Integral $\int_1^2 \frac1x dx$ with a Riemann sum. 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.
 A: 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):

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

A: 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?
