# Explicit forms of negapolygamma difference with arguments that differ by a half?

There are well known explicit formulas for negapolygamma expressions of the form $$\psi^{(-n)}(x)-\psi^{(-n)}(x-1)$$ for $n\in\mathbb{N}\gt1$

for example $$\psi^{(-2)}\left(x\right)-\psi^{(-2)}\left(x-1\right)=1-x+\left(x-1\right)\ln \left(x-1\right)+\frac{1}{2}\ln \left(2\pi \right)$$

Are there explicit expressions of the form $$\psi^{(-n)}(x)-\psi^{(-n)}(x-\frac12)$$

The negapolygamma function is related to the first derivative of the Hurwitz Zeta function so this is the same problem as finding an explicit form for $$\zeta'(-n+1,x)-\zeta'(-n+1,x\pm\frac12)$$

Where $\zeta'(s,x)=\frac{d}{dt}\zeta(t,x)|_{t=s}$

• So you are defining the polygamma function at negative values through the analytic extension of the derivative of the Hurwitz zeta function? – Simply Beautiful Art Aug 26 '17 at 13:26

For $\Re(s)>1$,

$$\zeta(s,q)=\sum_{n=0}^\infty\frac1{(n+q)^s}=\sum_{n=0}^\infty\frac{2^s}{(2n+2q)^s}$$

$$\zeta(s,q+0.5)=\sum_{n=0}^\infty\frac1{(n+q+0.5)^s}=\sum_{n=0}^\infty\frac{2^s}{(2n+2q+1)^s}$$

$$\zeta(s,q)+\zeta(s,q+0.5)=\sum_{n=0}^\infty\frac{2^s}{(n+2q)^s}=2^s\zeta(s,2q)$$

Differentiate both sides w.r.t. $s$ to get

$$\zeta^{(1,0)}(s,q)+\zeta^{(1,0)}(s,q+0.5)=2^s(\ln(2)\zeta(s,2q)+\zeta^{(1,0)}(s,2q))$$

which holds for all $s\ne1$.

$$\zeta(s,q)-\zeta(s,q+0.5)=\sum_{n=0}^\infty\frac{2^s(-1)^n}{(n+2q)^s}=2^s\Phi(-1,s,2q)$$
$$\zeta^{(1,0)}(s,q)-\zeta^{(1,0)}(s,q+0.5)=2^s(\ln(2)\Phi(-1,s,2q)+\Phi^{(0,1,0)}(-1,s,2q))$$