Areas under the graphs of $\frac{1}{x}$ and $\frac{1}{x^2}$ from $1$ to $\infty$

Question

A simple evaluation of the definite integral tells us that the area under the graph of $[\frac{1}{x}]^2$ from $1$ to $\infty$ is finite whereas that of $\frac{1}{x}$ for the same limits is infinite.

When we look at the graphs of these functions, we can see a striking similarity.

I may, after some reasonable and concrete explanation in the direction, accept that both have a finite area under their graphs (as they are asymptotic to the x-axis) or an infinite area (as they never practically touch the x-axis).

But accepting the fact that one is converging and the other is not seems absurd at this moment. Any intuition is really appreciated.

Answer 1

One may recall that, as $M \to +\infty$, we have $$ \begin{align} \int_1^M \color{blue}{\frac1{x^2}}\:dx&=\left[ -\frac1x\right]_1^M=1-\color{blue}{\frac1M} \to \color{blue}{1}, \\\\ \int_1^M \color{red}{\frac1{x}}\:\:dx&=\left[ \:\ln x\:\right]_1^M=\color{red}{\ln M} \to \color{red}{+\infty}. \end{align} $$

Thus your question might be equivalent to asking:

$$\text{Why intuitively} \, \ln M \to +\infty \, \text{as} \, M \to +\infty\,?$$

It is sufficient here to take $M:=2^{N+1}$ and consider the sum of rectangles under the curve of $\dfrac1x$, we get $$ \int_1^{2^{N+1}} \frac1{x}\:dx=\sum_{n=0}^{N}\int_{2^n}^{2^{n+1}} \frac1{x}\:dx\geq \sum_{n=0}^{N} \frac{2^{n+1}-{2^n}}{2^{n+1}}=\frac {N+1}2 \to \color{red}{+\infty} $$ as $N \to +\infty$.

Answer 2

Maybe it would help to consider that in cases like this if the integral of one function diverges and a similar function converges, then there is some cutoff in between where on one side it converges (gets smaller fast enough) and on one side it diverges (doesn't get smaller quite fast enough).

For the functions you asked about consider 1/x^p as p decreases to 1. The integral from 1 to $\infty$ converges until p=1

Oh, and I forgot to share the Wikipedia article on Gabriel's Horn The basic idea is that even though the area under the curve 1/x diverges, the volume inside its rotation around the x axis converges, but the surface area of the rotation diverges. I'm guessing that just made for more confusion, but hopefully the fun kind.