# Integral as a limit of a sum

How do I get $$\int_a^b \frac{1}{x}dx = \ln\left(\frac{b}{a}\right)$$ as a limit of sum. The constant width partition of the interval $(a,b)$ doesn't seem to work.

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What makes you think the fixed-width Riemann sum doesn't work in the limit? –  anon Jun 22 '12 at 14:14
Typically the logarithm is defined $\log t:= \int_1^t\frac{1}{x}dx$, so your question follows nearly by definition. –  nullUser Jun 22 '12 at 14:15
@Anon I tried it but couldn't get to the answer. –  Prakash Gautam Jun 22 '12 at 14:25
@PrakashGautam Write up what you tried below your question. –  rschwieb Jun 22 '12 at 14:29

Take $x_i = a \left( \frac{b}{a} \right)^{i/n}$ then:

$$\lim\limits_{n \to \infty} \sum_{i = 0}^{n-1} \frac{x_{i+1} - x_i}{x_i} = \lim\limits_{n \to \infty} \frac{ \left(\frac{b}{a}\right)^{\frac{1}{n}} - 1 }{\frac{1}{n}} = \ln \frac{b}{a}$$

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(+1), but I guess it is good to append that $$\log x = \lim\limits_{k \to 0} \frac{x^k-1}{k}$$ –  Pedro Tamaroff Jun 24 '12 at 22:43

The constant width partition works fine. No need for fancy partitions. To wit:

Let $x_k = a+\frac{k}{N}(b-a)$, with $k = 0,...,N$. We have, of course, $x_0=a$, $x_N = b$.

Using the Taylor series expansion of $x \mapsto \ln x$, and the mean value theorem, we obtain the bounds, for $x \geq y$:

$$\ln y + \frac{x-y}{y} - \frac{(x-y)^2}{2 y^2} \leq \ln x \leq \ln y + \frac{x-y}{y}.$$ Letting $y=x_k$, $x=x_{k+1}$, we get: $$\ln x_k + \frac{1}{N x_k} - \frac{1}{2 N^2 x_k^2} \leq \ln x_{k+1} \leq \ln x_k + \frac{1}{N x_k}.$$

Rearranging gives: $$- \frac{1}{2 N^2 x_k^2} \leq \ln x_{k+1} - \ln x_k - \frac{1}{N x_k} \leq 0.$$ Now sum the inequality over $k = 0,...,N-1$, to get: $$- \frac{1}{2 N^2} \sum_{k=0}^{N-1}\frac{1}{x_k^2} \leq \ln \frac{b}{a} - \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} \leq 0.$$

Since $x_k\geq a$, we obtain: $$- \frac{1}{2 N a^2} \leq \ln \frac{b}{a} - \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} \leq 0.$$

Now let $N \to \infty$, to obtain the desired result, $$\lim_{N \to \infty} \frac{1}{N}\sum_{k=0}^{N-1} \frac{1}{x_k} = \int_a^b \frac{1}{t} dt = \ln \frac{b}{a}.$$

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Well, fine is obviously not the same as "nicely". –  Pedro Tamaroff Jun 24 '12 at 22:42
No symbols where none intended. –  copper.hat Jun 24 '12 at 23:34
I don't follow. –  Pedro Tamaroff Jun 24 '12 at 23:36
Unlike mathematics, it loses meaning when explained; it is the last line of Beckett's idiosyncratic novel Watt. Seemed an appropriate response to your enigmatic comment... –  copper.hat Jun 24 '12 at 23:47
What I mean is that if one chooses an appropriate "fancy" partition the solution is much manageable (i.e. qoqozs solution or Amihai's), which I guess you're refering to. –  Pedro Tamaroff Jun 25 '12 at 0:09

Lets do it for $\int\limits_1^b\frac{1}{x}dx$:

