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This question relates to an attempt I made to answer the following question.

I wonder if the approach below counts as an alternative way to show that $$\int_1^{1+x}\frac1t\ \mathsf dt = \ln(1+x).$$ I don't think this constitutes a proof but shows that adding infinitesimally small rectangles following $f(x)=\dfrac{1}{x}$ equals $\ln(1+x)$. Apparently my approach is not correct since it got downvoted. I am still curious as to what is the link between the derivations below and $\ln(1+x)$, since as you may see here, both functions are equivalent as $n$ goes to infnity.

The result I obtain is a an infinite sum of infinitesimally small rectangles that mimics perfectly $\ln(1+x)$ for values where $x>-1$. The approach is as follows (copying from my attempt to answer the question on the link posted above):

Start by taking the area of a rectangle that has length $x$ and height $f(x)$. The area is $f(x)\cdot x$. This is too large and a bad approximation. So break the interval from $1$ to $1+x$ by two intervals of length $\dfrac{x}{2}$. Then the area would be, starting from the left:

$$f(1)\cdot \dfrac{x}{2} + f\left(1+\dfrac{x}{2}\right). \dfrac{x}{2} = \dfrac{x}{2} + \frac{2}{1+x}.\frac{x}{2} = x\left(\dfrac{1}{2}+\dfrac{1}{2+x}\right) $$

Then, in a third iteration, breaking $x$ into three intervals and calculating the area of the three rectangles, we would get:

\begin{align} & f(1)\cdot \dfrac{x}{3} + f\left(1+\frac{x}{3}\right)\cdot \dfrac{x}{3} + f\left(1+2\frac{x}{3}\right)\cdot \dfrac{x}{3}\\ = {} &\dfrac{x}{3} + \frac{3}{1+x}\cdot\frac{x}{3} + \frac{3}{1+2x}\cdot\frac{x}{3}\\ = {} &x\left(\dfrac{1}{3} + \dfrac{1}{3+x} + \dfrac{1}{3+2x}\right) \end{align}

A pattern starts to emerge. As we break the $x$ interval into smaller and smaller parts and sum the area of the smaller rectangles we approximate better the area under $f(x)=\frac{1}{x}$. This would give a formula as follows:

$$\sum_{k=0}^{n-1}\left(\frac{x}{n+xk}\right)$$

where $k$ represents the begining points of the bases of the rectangles and $n$ represents by how many chunks we break up $x$. By taking the limit of $n$ going to infinity I get:

$$\lim_{n\to\infty} \sum_{k=0}^{n-1}\left(\frac{x}{n+xk}\right)$$

you may notice that this expression seems to be equivalent to $\ln(1+x)$ on the link provided above. I don't know how to prove this mathematically, but if you plot both expressions on a graph you will notice that even with a small value of $n$ we approach quite well the $\ln(1+x)$ function. As $n$ increases both functions seem to converge everywhere.

I want to understand how the infinite sum above relates to $\ln(1+x)$ or if I have made a mistake in the derivations

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Gustavo
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1 Answers1

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There are several definitions for the integral (and in most normal cases they are all equivalent). One very popular one is the Riemann integral which can roughly be defined as (the full definition is a bit more complicated than this but for our purposes here this is not needed)

$$\int_a^b f(x)dx \equiv \lim_{n\to\infty}\sum_{i=0}^{n-1} f(x_i)\Delta x$$

where $\Delta x = \frac{b-a}{n}$ and $x_i = a + \frac{(b-a)i}{n}$. If we apply this to the integral $\int_1^{x+1}\frac{dt}{t}$ we get

$$\int_1^{x+1} \frac{dt}{t} \equiv \lim_{n\to\infty}\sum_{k=0}^{n-1} \frac{x}{n + xk}$$

which is the formula you have. Thus what you have written down is basically just the definition of the integral $\int_1^{x+1}\frac{dt}{t}$ and has nothing to do with the $\log$ function per ce.

In order to prove that $\log(1+x) = \int_1^{x+1}\frac{dt}{t}$ you need to start with a definition of $\log(1+x)$ and then continue to show that the integral you started with is equivalent to that definition. There are several such definition one could start with, for example

  • $\log(1+x) = \sum_{n=1}^\infty \frac{x^n(-1)^{n+1}}{n}$ valid for $x\in(-1,1]$.
  • The inverse function of the exponential function $e^x$.
Winther
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  • Thank you for your response. I thought about equating the infinite sum that I mentioned above to the Taylor approximation of $\ln(1+x)$ but from there I can't find a way to continue. Do you have any suggestions? – Gustavo Aug 10 '15 at 18:24
  • @Gustavo Expanding $\frac{1}{1+\frac{xk}{n}}$ is a geometrical series might work. There are many details to be filled in but something like this $\sum_{k=0}^{n-1}\frac{1}{n}\frac{x}{1+\frac{xk}{n}}$ = $\frac{x}{n}\sum_{k=0}^{n-1}\sum_{i=0}^\infty\left(\frac{-xk}{n}\right)^i$ = $\sum_{i=0}^\infty\left[\frac{1}{n}\sum_{k=0}^{n-1}(k/n)^i\right] (-1)^ix^{i+1}$ looks like it could work. – Winther Aug 10 '15 at 18:33
  • This would give you $\int_1^{x+1}\frac{dt}{t} = \lim_{n\to\infty}\sum_{i=0}^\infty\left[\frac{1}{n}\sum_{k=0}^{n-1}\left(\frac{k}{n}\right)^i\right] (-1)^ix^{i+1}= \sum_{i=0}^\infty \left\int_0^1t^idt\right^ix^{i+1} = \sum_{i=0}^\infty \frac{(-1)^ix^{i+1}}{i+1}$ where I have used the definition of the integral $\int_0^1t^idt$ – Winther Aug 10 '15 at 18:34
  • Yes, it works! thank you! Very helpful – Gustavo Aug 10 '15 at 20:15