Introduction
A few years ago I calculated this:
For $x\to\infty$
$$\sum_{n=1}^{x}n^{s-1}\ln(n)^m\propto (-1)^m\left[\color{blue}{\zeta^{(m)}(1-s)}+\frac{1}{s}\sum_{j=0}^{s}\binom{s}{j}\ B_{s-j}^{+}x^j B_{m}\left(a_0,...,a_{m-1}\right)\right]$$
Where:
- $a_\ell:=\ell!\left(H_j^{(\ell+1)}-H_{s-1}^{(\ell+1)}-\delta_\ell \ln(x)\right)$
- $B_n^{+}\text{ are the Bernoulli number with }B_1^+=\frac{1}{2}$
- $B_n(x_1,...,x_n)\text{ is the complete Bell polynomial}$
- $H_n^{(s)}\text{ are the generalized harmonic numbers}$
- $\delta_k\text{ is the Kroneker delta}$
From which, solving for $\zeta^{(m)}(1-s)$ it is possible to obtain an asymptotic expression.
Alternative formulation
The first formula involves too many complex terms to compute (such as the Bell polynomial depending on harmonic numbers and Kroneker delta), and also uses Bernoulli numbers $B^{+}_n$ while Bernoulli numbers $B_{2n}$ are usually used (since Bernoulli numbers with odd indices are $0$, it is inefficient to take this into account).
We get to write this formula:
$$\sum_{n=1}^{x}n^{s-1}\ln(n)^m\propto(-1)^m \left[\zeta^{(m)}(1-s)+... \right.\\\left.\frac{m!x^s}{s^{m+1}}\cdot\sum_{j=0}^{m}s^j \left(\frac{(-\ln(x))^j}{j!}\left(1+\frac{s}{2x}+\frac{j}{2x\ln(x)}\right)+\sum_{k=1}^{\left\lfloor\frac{s}{2}\right\rfloor} \frac{B_{2k}}{(2k)!x^{2k}}\sum_{l=0}^{j}\frac{(-\ln(x))^l}{l!}\left[2k\atop j-l\right]^{(s-2k+1)}\right)\right]$$ Where:
- $B_{2n}$ are the Bernoulli number
- $\displaystyle\left[n\atop k\right]^{(s)}$ are the generalized Stirling number of the first kind
I used a different notation than that of Dragoslav S. Mitrinović and Ružica S. Mitrinović's only for a visual reason, however the transformation is: $${}^pP_n^r=\displaystyle\left[n\atop r\right]^{(p)}$$ These numbers are not used much, so here a way to compute them:
$$\left[n+1\atop k\right]^{(p)}=\left[n\atop k-1\right]^{(p)}-(p+n)\left[n\atop k\right]^{(p)}\\ \left[n\atop k\right]^{(p)}=\sum_{j=0}^{n-k}(-1)^j\binom{k+j}{j}p^j\left[n\atop k+j\right]$$ In particular (using $0^0=1$): $$\left[n\atop k\right]^{(0)}=\left[n\atop k\right]\text{ are the Stirling Number of the first kind}$$
Question
I would like to see if it is possible to further simplify this formula, now the only "difficult" element to calculate are the generalized Stirling numbers of the first kind contained in the nested summation:
$$\sum_{l=0}^{j}\frac{(-\ln(x))^l}{l!}\left[2k\atop j-l\right]^{(s-2k+1)} \qquad\overset{\text{for simplicity}}{\mapsto}\qquad \sum_{k=0}^{n}\frac{z^k}{k!}\left[{m\atop n-k}\right]^{(p-m)}$$
Can anyone write this summation in terms of the most common (=more easily managed by a calculator) constants?