Riemann sum of $x\cdot \ln(x)$ I did not find any information regarding this Riemann sum anywhere:
Riemann sum of $f(x)=\begin{cases} 0& x=0 \\ x\cdot \ln(x)& \text{otherwise}\end{cases}$ in the interval $[0, 1]$.
I don't want the answer given to me, I'm only looking for a hint that could guide me in the right direction. I have already proved that the function is continuous in this interval, i.e can be integrated in it also.
 A: The sum can be written as
$$S_n = \frac{1}{n^2}\sum_{k=1}^nk \log k -\frac{\log n}{n^2}\sum_{k=1}^nk = \frac{1}{n^2}\sum_{k=1}^nk \log k -\frac{\log n}{n^2}\frac{n(n+1)}{2} \\ = \frac{1}{n^2}\sum_{k=1}^nk (\log k- H_k) -\frac{\log n}{n^2}\frac{n(n+1)}{2} + \frac{1}{n^2}\sum_{k=1}^nk H_k $$
where $H_n = \sum_{k=1}^n \frac{1}{k}$
Using summation by parts, we have
$$\frac{1}{n^2}\sum_{k=1}^nk H_k  = \frac{1}{n^2}H_n \sum_{k=1}^n k + \frac{1}{n^2}\sum_{k=1}^{n-1}\left(\sum_{j=1}^kj \right)(H_k - H_{k+1}) \\ = \frac{H_n}{n^2}\frac{n(n+1)}{2} - \frac{1}{4}\left(1 - \frac{1}{n}\right)  $$
Hence,
$$S_n = -\frac{1}{n^2}\sum_{k=1}^nk (H_k - \log k) + \frac{H_n -\log n}{n^2}\frac{n(n+1)}{2} - \frac{1}{4}\left(1 - \frac{1}{n}\right)$$
Using the fact that $H_n - \log n \to \gamma$ as $n \to \infty$,  we can show, with some additional work,  that the contribution of the first and second terms cancels in the limit as $n \to \infty$ and we have 
$$\lim_{n \to \infty} S_n = - \frac{1}{4} = \int_0^1 x \log x \, dx$$
Additional Work
Note that for any $\epsilon > 0$ there exists $N$ such that $\gamma - \epsilon < H_k - \log k < \gamma + \epsilon$ when $k > N$.
We have
$$\sum_{k=1}^nk (H_k - \log k) =  \frac{1}{n^2}\sum_{k=1}^Nk (H_k - \log k)+  \frac{1}{n^2}\sum_{k=N+1}^nk (H_k - \log k) $$
Since you don't want the full answer -- see if you can finish from here.
