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Riemann sum

Random Variable

Well-known member
MHB Math Helper
Jan 31, 2012
253
$\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
Feb 5, 2012
1,621
$\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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