Summation of a cubed logarithm of n.

Question

I understand that by the property of

$ \log(a) + \log(b) = \log(ab) $

we can derive

$\sum_{i=0}^n \log(i) = \log(n!) $

but how would I go about calculating

$\sum_{j=0}^{n^2} \log(j)^3 $ or $\sum_{j=0}^{n^2} \log(j)^3 - 1 $ ?

Answer 1

Note that this is not a solution to the exact problem, but to a similar problem.

One can solve the sum $\sum_{j=1}^n\log^2(j)$.

Let $f(n)=\sum_{j=1}^n\log^2(j)$ whenever $n\in\mathbb N$ so that we may have $f(n)=f(n-1)+\log^2(n)$

$$f(n)=f(n-1)+\log^2(n)$$

$$f'(n)=f'(n-1)+2\frac{\log(n)}n$$

$$f'(n)=f'(n-2)+2\left(\frac{\log(n-1)}{n-1}+\frac{\log(n)}n\right)$$

$$=f'(n-3)+2\left(\frac{\log(n-2)}{n-2}+\frac{\log(n-1)}{n-1}+\frac{\log(n)}n\right)\\\vdots\\f'(n)=f'(0)+2\sum_{k=1}^n\frac{\log(k)}k$$

$$f'(n)=f'(0)-2\gamma_1+2\gamma_1^\star(n+1)$$

where $\gamma_1$ is a Stieltjes constant and $\gamma_1^\star(n)$ is the generalized Stieltjes constant.

Thus,

$$f(n)=\int_0^nf'(0)-2\gamma_1+2\gamma_1^\star(x+1)dx$$

$$f(n)=f'(0)n-2\gamma_1n+\int_0^n\gamma_1^\star(x+1)dx$$

And if we let $n=1$,

$$0=f(1)=f'(0)-2\gamma_1+\int_0^1\gamma_1^\star(x+1)dx$$

$$f'(0)=2\gamma_1-\int_0^1\gamma_1^\star(x+1)dx=c_1$$

Call it $c_1$ for simplicity.

$$f(n)=\sum_{j=1}^n\log^2(j)=(c_1-2\gamma_1)n+\int_0^n\gamma_1^\star(x+1)dx$$