Finite sum of powers of $\log(j)$

Question

Is there any closed-form for:

$\sum_{j=1}^{n}[\log(j)]^k$, for integer $k$?

Perhaps we should start real simple, so how about this one: $\sum_{j=1}^{n}\log^2(j)$

The importance of having this closed-form is that it's related to the Zeta function through a method I devised.

Answer 1

I doubt that there is a closed form (except for $k=1$). That said, you can have good estimates. For instance, using integration by parts (for sums), $$\sum_{k=1}^n \log (k) \approx n\log(n)$$ $$\sum_{k=1}^n \log^2 (k) \approx n\log^2(n)$$

Answer 2

A long comment:

$$\sum _{j=1}^n \log ^k(j)=\\\sum _{j=1}^n \left(\underset{t\to 0}{\text{lim}}\frac{\partial ^kj^t}{\partial t^k}\right)=\\\underset{t\to 0}{\text{lim}}\frac{\partial ^k}{\partial t^k}\left(\sum _{j=1}^n j^t\right)=\\\underset{t\to 0}{\text{lim}}\frac{\partial ^kH_n^{(-t)}}{\partial t^k}=\\\underset{t\to 0}{\text{lim}}\frac{\partial ^k(-\zeta (-t,1+n)+\zeta (-t))}{\partial t^k}$$

For k=2:

$$\color{red}{\sum _{j=1}^n \log ^2(j)}=\\\underset{t\to 0}{\text{lim}}\frac{\partial ^2(-\zeta (-t,1+n)+\zeta (-t))}{\partial t^2}=\\\underset{t\to 0}{\text{lim}}\left(\zeta ''(-t)-\zeta ^{(2,0)}(-t,1+n)\right)=\color{red}{\\\frac{\gamma ^2}{2}-\frac{\pi ^2}{24}-\frac{1}{2} \log ^2(2 \pi )+\gamma _1-\zeta ^{(2,0)}(0,1+n)}$$

Mathematica code:

HoldForm[Limit[D[-Zeta[-t, 1 + n] + Zeta[-t], {t, 2}], t -> 0] == 
Limit[Derivative[2][Zeta][-t] - Derivative[2, 0][Zeta][-t, 1 + n], t -> 0] == 
EulerGamma^2/2 - Pi^2/24 - (1/2)*Log[2*Pi]^2 + StieltjesGamma[1] - Derivative[2, 0][Zeta][0, 1 + n]]

Where:

$H_n^{(-t)}$ is harmonic number of order t,

$\zeta ''(-t)$ is second derivative of Riemann zeta function.

$\zeta (-t,n+1)$ is generalized Riemann zeta function,

$\gamma$ is Euler's constant,

$\gamma _1$ is Stieltjes constant,

$\zeta ^{(2,0)}(0,n+1)$ is second derivative of generalized Riemann zeta function.