Why is $-\log(x)$ integrable over the interval $[0, 1]$ but $\frac{1}{x}$ not integrable?
I don't understand why some functions that contain a singularity in the domain of integration are integrable but others are not.
For example, consider $f(x) = -\log(x)$ and $g(x) = \frac{1}{x}$ on the interval $[0, 1]$. These functions look very similar when they are plotted but only $f(x)$ can be integrated.
- What is the precise mathematical reason(s) that makes some functions with singularities integrable while others are not?
- Are $\log$ functions the only functions with singularities that can be integrated or are there other types of functions with singularities that can be integrated?
It's just whether or not the area under the curve is finite or not. It's doesn't matter that there is an asymptote.
You might consider the area under the curves $y=e^{-x}$ and $y=1/x$ for $x>0$. These are really the same two curves you mention, just along the other axis.
It's akin to the idea that an infinite series may or may not converge; just because there are infinitely many terms in a series doesn't mean the series must diverge.
Think about it this way - what's the inverse?
$$y = \frac{1}{x}; x = \frac{1}{y}$$ $$y = -\log x; x = e^{-y}$$
Looking at it this way, it's clear that as $y$ shoots off to infinity, $x$ approaches zero much faster in one case than in the other.
The key is how fast the function is diverging.
Regarding your two examples, $-\log$ is going really fast close to the $y$-axis so it is integrable, but not $x\mapsto \frac {1}{x}$.
You have $$\int_a^1 -\log(x)\mathrm dx=a(1-\log(a))+1\xrightarrow[a\to 0^+]{} 1<\infty.$$ So this function is integrable.
You have $$\int_a^1 \frac 1x\mathrm dx=-1+\frac 1{a^2}\xrightarrow[a\to 0^+]{} +\infty.$$ So this function is not integrable.
Regarding your second question, $\log$ functions are absolutely not the only one. To convince yourself, take for instance $x\mapsto \frac 1{\sqrt{x}}$ on $(0,1)$.
Simple: $\quad -\log x=_0 o\Bigl(\dfrac1{\sqrt x}\Bigr)$ and the integral of $\dfrac 1{\sqrt x}$ on [0,1] is convergent.