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!
I do not know if there is a closed form.
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 $5671.26$.