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I tried to find the area of the reactangular hyperbola $$xy=1$$ from $x =0$ to any arbitrary $x$ by using integrals,I found the area to be infinity whereas when I found the area of $$y=\ln(x)$$ from zero to let's say $x=1$ then I found the area to be finite, that is 1 in this case. How to explain this when we know that both the functions tend to non finite values at $x=0$ or maybe I haven't calculated the area of hyperbola correctly , can someone please explain.

Bernard
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Nalin Yadav
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1 Answers1

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The area below the curve $f(x)=1/x$ from $0$ to $x_0$ is always infinite, because the area below the curve in the interval $[a,x_0]$ is given by:

$$\ln(x_0)-\ln(a)$$

Now, what happens when $a \to 0$?

You can apply a similar procedure for calculating the area behind $y=\ln(x)$, if you are not getting again an infinite value, it's because the function doesn't "explode" as "fast" as the hyperbola, making the analogous limit finite

You may want to compare this case with the following:

$$ \sum_{n=1}^{\infty} \frac{1}{n} = \infty $$ $$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}$$

Here you have the sum of two infinite sequences that both converge towards zero, but the second one does it faster, making the sum finite.

Going back to integrals, check the family of functions $f_p(x)=\frac{1}{x^p}$, you'll see that, for some values of $p$, the integral of $f_p$ between, let's say, $0$ and $1$ has a finite value, while for others it doesn't

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David
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    Yes I did it for ln(X) and I'm getting. Finite value ,what is the reason for that and what is the meaning of explode,how to find which function is more "explosive" ;) – Nalin Yadav Aug 01 '19 at 13:26
  • With "explode", I mean converging towards $0$ faster. See https://en.wikipedia.org/wiki/Zeros_and_poles This applies to complex analysis, but it's a similar story in the real numbers – David Aug 01 '19 at 13:28
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    Check out this post on why there cannot be a 'slowest rate of divergence' of a series. – Toby Mak Aug 01 '19 at 13:29
  • Both the functions i mentioned approach non finite values at zero but still there areas are different (finite for one and infinite for other) , doesn't that seem counterintuitive.is there any intuitive ,non mathematical explanation for that? – Nalin Yadav Aug 01 '19 at 13:32
  • @NalinYadav Think for instance of the $1000th$ term. $\frac{1}{1000}$ is a lot bigger than $\frac{1}{1000^2}$. There is an easy proof that shows that the first series diverges, based on the fact that $\frac{1}{3} + \frac{1}{4}>\frac{1}{2}$, also $\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8}>\frac{1}{2}$, and so on. You can therefore get as many times $\frac{1}{2}$ as you want, simply by taking enough terms. A similar argument is not possible in the second case – David Aug 01 '19 at 13:35
  • Sorry I've edited my earlier comment ,please explain that – Nalin Yadav Aug 01 '19 at 13:37
  • If you want an example of an easily-provable convergent series, think of $$\sum_{n=1}^{\infty}10^{-n}$$, which clearly equals $0.11111111.......$ The key point is, again, the rate at which sequences go towards zero. Intuitively, think that you are adding a lot of things that get smaller and smaller, so it's some kind of $0 \cdot \infty$ indetermination. If the things we are adding don't get smaller fast enough, there will end up being too many of them for the sum to stop increasing at some point – David Aug 01 '19 at 13:38
  • Is it important to understand sequences (especially convergence and divergence) to understand this problem? Because I don't know anything about convergence , divergence etc.,isn't there any other explanation for that? – Nalin Yadav Aug 01 '19 at 13:43
  • Not really, but they are useful for contrast! Sum of sequences are just a discrete version of this continuous problem (integrals are just continuous summation). You can just as well work with limits. You are adding infinitely-many infinitely-thin srips. If they are too tall, you will get infinite area, otherwise, it could settle for a finite value – David Aug 01 '19 at 13:44
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    So in case of hyperbola there are more tall strips(that is,the function explodes faster or approaching infinity faster) as compared to lnx which makes the area of hyperbola infinite? – Nalin Yadav Aug 01 '19 at 13:52
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    @NalinYadav That's right! Indeed, if you try to compare them, you will see that $1/x$ is much bigger than $log(x)$ for very small $x$. Try taking the limit of their ratio to check! – David Aug 01 '19 at 13:57