The following sum is not difficult to compute $$\sum_{i,j=1}^n \vert i-j\vert$$
using $\vert i-j\vert=\max(i,j)-\min(i,j).$
Now let $S_n:=\sum_{i,j=1}^n \vert i-j\vert^{\alpha}$ for $\alpha>0.$
Is it possible to compute $S_n$?
Or otherwise do we have $$\frac{S_n}{n}\to \sum_{k=-\infty}^\infty \vert k\vert^{\alpha}$$
For any $p, q \in \mathbb{Z}$, let $[p,q]$ be a short hand for $\{ p, p+1, \ldots, q-1, q \}$.
Notice for any $k \in \mathbb{Z}$,
$$\# \big\{\; (i,j) \in [1,n]^2 : |i-j| = k\;\big\} = \begin{cases} 2(n-k), & k \in [1,n-1]\\n,& k = 0\\0,& \text{otherwise}\end{cases}$$
This means for any function $f(x)$, we have $$\sum_{i,j=1}^n f(|i-j|) = n f(0) + 2\sum_{k=1}^{n-1}(n-k)f(k) $$
When $m \in \mathbb{Z}_{+}$, we can express the sum of $m^{th}$ power of integers in terms of Bernoulli poynomial:
$$\sum_{k=0}^n k^m = \frac{B_{m+1}(n+1) - B_{m+1}(0)}{m+1}$$
This means when $\alpha$ is a positive integer, we have
$$\sum_{i,j=1}^n |i-j|^\alpha = 2\left[n \frac{B_{\alpha+1}(n) - B_{\alpha+1}(0)}{\alpha+1} - \frac{B_{\alpha+2}(n) - B_{\alpha+2}(0)}{\alpha+2}\right]$$
For general $\alpha > 0$, I have no idea whether there are other closed form for the sum.