# Riemann sum

#### Random Variable

##### Well-known member
MHB Math Helper
$\displaystyle \int_{0}^{1} \ln \ x \ dx$ is not a proper Riemann integral since $\ln \ x$ is not bounded on $[0,1]$. Yet $\displaystyle \int_{0}^{1} \ln \ x \ dx = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n} \right)$. Is this because $\ln \ x$ is monotone on $(0,1]$?

#### Sudharaka

##### Well-known member
MHB Math Helper
$\displaystyle \int_{0}^{1} \ln \ x \ dx$ is not a proper Riemann integral since $\ln \ x$ is not bounded on $[0,1]$. Yet $\displaystyle \int_{0}^{1} \ln \ x \ dx = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n} \right)$. Is this because $\ln \ x$ is monotone on $(0,1]$?
Hi Random Variable,

Yes, $$\displaystyle\int_{0}^{1} \ln \ x \ dx$$ is an improper integral. However $$\displaystyle\int_{a}^{1} \ln \ x \ dx\mbox{ where }0<a<1$$ is a proper Riemann integral and by the Rectangle method we can obtain,

$\int_{a}^{1} \ln \ x \ dx = \lim_{n \to \infty}\left[\frac{1-a}{n}\sum_{k=1}^{n} \ln \left(\frac{k}{n} \right)\right]=(1-a)\lim_{n \to \infty}\left[\frac{1}{n}\sum_{k=1}^{n} \ln \left(\frac{k}{n} \right)\right]$

Note that, $$\displaystyle\lim_{n \to \infty}\left[\frac{1}{n}\sum_{k=1}^{n} \ln \left(\frac{k}{n} \right)\right]$$ exists since $$\displaystyle\int_{a}^{1} \ln \ x \ dx$$ is Riemann integrable. Therefore,

$\int_{0}^{1} \ln \ x \ dx=\lim_{a\rightarrow 0^{+}}\int_{a}^{1} \ln \ x \ dx =\lim_{n \to \infty}\left[\frac{1}{n}\sum_{k=1}^{n} \ln \left(\frac{k}{n} \right)\right]$

So, $$\displaystyle\int_{0}^{1} \ln \ x \ dx$$ is an improper integral and its value is equal to $$\displaystyle\lim_{n \to \infty}\left[\frac{1}{n}\sum_{k=1}^{n} \ln \left(\frac{k}{n} \right)\right]$$

Kind Regards,
Sudharaka.

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