I am aware that the area under the curve of $\frac{1}{x}$ is infinite yet the area under the curve of $\frac{1}{x^2}$ is finite.

Calculus and series wise, I understand what is going on, but I can't seem to get a good geometric intuition of the problem. Both curves can be shown to converge to $0$ (the curves themselves, not the area), and on the interval from $1$ to infinity, the two curves have nothing intrinsically different. Can someone please provide me with an good geometric intuition of what's going on? I can't find anything on the web, people seem to not want to explain it geometrically.

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    $\begingroup$ I guess, we generally don't see it. The curve $1/x$ is, in a sense, on the border between finite and infinite areas under, as any $1/x^{1+\varepsilon}$ already gives a finite improper integral. $\endgroup$ – Berci Jul 16 '15 at 22:44
  • $\begingroup$ if you break the integration interval, the series in each segment (even in very tiny intervals 1-1+e), you have harmonic seris which is divergent. i wish it help you to describe such a geometry $\endgroup$ – Cardinal Jul 16 '15 at 22:45
  • $\begingroup$ Does this suggest something interesting about infinity? since the only difference between the two curves is that the area under 1/x^2 decreases at a faster rate, so somehow if you decrease fast enough, you are finite... $\endgroup$ – Frank Jul 16 '15 at 22:59
  • $\begingroup$ @Frank Just curious ... do you have a similar curiosity regarding the divergence of the harmonic series and the convergence of the series of reciprocals of squares of natural numbers? $\endgroup$ – Mark Viola Jul 16 '15 at 23:07
  • $\begingroup$ There is no sensible way to apply the word "geometric intuition" to such a thing as an improper integral, really. Intuition is gotten from familiarity. Get familiar with improper integrals and then you will develop an intuition. (You say there is «nothing intrinsically different» between the two curves, but your question is precisely about a rather significant difference! (I doubt the word intrinsic means anything in this context)) $\endgroup$ – Mariano Suárez-Álvarez Jul 16 '15 at 23:10

I do not think there is any geometric intuition behind the convergence you mentioned. As much as there is no geometric intuition which can help solving Zeno's paradox of the Tortoise and Achilles.

Indeed, the convergence is due to how the integral is defined - in terms of Riemann sums (here I assume you are talking about Riemann integration).

Perhaps it is not a case that convergence of series were rigorously studied (e.g. by Abel) exactly when analytic methods took over geometric ones (see M. Kline, Mathematical Thought from Ancient to Modern Times, chapter 40).

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    $\begingroup$ With all due respect: To the tortoise paradox there is a very intuitive geometric resolution. You can show that halving the unit interval and then halving the half of it and so on results in an infinite partition of a finite segment. The problem for the great Greek was unsolvable because he could not imagine that infinite steps can be taken within a finite time interval. $\endgroup$ – zoli Jul 17 '15 at 2:54
  • $\begingroup$ @ zoli Yes, you are right. Perhaps I should have said space-time intuition? I don't know. I mentioned that because is related to the limit concept which is then related to the above post. $\endgroup$ – ZenoCosini Jul 17 '15 at 13:27

I made drawings of both functions.

With respect to 1/x^2 I was very strict: I calculated the area under the straight line connecting ( 1,1) and (2,1/4) and so on. This gives two series that are both convergent by comparison with the series 1, 1/2, 1/4 ....The area of this series is larger than the area we are interested in, so our area under the integral is finite, as expected.

With respect to 1/x I was very lenient, and I calculated the area connecting (1,1/2) and (2,1/2), (2,1/3) and (3, 1/3) and so on, which gives the harmonic series of which we know that it diverges. The area that we calculated is smaller than the area we are interested in, so the area of the integral here is infinite.Harmonic series was already suggested by Cardinal, thanks for that suggestion.

So even when you are very strict on the first, and very lenient on the second function, the first is clearly finite, and the other infinite.But in the end you use simple geometry- pre calculus I suppose- and combine it with knowledge with respect to convergence of series.

I now realize that Mark Viola hinted this already in his post "Frank just curious". His mysterious response was alluding to what would happen if you inspect the areas, thereby getting these series he made reference to. I guess he wanted to just give a hint;)!

It is common knowledge that both functions blow up at x=0. The function y=1/x diverges also at the tail, so I will leave it at that. By inspecting the areas of y=1/x^2 one gets a series that keeps adding terms, so the area is infinite if one includes x=0, as expected.

I took it from x=1 to make calculations easier and because I thought the question was with respect to the tail of both functions.Since it is about the tail , choosing the left limit of the integration seems arbitrary.

My response is late: a student at the calculus course I am mentoring asked the question, now being about x going from 1 to infinity, I saw the page of this site, and I didn't notice any posts that did the same thing as I did. It was very simple, but I thought that this was meant by "geometrical" in the original question. I posted because I liked to have some feedback on the content, not because I wanted a medal;)! As to your comment hardmath: I thought it was clearly understood that the question was not about the blow up of both functions. Are there follow up answers that I didn't see, or did the discussion pass away some four years ago? If the last is true, is it senseless to add a post after a long time? It seemed to me that the original post actually wanted to point out that things were questionable because they were not intuitive, brought philosophy into the discussion, and after that I saw some reference to space time. After evaluating the level of some of these comments- only a few of them- I thought that simple calculations would at least meet that level;)! Kind regards, Ad van Straeten

P.S. In fact this was a very simple answer to a very simple question. Both are not consistent with the advanced level of the other discussions at this site. But it is still an answer that was not provided yet.

  • $\begingroup$ Your geometric approach does not account for the area under the curve near $x=0$, though perhaps you've thought about this without making any direct remarks. In general it is good practice before responding to a three+ year old Question to read the previous Accepted or upvoted Answers, and to highlight what new information is being added. $\endgroup$ – hardmath Mar 11 '19 at 22:25
  • $\begingroup$ Sorry for that! I had a good look at the question as was asked at the page that I saw, and I guessed the question was about the tail. And I only saw this page and thought that these were the comments made: didn't suspect that there were other posts somewhere else. The whole thing explodes to the left, sorry for that! $\endgroup$ – Ad van Straeten Mar 11 '19 at 23:28

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