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I am trying to find a Riemann Sum expression for the following improper integral using right end points.

$$\int_0^1 \ln(x)dx$$

The problem doesn't resemble in the details, any of the other practice problems I've worked on, and so I'm having trouble constructing a frame of reference from which to tackle it.

The most I've gotten down is a rough "plug in the values" attempt using the general form of a Riemann Sum in n-intervals:

$$\lim_{n->\infty}\sum_{i=0}^n\ln(x_i*)\frac{1}{n}$$

Am I on the right track? What else do I need to consider?

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We can use Stirling's Formula

$$n!=\sqrt{2\pi n}\left(\frac ne\right)^n\left(1+O\left(\frac1n\right)\right)$$

to evaluate the Riemann sum $\sum_{k=1}^n \log(k/n)\frac1n$ for $\int_0^1 \log(x)\,dx=-1$.

Proceeding, we have

$$\begin{align} \sum_{k=1}^n \log(k/n)\frac1n&=\frac{\log(n!)}{n}-\log(n)\\\\ &=\frac1n\log\left(\sqrt{2\pi n}\left(\frac ne\right)^n\left(1+O\left(\frac1n\right)\right)\right)\\\\ &=\log(n)-1+\frac{\log(2\pi n)}{2n}+O\left(\frac1{n^2}\right)-\log(n)\\\\ &=-1+\frac{\log(2\pi n)}{2n}+O\left(\frac1{n^2}\right)\\\\ &\to -1\,\,\text{as}\,\,n\to \infty \end{align}$$

as expected!

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