Solve using Riemann sums: $\int_{a}^{b}\frac{1}{x}dx$

Question

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?

Answer 1

$\newcommand{\d}{\,\mathrm{d}}$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 $\ln(y):=\int_1^y\frac{1}{x}\d x$.

Answer 2

Here's a proof involving the mean value theorem.

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Using the mean value theorem, we can write $$\ln(b) - \ln(a) = \sum_{i=1}^n\left(\ln\left(a+\tfrac{(b-a)i}{n}\right)-\ln\left(a+\tfrac{(b-a)(i-1)}{n}\right)\right) = \sum_{i=1}^n\frac{1}{x_i}\cdot \frac{b-a}{n}$$ for some $x_i$ satisfying $a + \tfrac{(b-a)(i-1)}{n} < x_i < a+\tfrac{(b-a)i}{n}$ for each $i$.

Then using the mean value theorem again, there is some $y_i$ satisfying $x_i < y_i < a+\tfrac{(b-a)i}{n}$ such that $$\left|\frac{1}{a+\tfrac{(b-a)i}{n}}-\frac{1}{x_i}\right| = \left|-\frac{1}{y_i^2}\left(a+\tfrac{(b-a)i}{n}-x_i\right)\right| \leq \frac{1}{a^2}\cdot \frac{b-a}n.$$

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Then starting from where you've written,

$$\begin{align*}\left|\sum_{i=1}^{n}\left(\frac{1}{a+\frac{(b-a)i}{n}}\right)\frac{b-a}{n} - (\ln(b)-\ln(a))\right| &= \left|\sum_{i=1}^n\left(\frac{1}{a+\tfrac{(b-a)i}{n}}-\frac{1}{x_i}\right)\frac{b-a}{n}\right| \\ &\leq \sum_{i=1}^n\left|\left(\frac{1}{a+\tfrac{(b-a)i}{n}}-\frac{1}{x_i}\right)\right|\frac{b-a}{n} \\ &\leq \sum_{i=1}^{n}\frac{1}{a^2}\cdot \frac{b-a}{n} \cdot \frac{b-a}{n} \\ &= \frac{1}{n} \cdot \frac{(b-a)^2}{a^2} \end{align*}$$

Since $\lim_{n\to\infty} \frac{1}{n} \cdot \frac{(b-a)^2}{a^2} = 0$, it follows that $$\lim_{n\to\infty}\left|\sum_{i=1}^{n}\left(\frac{1}{a+\frac{(b-a)i}{n}}\right)\frac{b-a}{n} - (\ln(b)-\ln(a))\right| = 0$$ and then $$\lim_{n\to\infty} \sum_{i=1}^{n}\left(\frac{1}{a+\frac{(b-a)i}{n}}\right)\frac{b-a}{n} = \ln(b)-\ln(a).$$