Balanced polygamma function

In mathematics, the generalized polygamma function or balanced negapolygamma function is a function introduced by Olivier Espinosa Aldunate and Victor Hugo Moll. 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:

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

or alternatively,

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

where ψ(z) is the polygamma function and ζ(z,q), is the Hurwitz zeta function.

The function is balanced, in that it satisfies the conditions

$f(0)=f(1)\text{and}\underset{0}{\overset{1}{\int }}f(x)dx=0$.

Relations

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

$\begin{array}{rl}\psi (x)& =\psi (0,x)\\ {\psi }^{(n)}(x)& =\psi (n,x)n\in \mathbb{\mathbb{N}}\\ \mathrm{\Gamma }(x)& =\exp (\psi (-1,x)+\frac{1}{2}\ln 2\pi )\\ \zeta (z,q)& =\frac{\mathrm{\Gamma }(1-z)}{\ln 2}(2^{-z}\psi (z-1,\frac{q+1}{2})+2^{-z}\psi (z-1,\frac{q}{2})-\psi (z-1,q))\\ {\zeta }^{\prime }(-1,x)& =\psi (-2,x)+\frac{x^2}{2}-\frac{x}{2}+\frac{1}{12}\\ B_n(q)& =-\frac{\mathrm{\Gamma }(n+1)}{\ln 2}(2^{n-1}\psi (-n,\frac{q+1}{2})+2^{n-1}\psi (-n,\frac{q}{2})-\psi (-n,q))\end{array}$

where B_n(q) are the Bernoulli polynomials

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

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

Special values

The balanced polygamma function can be expressed in a closed form at certain points (where A is the Glaisher constant and G is the Catalan constant):

$\begin{array}{rlrl}\psi (-2,\frac{1}{4})& =\frac{1}{8}\ln 2\pi +\frac{9}{8}\ln A+\frac{G}{4\pi }& & \\ \psi (-2,\frac{1}{2})& =\frac{1}{4}\ln \pi +\frac{3}{2}\ln A+\frac{5}{24}\ln 2& \\ \psi (-3,\frac{1}{2})& =\frac{1}{16}\ln 2\pi +\frac{1}{2}\ln A+\frac{7\zeta (3)}{32{\pi }^2}\\ \psi (-2,1)& =\frac{1}{2}\ln 2\pi & \\ \psi (-3,1)& =\frac{1}{4}\ln 2\pi +\ln A\\ \psi (-2,2)& =\ln 2\pi -1& \\ \psi (-3,2)& =\ln 2\pi +2\ln A-\frac{3}{4}\\ \end{array}$