I was trying to find an upper bound for
$$ \frac{d^n}{ds^n} \Gamma(s)^{n}|_{s=1}$$
yet, I only get the bound for the nth derivative of gamma, as follow:
First, the integral of the nth derivative of gamma is
$$\frac{d^n}{ds^n} \Gamma(s) = \int_0^{\infty} t^{s-1} e^{-t} (\log t)^n dt$$
Note that this integral, if it changes sign, does so at $t=1$. So let us break it into two integrals, one ranging from 0 to 1, the other from 1 to $\infty$.
$$I_n = \int_0^{1} e^{-t} (\log t)^n dt + \int_1^{\infty} e^{-t} (\log t)^n dt = K_n + L_n$$
Consider first $K_n$. We have that, in the $[0,1]$ interval, $\exp(-t)\leq 1$. Therefore we can bound
$$K_n = \int_0^{1} e^{-t} (\log t)^n dt\leq \int_0^{1} (\log t)^n dt = (-1)^n n!.$$
For $L_n$ we can do a change of variables $t \rightarrow t+1$ to write
$$L_n = \frac{1}{e} \int_0^{\infty} \exp(-t)[\log (1+t)]^n dt.$$
Now note that $\log(1+t) \leq \sqrt{t}$ for all positive $t$. Therefore we can write
$$L_n \leq \frac{1}{e} \int_0^{\infty} \exp(-t)t^{n/2} dt= \frac{1}{e} \Gamma\left(1+\frac{n}{2}\right).$$
There, we can write
$$ \left|\frac{d^n}{ds^n} \Gamma(s)\right|_{s=1} \leq n!+\frac{1}{e} \left(\frac{n}{2}\right)!$$
I've like to have some ideas of how to relate this bound with the bound I needed, or some way of writing $ \frac{d^n}{ds^n} \Gamma(s)^{n}|_{s=1}$ in terms of polygamma function if possible.
Best Answer
You can just use the general Leibniz rule $$ (f_1\dots f_m)^{(n)}=\sum_{k_1+k_2+\dots+k_m=n}\binom{n}{k_1,k_2,\dots,k_n}\prod_{i=1}^m f_i^{(k_i)} $$ and get $$ \left.\frac{\mathrm{d}^n}{\mathrm{d}s^n}[\Gamma(s)^n]\right\rvert_{s=1} = \sum_{k_1+k_2+\dots+k_n=n}\binom{n}{k_1,k_2,\dots,k_n}\prod_{i=1}^n \left.\frac{\mathrm{d}^{k_i}}{\mathrm{d}s^{k_i}}\Gamma(s)\right\rvert_{s=1}. $$
There is an error in your derivation. Since $\log t<0$ for $t\in(0,1)$, and you bound $0<e^{-t}<1$, your bound for $K_n$ doesn't work with the $(-1)^n$. If you put everything in modulus sign you get $\lvert K_n\rvert\leq n!$ which seems to be what you were doing at the end.