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How do you find the $$ \int \dfrac{1}{x} dx$$ by using the idea of a limit of a Riemann sum on the interval [1,2]?

I tried splitting the interval into a geometric progression and evaluating the Riemann sum, but i cant simplify the expression at this stage.

Olivier Oloa
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Pav
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2 Answers2

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You may write

$$ \begin{align} \int_1^2\frac1xdx&=\lim_{n \to \infty}\frac1n\sum_{k=0}^nf\left(1+\frac{k}{n}\right)\\\\ &=\lim_{n \to \infty}\frac1n\sum_{k=1}^n\frac1{1+\frac{k}{n}}\\\\ &=\lim_{n \to \infty}\sum_{k=1}^n\frac1{n+k}\\\\ &=\lim_{n \to \infty}\left(\sum_{k=1}^{2n}\frac1{k}-\sum_{k=1}^n\frac1{k}\right)\\\\ &=\lim_{n \to \infty}\left(H_{2n}-H_n\right)\\\\ &=\lim_{n \to \infty}\left(\left(H_{2n}-\ln (2n)-\gamma\right)-\left(H_{n}-\ln (n)-\gamma\right)+\color{blue}{\ln 2}\right)\\\\ &=\color{blue}{\ln 2} \end{align} $$ where we have used the definition of the Euler-Mascheroni constant $\gamma=0.5772\cdots$, $$ \gamma= \lim_{n\to \infty}\left(H_n-\ln n\right). $$ Remark. As $n \to \infty$, we have the asymptotic formula (6.3.18):

$$ H_n=\ln n+\gamma-\frac1{2n}+O\left(\frac1n \right). $$

Olivier Oloa
  • 120,989
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You can divide the interval $[1,2]$ into $n$ parts in geometric progression: set $r=\sqrt[n]{2}$ and consider $$ x_0=1<x_1=r<x_2=r^2<\dots<x_{n-1}=r^{n-1}<x_n=r^n=2 $$ The Riemann sum corresponding to taking the right extreme of the intervals is $$ \sum_{k=1}^n(x_k-x_{k-1})f(x_k)= \sum_{k=1}^n(r^k-r^{k-1})\frac{1}{r^k}=n\left(1-\frac{1}{r}\right) $$ If you use the left extreme of the intervals, you get $$ \sum_{k=1}^n(x_k-x_{k-1})f(x_{k-1})= \sum_{k=1}^n(r^k-r^{k-1})\frac{1}{r^{k-1}}=n(r-1)=n(\sqrt[n]{2}-1) $$ Just to make a try, let $n=2^{10}$; then we can easily compute $\sqrt[n]{2}$ by just pressing ten times the square root key on a pocket calculator, getting $1.000677130693066$; subtract $1$ and multiply by $2^{10}=1024$, getting $0.69338182970368$.

With the “lower Riemann sum” we obtain $0.6929126372864$.

If, with the same calculator, I ask for the natural logarithm of $2$, I get $0.693147180559945$; the average between the above values is $0.69314723349504$.

Pretty good, isn't it?

egreg
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