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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.

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    $\begingroup$ The fact is, $(\frac{1}{x})^2$ gets smaller fast enough that even adding its subgraph area you don't get too big, whereas $\frac{1}{x}$ isn't fast enough. $\endgroup$
    – Riccardo Orlando
    Apr 30, 2016 at 11:09
  • $\begingroup$ I can see that (1/x)^2 has a steeper slow than (1/x) and that is seems to converge faster. But how can we intuitively conclude about the 'speed of convergence' to be fast enough to give a finite area? $\endgroup$
    – Dhruv Gupta
    Apr 30, 2016 at 11:19
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    $\begingroup$ Well, $\frac{1}{x}$ is the edge. Anything faster converges, anything slower doesn't (as a rule of thumb). $\endgroup$
    – Riccardo Orlando
    Apr 30, 2016 at 11:21
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    $\begingroup$ $\int x^\alpha = x^{\alpha +1}$: as you can see, for $x \to +\infty$, $x^{\alpha +1} \to 0 \iff \alpha +1 \leq 0$, that is, $\alpha \leq -1.$ $\endgroup$
    – Riccardo Orlando
    Apr 30, 2016 at 11:32
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    $\begingroup$ "But accepting the fact that one is converging and the other is not seems absurd at this moment.", it is not. And it is just absolutely common that intuition can play tricks on us. $\endgroup$
    – Vim
    Apr 30, 2016 at 12:19

2 Answers 2

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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 $N \to +\infty$.

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  • $\begingroup$ I appreciate you writing the answer. I can do the calculations behind it. My problem is with the intuition. Both the graphs are asymptotic to the x axis. Yet one has finite area and the other doesnt. You misunderstood the question. $\endgroup$
    – Dhruv Gupta
    Apr 30, 2016 at 11:15
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    $\begingroup$ If you sum the areas of the rectangles under $x \mapsto 1/x$ this tends to $\infty$, it thus diverges, as I proved above. If you do the same with the curve $ x \mapsto 1/x^2$ then the sum converges. So there is a difference. I know that you would like an intuitive answer. That's why I speak in terms of 'rectangles', because I think the area of a rectangle might be considered something intuitive. $\endgroup$
    – Olivier Oloa
    Apr 30, 2016 at 11:29
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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.

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  • $\begingroup$ Actually, the addition to the answer was helpful. It clarifies the core concept in a beautiful manner. Considering the graph of (1/x), if we take a point on the x-axis and rotate it, we end up with the circle having an area of π(1/x)^2. We can multiply this with a small width 'dx' and integrate over the domain to obtain the volume. Since the original function has now been transformed into π(1/x)^2 ( p > 1) , it is converging. $\endgroup$
    – Dhruv Gupta
    May 2, 2016 at 15:05

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