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I'm sure this is elementary, but:

If a 1-D function is bounded everywhere except at a point P (measure zero), then the integral exists. So why should it matter if P is at the end of the integration interval?

Presumably the same is true of a line integral in, say, R3 --- which would seem to imply that the path for a closed line integral can include a countable number of discontinuities. But what happens if the closed integral is calculated in such a way that the endpoints a=b are where the function is unbounded?

Taking this one more step, the closed line integral, by Stokes' theorem, can be written as a surface integral of the curl. As long as the regions is simply connected. But if, say, simply connectedness fails at the origin, why should it matter -- isn't that just a set of measure zero for the surface integral?

Clearly, I'm confused and/or missing something basic.

thanks!

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How do you mean unbounded? You could just redefine the function in all "bad" points if they remain to be a null set. Note that a countable set is of measure zero, but it can be way large. For example the Cantor set. –  Jonas Teuwen Oct 30 '11 at 20:48
    
A function cannot be unbounded at a point. Whenever you think about a function, like say $\tan(x)$ which is not bounded "at" $\frac{\pi}{2}$, it is actually not bounded on any interval containing or touching $\frac{\pi}{2}$. –  N. S. Oct 30 '11 at 20:51
    
Maybe by terminology is poor. My apologies. So let me try an example: $\int_{-a}^a \frac{dx}{x^2}$ exists, but --- at least formally --- is not the same as $2\int_0^a \frac{dx}{x^2}$ because the latter integral doesn't exist. The difference between the two expressions is the origin, a set of measure zero, at the endpoint of the second integral. –  gilonik Oct 30 '11 at 22:33
    
What do you mean by "simply connectedness fails at the origin"? You mean at "the endpoints a=b"? Can you give me an example? –  savick01 Oct 30 '11 at 22:35
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@Jonas, the way I read your comment, it presents the Cantor set as an example of a countable set. Maybe you didn't mean it to be read that way. Anyway, the Cantor set is not countable. –  Gerry Myerson Oct 30 '11 at 23:49

2 Answers 2

It does matter because when you calculate the integral of $\int_0^1 x^{-2} dx$, you actually calculate a different KIND of integral than the integral say $\int_0^1 x dx$.

If you go back to definitions: a 'normal' Riemann integral is a limit of Riemann sums. When you want to define $\int_0^1 x^{-2} dx$ you can't use this definition, because any sum can be arbitrarily huge - it is a matter of taking an appropriate midpoint near $0$, which generates arbitrarily big value. So you say: I can calculate a limit of integrals $\lim_{\varepsilon \to 0}\int_\varepsilon^1 x^{-2} dx$. But look: it is kinda trick, that you take the limit. If you want, you can say: I want the number: $\lim_{\varepsilon \to 0} \int_{-1}^{-\varepsilon} x^{-2} dx + \int_\varepsilon^1 x^{-2} dx$, to be the value of the integral $\int_{-1}^1 x^{-2} dx$.

That really shows weakness of the Riemann integral, so people usually use different definitions (like Lebesgue integral), to be able to calculate the integral of the characteristic function of $\mathbb{Q}$ or things like that. They prove that it is equal to traditional Riemann integral if Riemann integral is defined. However there are some cases where Riemann integral in the second version (with $\lim$) is defined and Lebesgue integral is not: that is because for example $\int_1^\infty{}\frac{sin(x)}{x}$ for Riemann is the limit when you go to infinity and Lebesgue looks at the whole halfline and he sees infinitely big surface (I mean "the surface below the graph") above zero and below zero and he does know how much $\infty - \infty$ is.

At the beginning you say bounded and then you switch to discontinuous, I don't get it. You mean that the path is not continuous or the function which is integrated along the path? I haven't heard about discontinuous path (see: path), but integrals on $\mathbb{R}$ and line integrals have indeed a lot in common.

Simple connectedness - it is a term from topology not measure theory and in topology single points sometimes really matter. For a topologist a circle without one point is an interval which is of course completely different from a circle (it is simply connected for example, but you can also notice that if you take one point out of a circle it is still connected, what is not true in case of interior points of an interval). For a topologist a disk without one point is in some way more similar to the circle than to the whole disk. EDIT: So topology sees no difference if in your formula there is a singularity at zero or at some disk (with positive measure) around it. END

Going back to simple connectedness (s.c): taking one point from an open disk spoils its s.c. And it does matter if you - for example - want to say that particular line integrals are equal, because the endpoints are equal - if you have s.c, you can use a homotopy between those paths to prove such a statement and you have a problem, when there is no homotopy (because of no s.c.). In that case an infinite value at one single point may spoil everything. Well, such is life.

