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From a mathematical standpoint, I understand and I can solve the following: $$ \lim_{M\rightarrow\infty} \int_1^M \left({1 \over x}\right) \rightarrow \infty $$ Additionally, $$ \lim_{M\rightarrow\infty} \int_1^M \left({1 \over x^2}\right) \rightarrow 1 $$

This all makes mathematical sense to me. It's the geometric parts that confuse me. The family of of $ 1/x^p $ graphs look very similar to me, so it makes me wonder why $ 1/x $ doesn't converge to some value as well.

Especially considering the fact when you rotate $ 1/x $ and calculate the volume of that shape; it converges to some value. Again, mathematically, this makes sense because:

$$ \lim_{M\rightarrow\infty}\int_1^M \left({1 \over x}\right) dx > \lim_{M\rightarrow\infty} \int_1^M \pi\left({1 \over x^2}\right) dx $$

But the geometric implications of this are that a cross-section of such an object has an infinite area but the volume is some finite value.

My questions:

  1. Using an intuitive or geometric explanation, why doesn't $ 1\over x $ converge to some value?
  2. Why is the volume described above finite while the cross-section is infinite?

Edit: Changed $[]$ to $()$

Botond
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npengra317
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    Unfortunately, intuitition is not very helpful here. It is difficult to visualize whether a function tends to $0$ fast enough. Take for example $\frac{1}{x^{1.00001}}$. I doubt you can distinguish it visually from $\frac{1}{x}$ – Peter Feb 08 '18 at 20:50
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    I noted that as well. Which only raises my curiosity even further. What's the "difference" (for a lack of a better term) between $ 1/x $ and $ 1/x^{1.000 ... 0001} $? Is the difference in area infinite? I struggle to understand that. – npengra317 Feb 08 '18 at 20:53
  • What do you mean by $[{1 \over x}]$? – copper.hat Feb 08 '18 at 20:54
  • @copper.hat I'm unsure what you're asking. I mean "one divided by a number x" – npengra317 Feb 08 '18 at 20:55
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    Maybe derive some intuition from the series $\sum_n {1 \over n}$ and $\sum_n {1 \over n^2}$? – copper.hat Feb 08 '18 at 20:55
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    Some folks use that notation for floor or ceiling. – copper.hat Feb 08 '18 at 20:55
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    On the other hand, @npengra317, you have no problem visualising the 2D object having infinite width but finite area like the one below the curve $\frac{1}{x^2}$? I guess this should violate intuition in exactly the same way - yet we have all got used to it via years of maths' training. –  Feb 08 '18 at 20:56
  • @user8734617 You misunderstand my question. I can "imagine" an infinite area. I just don't understand the concept of creating a finite volume with an infinite area. – npengra317 Feb 08 '18 at 20:57
  • I think you misunderstood my comment. You have a case of one measure (area of cross-section) infinite and another (volume) finite. I am asking about something analogous, in 2 dimensions - width infinite but area finite. Does that not puzzle you in the same way? If not, why? –  Feb 08 '18 at 20:59
  • It is also hopeless to visualize that $e^x$ eventually dominates $\large x^{(10^6)}$ – Peter Feb 08 '18 at 21:02
  • That's a very good point. I'm not sure how I didn't think of that. I believe you answered my question; consider the relationship between the finite volume and infinite area the same way I consider the relationship between the finite area and infinite width. Though, truth be told I am still quite puzzled by that relationship too. – npengra317 Feb 08 '18 at 21:03
  • @npengra317 Wait: isn't for you $;\left\lfloor\frac1x\right\rfloor;$ the integral part of $;\frac1x;$ ? Because it is for most mathematicians... – DonAntonio Feb 08 '18 at 21:13
  • @DonAntonio Can you elaborate what you mean? – npengra317 Feb 08 '18 at 21:16
  • The function $;f(x):=\lfloor x\rfloor;,;;x\in\Bbb R$, is the integral part of the real number $;x;$ . Thus, for example, $;\lfloor 0.983\rfloor=0;,;;\lfloor 3.45\rfloor=3;,;;\lfloor 110\rfloor =110;$ , etc. Now, if this were the case, your question is completely trivial...But I'm just asking as those are not the usual parentheses used. – DonAntonio Feb 08 '18 at 21:22
  • Ok, I see you've changed the parentheses...fine. – DonAntonio Feb 08 '18 at 21:22
  • @DonAntonio Understood, I've changed them. Our textbook utilized brackets in integration and it's become my habit to use them as well. – npengra317 Feb 08 '18 at 21:25
  • @DonAntonio Might be pedantic, but weren't the paranthesis closed in both directions ? The integral part is only closed at the bottom. – Peter Feb 08 '18 at 21:31
  • @Peter Good point...I don't remember. – DonAntonio Feb 08 '18 at 21:41
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    @npengra317 Another example, the length of the curve defined by $\sin(1/x^2)$ for $x \in (0, 1)$ is infinite, however it is contained in a very small area. – WeakestTopology Feb 08 '18 at 22:07
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    This is known as Gabriel's horn; or also, i believe, as Torricelli's trumpet... –  Feb 09 '18 at 00:03

2 Answers2

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Here's something that may be helpful geometrically, borrowed from a classic argument about the harmonic series.

One argument for the divergence of the harmonic series goes as follows:

$$ \begin{align} &\frac12 + \frac13 + \frac14 + \frac15 + \frac16 + \frac17 + \frac18 + \dots \\ > &\frac12 + \frac14 + \frac14 + \frac18+ \frac18+ \frac18+ \frac18 + \dots \\ = &\frac12 + \left(\frac14 + \frac14\right) + \left(\frac18+ \frac18+ \frac18+ \frac18\right) + \dots \\ = &\frac12 + \frac12 + \frac12 + \dots \end{align} $$

Let's translate this into our integral. Imagine the area under the curve $\frac1x$. Fit a rectangle of area $\frac12$ between the points $(1,0),(2,0),(1,\frac12),(2,\frac12)$ -- this inside the area of the curve. Fit the next rectangle of area $\frac12$ between the points $(2,0),(4,0),(2,\frac14),(4,\frac14)$. In general the $i$th rectangle will be placed between the points $(2^{i-1},0),(2^i,0),(2^{i-1},\frac1{2^i}),(2^i,\frac1{2^i})$. No two rectangles overlap, and each rectangle has area $\frac12$. Since you can fit infinite rectangles of equal area under the integral, it must diverge.

BallBoy
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Suppose $\lim_{M\rightarrow\infty} \int_1^M \dfrac{dx}{x} $ exists. Letting $L$ be this limit, $\lim_{M\rightarrow\infty} \int_1^M \dfrac{dx}{x} =L$ so that, for any $c>0$ there is a $M(c)$ such that $0 \lt L-\int_1^M \dfrac{dx}{x} \lt c$ for $M > M(c)$. Choosing such an $M$, we also have $0 \lt L-\int_1^{2M} \dfrac{dx}{x} \lt c$ so that $0 \lt L-\int_1^{2M} \dfrac{dx}{x} =L-\int_1^{M} \dfrac{dx}{x}-\int_M^{2M} \dfrac{dx}{x} $ or $\int_M^{2M} \dfrac{dx}{x} \lt L-\int_1^{M} \dfrac{dx}{x} \lt c$.

But $\int_M^{2M} \dfrac{dx}{x} \gt \dfrac{M}{2M} =\dfrac12$.

This is a contradiction for $c < \dfrac12$.

This is, of course, a restating of the standard elementary proof that the harmonic sum diverges.

marty cohen
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