi(x)=\psi(0,x)\,$
• $\psi^{(n)}(x)=\psi(n,x)\,\,\,(n\in\mathbb{N})$
• $\Gamma(x)=e^{\psi(-1,x)+\frac 12 \ln(2\pi)}\,\,\,$
• $\zeta(z,q)=\frac{\Gamma (1-z) \left(2^{-z} \left(\psi \left(z-1,\frac{q}{2}+\frac{1}{2}\right)+\psi \left(z-1,\frac{q}{2}\right)\right)-\psi(z-1,q)\right)}{\ln(2)}$
• $\zeta'(-1,x)=\psi(-2, x) + \frac{x^2}2 - \frac{x}2 + \frac1{12}$
• $B_n(q) = -\frac{\Gamma (n+1) \left(2^{n-1} \left(\psi\left(-n,\frac{q}{2}+\frac{1}{2}\right)+\psi\left(-n,\frac{q}{2}\right)\right)-\psi(-n,q)\right)}{\ln (2)}$

where $B_n(q)$ are Bernoulli polynomials

• $K(z)=A e^{\psi(-2,z)+\frac{z^2-z}{2}}$

where K(z) is K-function and A is the Glaisher constant.

## Special values

The balanced polygamma function can be expressed in a closed form at certain points:

• $\psi\left(-2,\frac14\right)=\frac18\ln(2\pi)+\frac98\ln A+\frac{G}{4\pi},$ where $A$ is the Glaisher constant and $G$ is the Catalan constant.
• $\psi\left(-2, \frac12\right)=\frac14\ln\pi+\frac32\ln A+\frac5{24}\ln2$
• $\psi(-2,1)=\frac12\ln(2\pi)$
• $\psi(-2,2)=\ln(2\pi)-1$
• $\psi\left(-3,\frac12\right)=\frac1{16}\ln(2\pi)+\frac12\ln A+\frac{7\,\zeta(3)}{32\,\pi^2}$
• $\psi(-3,1)=\frac14\ln(2\pi)+\ln A$
• $\psi(-3,2)=\ln(2\pi)+2\ln A-\frac34$