Consider the following partition of our interval $\,[1,b]$:

$$P_n=\{1, \sqrt[n]{b},\sqrt[n]{b^2},\dots,\sqrt[n]{b^n}=b\}$$

Now, we need to calculate the upper and lower Darboux sums in respect to the $P_n$'s. Notice that $\frac{1}{x}$ is monotonic decreasing on the interval so the $\sup$ and the $\inf$ are obtained on the edges:

$$\mathcal{U}\left(\frac{1}{x},P_n\right)=\sum_{k=1}^n b^{-\frac{k}{n}}(b^{\frac{k}{n}}-b^{\frac{k-1}{n}})=\sum_{k=1}^n 1-b^{-\frac{1}{n}}=n(\sqrt[n]{b}-1)$$

define $g(x)=b^x$:

$$n(\sqrt[n]{b}-1)=\frac{g(\frac{1}{n})-g(0)}{\frac{1}{n}-0}\to_{n\to\infty}g'(0)=b^x\ln b|_{x=0}=\ln b$$

I'll leave it to you to figure out the lower sums as it's practically the same procedure.

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Well, how do you define $\ln b$? This seems circular. THe derivative of $b^x$ is defined using that of $e^x$, which might or might not be defined in terms of the $\log x$. –  Pedro Tamaroff Jun 22 '12 at 16:05
Why can't I define the logarithm as the inverse of the $\exp$? –  Amihai Zivan Jun 23 '12 at 17:36
I said "might or might not" in the sense you can choose that method, but there are others available. –  Pedro Tamaroff Jun 24 '12 at 0:06

$$\int _a^b {f(t) dt}=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n f(\frac{k(b-a)}{n}+a)$$

$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n \frac{n}{k(b-a)+na}=\lim_{n\to\infty} \frac{(b-a)}{1}\sum \limits_{k=1}^n \frac{1}{k(b-a)+na}$$

$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \frac{(b-a)}{n}\sum \limits_{k=1}^n \frac{n}{k(b-a)+na}=\lim_{n\to\infty} \frac{(b-a)}{(b-a)}\sum \limits_{k=1}^n \frac{1}{k+\frac{na}{b-a}}$$

$$\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{(b/a)-1}}$$

$b/a=x$

$$f(x-1)=\int _a^b \frac{1}{t} dt=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x-1}}$$

$$f(x)=\lim_{n\to\infty} \sum \limits_{k=1}^n \frac{1}{k+\frac{n}{x}}$$

$$f(x)=\lim_{n\to\infty}\frac{x}{n} \sum \limits_{k=1}^n \frac{1}{1+\frac{kx}{n}}=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n (1-\frac{kx}{n}+\frac{k^2x^2}{n^2}-\frac{k^3x^3}{n^3}+....)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....$$

$$f(x)=\lim_{n\to\infty} \frac{x}{n} \sum \limits_{k=1}^n 1 -\lim_{n\to\infty} \frac{x^2}{n^2} \sum \limits_{k=1}^n k+\lim_{n\to\infty} \frac{x^3}{n^3} \sum \limits_{k=1}^n k^2-\lim_{n\to\infty} \frac{x^4}{n^4} \sum \limits_{k=1}^n k^3+.....=\lim_{n\to\infty} \frac{x}{n} n -\lim_{n\to\infty} \frac{x^2}{n^2} (\frac{n^2}{2}+\frac{n}{2})+\lim_{n\to\infty} \frac{x^3}{n^3} (\frac{n^3}{3}+\frac{n^2}{2}+\frac{n}{6})-\lim_{n\to\infty} \frac{x^4}{n^4}(\frac{n^4}{4}+\frac{n^3}{2}+\frac{n^2}{4})+....$$

$$\sum \limits_{k=1}^{n} k^m=\frac{n^{m+1}}{m+1}+a_mn^m+....+a_1n=\frac{n^{m+1}}{m+1}+\sum \limits_{j=1}^m a_jn^j$$ where $a_j$ are constants.where aj are constants. More information about summation http://en.wikipedia.org/wiki/Summation

After solving limits. We get:

$$f(x)=\frac{x}{1} -\frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+ ....=\sum \limits_{k=1}^{\infty} (-1)^{k+1} \frac{x^k}{k}=\ln(x+1)$$

$$f(x-1)=\ln(x)$$

$b/a=x$

$$\int _a^b \frac{1}{t} dt=f(x-1)=\ln(x)=\ln(b/a)$$

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