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A Riemann type sum
https://www.mathimatikoi.org/forum/viewtopic.php?f=4&t=1348
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Author:  Riemann [ Sat May 18, 2019 9:09 pm ]
Post subject:  A Riemann type sum

Justify the fact that

$$\frac{1}{m} \sum_{j=1}^{m} \left(\frac{2j-1}{m}-1\right) \log \left(\frac{j}{m}\right) \xrightarrow[m \rightarrow +\infty]{} \int_{0}^{1} (2x-1)\log x \, \mathrm{d}x =\frac{1}{2}$$

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