I'm trying to simplify


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


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.

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

  • 1
    $\begingroup$ I think you've mixed up $x$ and $z$ in your integral formula. $\endgroup$ – Robert Israel Nov 21 '18 at 15:29
  • $\begingroup$ @RobertIsrael True, a lot to edit. Thnx $\endgroup$ – 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:

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


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

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.


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