Computing the value of logarithmic series: $Q(s,n) = \ln(1)^s + \ln(2)^s + \ln(3)^s + \cdots+ \ln(n)^s $ Given a series of the type:
$$Q(s,n) = \ln(1)^s + \ln(2)^s + \ln(3)^s + \cdots+ \ln(n)^s $$
How does one evaluate it?
Something I noticed was:
$$Q(1,n) = \ln(1) + \ln(2) + \ln(3)+ \cdots+\ln(n) = \ln(1\cdot 2\cdot 3 \cdots n) = \ln(n!) $$
I also noticed that:
$$\int^{n}_{1}\ln(x)^s\, dx\quad\sim\quad\sum^{n}_{i = 1}\ln(i)^s$$
But I am really interested in an exact formula or at least one whose difference from the actual value progressively decreases as opposed to merely whose ratio from the actual progressively decreases.
 A: I hope you don't mind if I use $\log$ as you use $\ln$ (it is more standard in analytic number theory).
Since $\log$ is monotonic increasing:
$$\int_{1}^{n}\log^s x \ dx < Q(s,n) < \int_{1}^{n+1}\log^s x \ dx$$
(using left/right endpoints and $Q(s,1)=0$).
This shows that $Q(s,n)=\int_1^n\log^s x + O(\log^s n)$, which is already a pretty good asymptotic formula; this error term is massively dwarfed by the main term (even though it is not a decreasing function as you have asked for - this requirement might be too strict).
Evaluating integrals, we obtain:
$$Q(s,n)=\int_{1}^{n}\log^s x \ dx+O(\log^s n)=n\left(\sum_{k=0}^{s}(-1)^{s-k}\frac{s!}{k!}\log^k n\right)+O(\log^s n)\qquad (*)$$
when $s$ is an integer, and an analogous expression in terms of the incomplete gamma function otherwise.
Suffice it to say, as far as you are probably concerned, $$Q(s,n)=n\log^s(n)+O\left[n\log^{s-1}(n)\right]$$The full version is $(*)$.
A: You may use Euler-Maclaurin formula to get $Q(x,s)=\sum_{n\leq x}(\log n)^s$. That would be $x(\log x)^s-s\int_{1}^{x}(\log t)^{s-1} \ dt+O((\log x)^s)$. It should be a fine approximation for your work!
