A Riemann type sum
Posted: Sat May 18, 2019 9:09 pm
Justify the fact that:
$$\frac{1}{m} \sum_{j=1}^{m} \left(\frac{2j-1}{m}-1\right) \log \left(\frac{j}{m}\right) \rightarrow \int_{0}^{1} (2x-1)\log x \, \mathrm{d}x =\frac{1}{2}$$
$$\frac{1}{m} \sum_{j=1}^{m} \left(\frac{2j-1}{m}-1\right) \log \left(\frac{j}{m}\right) \rightarrow \int_{0}^{1} (2x-1)\log x \, \mathrm{d}x =\frac{1}{2}$$