There are actually no theorems in my answer, but I hope it helps a bit.

P.S. If you calculate the 'limit' Riemann integral and normal Riemann integral exists, they are equal, so you can use only the 'limit' one, but it is somehow pointless.

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Re: Stokes' theorem -- It's my understanding that the singularity at the origin renders the surface integral of the curl undefined. If so, it's not clear to me what that statement has to do with homotopy -- and I'm left wondering why a set of measure zero results in a non-existent integral. So then I guess what you're saying is that the surface integral does exist, but homotopy considerations show that instead it's Stokes' theorem which fails. Yes? –  gilonik Oct 31 '11 at 0:08
    
Yes, it is about Stokes not about the integral. But it was a general speach about simple connectedness as a topological term (which is sometimes applied to integrals). Actually Stoke's theorem may be applied to sets like an annulus. In this case you have to remember, that its boundary consists of two parts (and the integrals have to be counted "in the right direction"). –  savick01 Oct 31 '11 at 0:39
    
Could Stokes be patched up, at least in the $\frac{\hat{\phi}}{r}$ example, but using a delta function? It sounds just like the divergence theorem applied to $\frac{\hat{r}}{r}$... –  gilonik Oct 31 '11 at 1:03
    
@gilonik I have a counterexample! If you add a constant to the function, then closed integral doesn't change, surface integral (of the derivative) doesn't change, but the value coming from delta function does change! –  savick01 Oct 31 '11 at 1:52
    
@gilonik My penultimate comment about infinities was a bullshit, since I mixed up values of a derivative and a function. Deleted. Anyway, trying to define value coming from delta function requires some decision on what to do with $dx$ and $dy$ (since those are good for 2D and you use delta for 1D), but it seems that any decission leads to problems with constants. By constant I mean $+5dx$ or $+5dx + 5dy$, whatever. –  savick01 Oct 31 '11 at 10:38

EDIT: as pointed out in the comments, even the principal value interpretation doesn't work for this particular function. I had in mind the situation where a principal value exists because positive and negative contributions cancel out, e.g., for ${\rm p.v.}\int_{-1}^1(1/x)\,dx=0$. So much of what follows is wrong and/or not relevant to the question at hand.

I would say that $\int_{-1}^1(1/x^2)\,dx$ doesn't exist, and neither does $\int_0^1(1/x^2)\,dx$. The first integral differs from the second one in that there is something called the "principal value" defined by $${\rm p.v.}\int_{-1}^1{1\over x^2}\,dx=\lim_{a\to0}\left(\int_{-1}^{-a}{1\over x^2}\,dx+\int_a^1{1\over x^2}\,dx\right)$$ and the principal value does exist, while no principal value exists for the second integral. But the principal value isn't the integral; it's a generalization of the integral.

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...which in common parlance is what we actually call the "integral". And if we can do this (i.e., find the p.v.) for any set of measure zero, then this is why integration through a countable number of singular points is not a problem. Fabulous! –  gilonik Oct 31 '11 at 0:12
    
Ehm... $\left(\int_{-1}^{-a} \frac1{x^2}dx + \int_a^1 \frac1{x^2}dx\right) = \frac 2a -2.$ I don't see how $\int_{-1}^1 \frac{dx}{x^2}$ has a principal value? –  kahen Oct 31 '11 at 0:35
    
@gilonik How do you want to integrate through $\mathbb{Q}$ in that way? –  savick01 Oct 31 '11 at 0:46
    
@kahen Oh come on, can't you see this tiny "1/" before 2 in the exponent? Yes, you've got us! It's probably time to sleep... –  savick01 Oct 31 '11 at 0:49
    
Clearly I didn't stop to think. This p.v. argument would only work for an odd function around a symmetric interval.... I guess I'm being dense, but I'm just looking for an intuitive understanding of why $\int_{-a}^b \frac{dx}{x^2}$ is OK but $\int_0^b \frac{dx}{x^2}$ is not. Somehow that zero does not matter in the first, but does in the 2nd.... –  gilonik Oct 31 '11 at 0:58

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