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I don't understand why some functions that contain a singularity in the domain of integration are integrable but others are not.

For example, consider $f(x) = -\log(x)$ and $g(x) = \frac{1}{x}$ on the interval $[0, 1]$. These functions look very similar when they are plotted but only $f(x)$ can be integrated.

  1. What is the precise mathematical reason(s) that makes some functions with singularities integrable while others are not?
  2. Are $\log$ functions the only functions with singularities that can be integrated or are there other types of functions with singularities that can be integrated?
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    $\begingroup$ How about $1/x^p$, $0<p<1$? $\endgroup$ – user175968 Jan 2 '17 at 11:46
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    $\begingroup$ The integral in $[0,1]$ is the same that the integral in $(0,1)$. The reason is not the singularity, the reason is the divergence or convergence of the integral. $\endgroup$ – Masacroso Jan 2 '17 at 11:57
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It's just whether or not the area under the curve is finite or not. It's doesn't matter that there is an asymptote.

You might consider the area under the curves $y=e^{-x}$ and $y=1/x$ for $x>0$. These are really the same two curves you mention, just along the other axis.

It's akin to the idea that an infinite series may or may not converge; just because there are infinitely many terms in a series doesn't mean the series must diverge.

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  • $\begingroup$ I don't see how these are the "same two curves"..one of them, $y=e^{-x}$, has a precise value at $x=0$ and so it can be integrated on interval $[0, \infty)$, whereas $y=1/x$ gets arbitrarily large near zero and can't be integrated on that interval so the area under this curve is undefined. I don't understand what point you are making in your post? $\endgroup$ – ManUtdBloke Jan 6 '17 at 22:00
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    $\begingroup$ @eurocoder : You misunderstand--I mean the two curves I mention are essentially the same two curves you ask about, just along the other axis. I added this to my answer to clarify. $\endgroup$ – MPW Jan 7 '17 at 4:56
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    $\begingroup$ They're note essentially the same for the purposes of integration, $\log x$ is not integrable on $(0,\infty)$ but $e^{-x}$ is. Perhaps you mean something else, but it's not clear from your wording what notion of "same" you are using and why it's relevant to the question at-hand. $\endgroup$ – Adam Hughes Jan 8 '17 at 9:26
  • $\begingroup$ @AdamHughes : Of course, I'm talking about finding the area between the alternate curves and the other axis, as I already said. They are exactly the same thing. $\endgroup$ – MPW Jan 25 '17 at 14:16
  • $\begingroup$ @MPW maybe as curves I different coordinate systems, but integration users a consistent orientation, so this doesn't amount to much mathematically to say they're the same in an integration context if they're not even being considered on the same ground for integration's necessary context. $\endgroup$ – Adam Hughes Jan 25 '17 at 14:20
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Think about it this way - what's the inverse?

$$y = \frac{1}{x}; x = \frac{1}{y}$$ $$y = -\log x; x = e^{-y}$$

Looking at it this way, it's clear that as $y$ shoots off to infinity, $x$ approaches zero much faster in one case than in the other.

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  • $\begingroup$ Ok so one approaches zero much faster but how fast precisely does a function have to approach zero to make it integrable? ..what is the minimum 'speed' a function needs to have for it to be integrable? $\endgroup$ – ManUtdBloke Jan 6 '17 at 22:08
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    $\begingroup$ Well, there isn't a minimum exactly... it has to approach quickly enough that the inverse converges at infinity. So, for instance, $\frac{1}{x^{1-\epsilon}}$ would converge at zero, since $\frac{1}{x^{1+\epsilon}}$ converges at infinity. $\endgroup$ – user361424 Jan 7 '17 at 4:09
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The key is how fast the function is diverging.

Regarding your two examples, $-\log$ is going really fast close to the $y$-axis so it is integrable, but not $x\mapsto \frac {1}{x}$.

  • You have $$\int_a^1 -\log(x)\mathrm dx=a(1-\log(a))+1\xrightarrow[a\to 0^+]{} 1<\infty.$$ So this function is integrable.

  • You have $$\int_a^1 \frac 1x\mathrm dx=-1+\frac 1{a^2}\xrightarrow[a\to 0^+]{} +\infty.$$ So this function is not integrable.

Regarding your second question, $\log$ functions are absolutely not the only one. To convince yourself, take for instance $x\mapsto \frac 1{\sqrt{x}}$ on $(0,1)$.

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  • $\begingroup$ How did you get a result of $1$ for the first integral? $\endgroup$ – ManUtdBloke Jan 6 '17 at 22:08
  • $\begingroup$ Also, wolfram alpha evaluates $a (1 - \log a)$ as $a \to 0$ to be $0$. $\endgroup$ – ManUtdBloke Jan 6 '17 at 22:43
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    $\begingroup$ @eurocoder Oups, my mistake : I forgot to write the $+1$. Thanks for the correction. $\endgroup$ – E. Joseph Jan 8 '17 at 9:21
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Simple: $\quad -\log x=_0 o\Bigl(\dfrac1{\sqrt x}\Bigr)$ and the integral of $\dfrac 1{\sqrt x}$ on [0,1] is convergent.

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  • $\begingroup$ Where is the $o(1/\sqrt{x})$ coming from and how does convergences of that function on the interval imply convergence of $-\log{x}$? $\endgroup$ – ManUtdBloke Jan 6 '17 at 22:05
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    $\begingroup$ The $o(1/\sqrt{x})$ corresponds to the high school limit $\sqrt x\log x\xrightarrow[x\to 0^+]{}0$. The convergence of the integral is a well-known theorem in asymptotic analysis. $\endgroup$ – Bernard Jan 6 '17 at 22:31
  • $\begingroup$ But the problem I have is that why are some functions with singularities integrable and orthers aren't. So you are saying $\frac{1}{\sqrt{x}}$ is integrable, but we know $\frac{1}{x}$ is not. Why is this? It seems the exponent on $x$ in the denominator determines whether a function will be integrable. So if that is the case 'where' precisely (for what value of exponent) does a function fail to be integrable? $\endgroup$ – ManUtdBloke Jan 10 '17 at 8:17
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    $\begingroup$ Since the anti derivative of$\dfrac1{x^r}$ is $-\dfrac1{(r-1)x^{r-1}}$ the answer is $r-1<0$ (there must be no denominator), i.e. $r<1$. $\endgroup$ – Bernard Jan 10 '17 at 9:43
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    $\begingroup$ Actually, the ‘theorem’ (a great word for a simple observation) is valid for non-negative functions $f$ and $g$ such that $f=O(g)$: if the integral of $g$ converges, the integral of $f$ does too (obvious since the integral preserves inequalities). As $f=o(g)\implies f=O(g)$, the result follows. $\endgroup$ – Bernard Feb 21 '19 at 14:58

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