# How to find the sum of n natural logarithms that are cubed?

I seem to be struggling with a mathematical problem. I need to evaluate this: I know that the sum of all natural logarithms is ln(n!) and that the sum of all natural cubes is (n(n+1)/2)^2 but these don't seem to be of any help in this situation. I would really appreciate some help!

• In general: $\sum _{i=1}^n \log ^m(i)=\underset{x\to 0}{\text{lim}}\frac{\partial ^mH_n^{(-x)}}{\partial x^m}$ where:$H_n^{(-x)}$ is Harmonic Number of order x. Mar 14 '20 at 13:06
• @MariuszIwaniuk and what is ∂ then? Mar 14 '20 at 13:23
• Symbol of derivative,I yours case m-th derivative.Answer from CAS like Mathematica: $$\zeta ^{(3,0)}(0,n+1)-\frac{3 \gamma _2}{2}-3 \gamma _1 (\gamma +\log (2 \pi ))+\zeta (3)+\frac{1}{8} \left(-8 \gamma ^3+4 \log ^3(2 \pi )-12 \gamma ^2 \log (2 \pi )+\pi ^2 \log (2 \pi )\right)$$ Mar 14 '20 at 13:48
• "sum of all natural logarithms" ? You mean $$\sum_{j=1}^n \ln(j)=\ln(n!)$$ Similar for the sum of the cubes. Mar 14 '20 at 14:26
• @Peter your writing is correct but in my case the logarithm itself is cubed and not the j so I can't add them like that. Mar 14 '20 at 14:38

If you want to get at least an order of magnitude, consider $$\int_{i=1}^n \log^3(i) \,di < \sum_{i=1}^n \log^3(i) <\int_{i=1}^{n+1} \log^3(i) \,di$$ with $$\int \log^3(i) \,di= -6 i+i \log ^3(i)-3 i \log ^2(i)+6 i \log (i)$$
For $$n=100$$, the left integral is $$5573.28$$, the sum is $$5622.17$$ and the right integral is $$5671.26$$.