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What is so special about $\alpha=-1$ in the integral of $x^\alpha$?

We all know the rule to find primitives (=antiderivatives) of powers of $x$:

$$ \int x^\alpha dx = \frac{x^{\alpha +1}}{\alpha +1}+C$$

whenever $\alpha \in \mathbb{R} \setminus \{-1\}$. In this case we know a primitive is given by the natural logarithm:

$$ \int \frac{1}{x} dx = \ln(x)+C$$

Doesn't this upset you? I mean, for every other power the rule is "algebraic" in some sense: you just get a different power and a coefficient. But for $\alpha = -1$ you get something trascendental! where does this "exception" come from? I can observe that the order of the pole at the origin is the same as the dimension of the space I'm integrating on...but I'm not satisfied enough with this.

I was not aware of this previous question

What is so special about $\alpha=-1$ in the integral of $x^\alpha$?

when I asked mine, I apologize. Anyway, all the answers to that question are "just" a collection of ways to see and say $\it{that}$ the case $\alpha =-1$ is different, but nothing substantial came out of that discussion, from my point of view. I was looking for something deeper, that could shed some light on $\it{why}$ the simples power has the hardest primitive.

To give you a taste of what I was thinking about, let me talk about something that maybe has nothing to do with this. If one wonders about the square-root function, he sees it is not well defined, because for example both $2$ and $-2$ are square roots of $4$. Saying that things just go like this, or that one has to choose a branch of the square-roots, is something you can do. But a better explanation would be talking about Riemann surfaces and considering the function not on $\mathbb{C}$ but on the sphere $\mathbb{CP}^1$. I was hoping something "like this" (or like something else, I don't know) was lurking beneath the surface...something more interesting that a bunch of computations.

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  • You do mean the anti-derivative by $\int_\mathbb{R}$, not the integral over the real line, right? – Sasha Jan 31 '13 at 18:07
  • yes, right! I mean the antiderivative. Maybe I should omit the subscript. – Lor Jan 31 '13 at 18:13
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    For $n \neq -1$, the holomorphic 1-form $z^n , dz$ is naturally defined on a simply connected domain (either $\Bbb{C}$ or $\hat{\Bbb{C}} \backslash 0$), and so it possesses a globally well-defined primitive on its domain of definition. In contrast, $\frac{dz}{z}$ has poles at both $0$ and $\infty$, so its natural domain is the punctured plane $\Bbb{C} \backslash 0$, which is not simply connected. So its primitive may be (and in fact is) defined only locally, and so it's unsurprising (at least to me) that it's a different sort of function than all the others. – Micah Jan 31 '13 at 19:32
  • @Micah : that's enlightening. Its natural domain is an open set that si omotopically equivalent to a circle, and we get a complex logarithm, which has a lot do with argument and "turning around". – Lor Feb 02 '13 at 09:42

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