# Polygamma expression for $\frac{\Gamma^{(k)}(z)}{\Gamma(z)}$?

I'm trying to simplify

$$\frac{\Gamma^{(k)}(z)}{\Gamma(z)}$$

for $$k=1,2,\cdots$$, using polygamma notation

Try

I've calculated a few, using

$$\Gamma^{(k)}(z) = \int_0^\infty (\log x)^k x^{z-1} e^{-x} dx$$

but I'm not sure if there is any way to generalize.

\begin{aligned} \frac{\Gamma^{(1)}(z)}{\Gamma(z)} &= \psi^{(0)}(z) \\ \frac{\Gamma^{(2)}(z)}{\Gamma(z)} &= \psi^{(1)}(z) +\left(\psi^{(0)}(z)\right)^2 \\ \frac{\Gamma^{(3)}(z)}{\Gamma(z)} &= \psi^{(2)}(z) + 3 \psi^{(1)}(z) \psi^{(0)}(z)+\left(\psi^{(0)}(z)\right)^3 \\ \frac{\Gamma^{(4)}(z)}{\Gamma(z)} &= \psi^{(3)}(z) + 4 \psi^{(2)}(z) \psi^{(0)}(z)+ 6 \psi^{(1)}(z) \left(\psi^{(0)}(z)\right)^2+ 3 \psi^{(1)}(z)^2 +\left(\psi^{(0)}(z)\right)^4 \\ \end{aligned}

• I think you've mixed up $x$ and $z$ in your integral formula. – Robert Israel Nov 21 '18 at 15:29
• @RobertIsrael True, a lot to edit. Thnx – Moreblue Nov 21 '18 at 15:31

Of course the exponential generating function is

$$\sum_{n=0}^\infty \frac{s^n}{n!} \frac{\Gamma^{(n)}(z)}{\Gamma(z)} = \frac{\Gamma(z+s)}{\Gamma(z)}$$ so basically you want the Taylor coefficients of $$\Gamma$$ around $$z$$.

Now $$\ln(\Gamma)$$ has a nice series:

$$\ln(\Gamma(z+s)) = \ln(\Gamma(z)) + \sum_{k=1}^{\infty} \frac{\Psi^{(k-1)}(z)}{k!} s^k$$

so

$$\frac{\Gamma(z+s)}{\Gamma(z)} = \exp \left(\sum_{k=1}^\infty \frac{\Psi^{(k-1)}(z)}{k!} s^k \right) = \prod_{k=1}^\infty \exp\left(\frac{\Psi^{(k-1)}(z)}{k!} s^k\right)$$

and the coefficient of $$s^n$$ here is

$$\sum_{\sum_k k m_k = n} \prod_{k=1}^\infty \frac{(\Psi^{(k-1)}(z))^{m_k}}{(k!)^{m_k} m_k!}$$

the sum being over all sequences $$m = (m_1, m_2, \ldots)$$ of nonnegative integers with $$\sum_k k m_k = n$$. These correspond to partitions of $$n$$, where $$m_k$$ is the number of occurences of $$k$$ in the partition. Multiply by $$n!$$ to get $$\Gamma^{(n)}(z)/\Gamma(z)$$. Thus for $$n=3$$, the partitions of $$3$$ are $$1+1+1$$, $$1+2$$ and $$3$$, corresponding to the terms $$\Psi^{(0)}(z)^3$$, $$3 \Psi^{(0)}(z) \Psi^{(1)}(z)$$ and $$\Psi^{(2)}(z)$$ respectively.