Jtdylktuy

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.

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

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

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

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.