Balanced polygamma function

The generalized polygamma function is defined as follows:

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

or alternatively,

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

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

${\displaystyle f(0)=f(1)\quad {\text{and}}\quad \int _{0}^{1}f(x)\,dx=0}$.

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

${\displaystyle {\begin{aligned}\psi (x)&=\psi (0,x)\\\psi ^{(n)}(x)&=\psi (n,x)\qquad n\in \mathbb {N} \\\Gamma (x)&=\exp \left(\psi (-1,x)+{\tfrac {1}{2}}\ln 2\pi \right)\\\zeta (z,q)&={\frac {\Gamma (1-z)}{\ln 2}}\left(2^{-z}\psi \left(z-1,{\frac {q+1}{2}}\right)+2^{-z}\psi \left(z-1,{\frac {q}{2}}\right)-\psi (z-1,q)\right)\\\zeta '(-1,x)&=\psi (-2,x)+{\frac {x^{2}}{2}}-{\frac {x}{2}}+{\frac {1}{12}}\\B_{n}(q)&=-{\frac {\Gamma (n+1)}{\ln 2}}\left(2^{n-1}\psi \left(-n,{\frac {q+1}{2}}\right)+2^{n-1}\psi \left(-n,{\frac {q}{2}}\right)-\psi (-n,q)\right)\end{aligned}}}$

where B_n(q) are the Bernoulli polynomials

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

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

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):

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