Recurrence relation for the polygamma function of negative order?

I know the recurrence relation for the Polygamma function is

$$\psi^{(m)}(x+1)=\psi^{(m)}(x)+\frac{(-1)^mm!}{x^{m+1}}$$

Does such a recurrence formula exist for negative integer $m$?

I am using the integral definition

$$\psi^{(-n)}(x)=\frac{1}{(x-2)!}\int_0^x (x-t)^{n-2}\ln(\Gamma(t))dt$$ for $n$ a positive integer, which I assume is equal to the $(n-1)$th integral of $\ln{\Gamma(x)}$.

• Well, what definition of "negapolygamma" are you using? – J. M. is a poor mathematician Aug 17 '17 at 22:59
• Note: my answer has been updated to provide a more explicit recursive relation of the polygamma function on negative orders, using the provided definition. – Simply Beautiful Art Aug 14 '18 at 1:57

Using your definition, we hence have

\begin{align}\psi^{(-n)}(x+1)&=\psi^{(-n)}(x)+\frac1{(n-2)!}\int_0^x(x-t)^{n-2}\ln(t)~dt+\sum_{k=0}^{n-2}\frac{\psi^{(k-n)}(1)}{k!}x^k\\&=\psi^{(-n)}(x)+\frac{x^{n-1}[\ln(x)-H_{n-1}]}{(n-1)!}+\sum_{k=0}^{n-2}\frac{\psi^{(k-n)}(1)}{k!}x^k\end{align}

where $H_n=\sum_{k=}^n\frac1k$ is the harmonic number.