Expression for $\beta =1$ in $\sum_r r^\beta \ln r$
I recently had the following idea to use the below identity:
$$ (1!2! 3! \dots n!) (12^2 3^3 4^4 \dots n^n) = n!^{n+1}$$
Dividing both sides by $n^{n(n+1)/2}$ and using $1+2+3 \dots + n = \frac{n(n+1)}{2}$:
$$ (1!2! 3! \dots n!) (\frac{1}{n}(\frac{2}{n})^2 (\frac{3}{n})^3 (\frac{4}{n})^4 \dots (\frac{n}{n})^n) = \frac{n!^{n+1}}{n^{n(n+1)/2}}$$
Raising both sides to power $1/n$:
$$ (1!2! 3! \dots n!)^{\frac{1}{n}} ((\frac{1}{n})^{\frac{1}{n}}(\frac{2}{n})^{\frac{2}{n}} (\frac{3}{n})^{\frac{3}{n}} (\frac{4}{n})^{\frac{4}{n}} \dots (\frac{n}{n})^{\frac{1}{n}}) = \frac{n!^{1+1/n}}{n^{(n+1)/2}}$$
Taking $\ln$ both sides:
$$ \sum_{r=1}^n \ln(r!) \frac{1}{n} + \sum_{r=1}^n \frac{r}{n} \ln(\frac{r}{n}) = \ln(\frac{n!^{1+1/n}}{n^{(n+1)/2}}) $$
Dividing both sides with $1/n$:
$$ \sum_{r=1}^n \ln(r!) \frac{1}{n^2} + \sum_{r=1}^n \frac{r}{n} \ln(\frac{r}{n})\frac{1}{n} = \frac{1}{n} \ln(\frac{n!^{1+1/n}}{n^{(n+1)/2}}) $$
In the limit $n \to \infty $ then $ \sum_{r=1}^n \frac{r}{n} \ln(\frac{r}{n})\frac{1}{n} \to \int_0^1 x \ln x dx = -1/4 $. Hence,
$$ \sum_{r=1}^n \ln(r!) \frac{1}{n^2} - \frac{1}{4}\sim \frac{1}{n} \ln(\frac{n!^{1+1/n}}{n^{(n+1)/2}}) $$
Thus,
$$ \sum_{r=1}^n \ln(r!)\frac{1}{n^2} \sim (1+ \frac{1}{n} ))( n \ln(n) -n + O(\ln n)) - \frac{(n+1)}{2}\ln(n) + \frac{1}{4} $$
Using Stirling's approximation on the R.H.S:
$$ \sum_{r=1}^n \ln(r!)\frac{1}{n^2} \sim \frac{n-1}{2} \ln n + \ln n -\frac{3}{4} + O(\frac{\ln n}{n}) $$
Using Stirling's approximation on the L.H.S:
$$ \sum_{r=1}^n(r \ln r + r + O(\ln r))\frac{1}{n^2} \sim \frac{n-1}{2} \ln n + \ln n -\frac{3}{4} + O(\frac{\ln n}{n}) $$
Simplifying both sides:
$$ \sum_{r=1}^n r \ln r \sim n^2 \frac{n-1}{2} \ln n + n^2\ln n -\frac{(5n+2)n}{4} + O(n \ln n + \ln n!) $$
Question
I did not consider the error when converting to an integral when I did the step "In the limit $n \to \infty $ then $ \sum_{r=1}^n \frac{r}{n} \ln(\frac{r}{n})\frac{1}{n} \to \int_0^1 x \ln x dx = -1/4 $" .. What is the error? Is there a list of asymptotic expressions for $\sum r^\beta \ln r$ without using $ \sum r^\beta \leq \sum r^\beta \ln r \leq \sum r^{\beta + 1}$?
By the Euler-Maclaurin summation formula,
$$\sum_{k=1}^nk^\beta\ln k\approx\frac{n^{\beta+1}}{\beta+1}\log n-\frac{n^{\beta+1}-1}{(\beta+1)^2}+\frac{n^\beta\ln n}2+\frac{\beta n^{\beta-1}\ln n-n^{\beta-1}+1}{12}+\cdots$$