To solve this integral using Riemann sums: $$\int_{a}^{b}\frac{1}{x}dx$$ first we find $\Delta x_i=\frac{b-a}{n}$, and frome there we find $x_i=a+\frac{(b-a)i}{n}$. After plugging this into a formula, we get:
$$\lim_{n\to\infty}\sum_{i=1}^{n}\left(\frac{1}{a+\frac{(b-a)i}{n}}\right)\frac{b-a}{n}=\lim_{n\to\infty}\sum_{i=1}^{n}\left(\frac{b-a}{an+(b-a)i}\right)$$
But this is the part where i get stuck.
I've even tried separating this into two integrals so that we get:
$$\int_{a}^{b}\frac{1}{x}dx=\int_{0}^{b}\frac{1}{x}dx-\int_{0}^{a}\frac{1}{x}dx$$, but then when i try solving it using the definition of Riemann sum, I get:
$$\lim_{n\to\infty}\sum_{i=1}^{n}(\frac{n}{bi}\frac{b}{n})-\lim_{n\to\infty}\sum_{i=1}^{n}(\frac{n}{ai}\frac{a}{n})=\lim_{n\to\infty}\sum_{i=1}^{n}(\frac{1}{i}-\frac{1}{i})=0$$
Obviously this isn't the correct solution.
Using the Newton-Leibniz formula we find that the right answer should be $\ln a-\ln b$, but how to find that solution using Riemann sums?
Best Answer
$\newcommand{\d}{\,\mathrm{d}}$When you try to "solve it using the definition of Riemann sums": for starters, the way in which Riemann defined his integral was a bit more general than simply looking at sums of $f(i/n)/n$.
Leaving that aside, there is a mistake in using the approach of $\int_0^b-\int_0^a$. $$\lim_{n\to\infty}\sum_{i=1}^n\frac{n}{ai}\cdot\frac{a}{n}=\lim_{n\to\infty}\sum_{i=1}^n\frac{1}{i}=\infty$$So you have calculated $\infty-\infty$ which is an indeterminate form. $\int_0^a\frac{1}{x}\d x$ does not exist, so you cannot make manipulations with it.
To the problem: the following approach is another example of this technique.
Say $0<a<b$ are fixed. For any $n\in\Bbb N$ notice that $P_n:=\{aq^j:0\le j\le n\}$ where $q:=(b/a)^{1/n}$ forms a partition of $[a,b]$ and we can decide to tag this partition on the left, at every point $aq^j$ for $0\le j<n$. Then, "by definition of a Riemann sum/integral" (this is where the more general definition is useful!!), as we know $x\mapsto1/x$ is continuous hence integrable and the mesh of this partition $P_n$ is $aq^{n-1}(q-1)\to0$ which decays to zero as $n\to\infty$, we can say: $$\int_a^b\frac{1}{x}\d x=\lim_{n\to\infty}\sum_{j=0}^{n-1}\left(\frac{1}{aq^j}\right)(aq^{j+1}-aq^j)=\lim_{n\to\infty}\sum_{j=0}^{n-1}(q-1)=\lim_{n\to\infty}n(q-1)$$
To evaluate this limit, let's rewrite it as: $$\lim_{n\to\infty}\frac{(b/a)^{1/n}-1}{1/n}$$
How to approach this depends on how you want to define the natural logarithm. But if you have a working definition, you can evaluate this limit as $\ln(b/a)$ quite easily and that is indeed the correct answer.
This might be slightly circular though, since it is common to define $\ln(y):=\int_1^y\frac{1}{x}\d x$.