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I have been puzzled by the following since I have learned about the logarithm, though I never tried to express what puzzled me more precisely. Here is an attempt.

$\int x^n = \frac{x^{n+1}}{n+1}$ if $n \neq -1$

and

$\int x^{-1} = \ln(x)$

This brings up the following question. Is there a sense in which the limit as $n \to -1$ of $x^{n+1}/(n+1)$ is $\ln(x)$? Here is a non-rigorous attempt to make sense of this. Let us replace $n+1$ by a complex variable $z$. We are interested in the limit, as $z$ goes to $0$, of the function

$f(x,z) = \frac{x^z}{z} = \frac{e^{z \ln(x)}}{z}$

For $z$ close to $0$, I expect something like

$f(x,z) = \frac{1}{z} + \ln(x) + O(z)$

So somehow, Mathematics decides to completely ignore the term $1/z$, which would give $\infty$, and focus on the "constant" term $\ln(x)$. This reminds me of divergent series and QFT. I used to frown at Physicists when they would throw away an "infinite" term, and keep a finite term, thinking that this did not really happen in the Mathematical world, albeit with some very important exceptions consisting of people like Euler, Ramanujan and Witten (who is both a Mathematician and a Physicist). But here is an example of that at the heart of basic calculus.

Can someone explain to me in a precise and rigorous way what is going on? Thank you.

Edit: I guess one can get something close to the interchange of an indefinite integral with a limit by modifying the meaning of limit in this case. Namely, you would like to take the limit as $z$ goes to $0$ of $f(x,z)$ but modulo functions of $z$ alone (that is to say, elements of the kernel of the partial $x$-derivative). This actually makes sense (I think) for linear operators depending on a parameter. To be able to interchange the limit with respect to the parameter with the general solution "operator", one has to take the limit of the entire family of solutions with respect to that parameter. These are some elementary thoughts, but I think my curiosity is more or less satisfied.

Malkoun
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  • This has been asked here several times. I can't find the question I'm thinking about, but this answer should do too: http://math.stackexchange.com/a/45395/1242 – Hans Lundmark Mar 29 '17 at 19:10
  • @HansLundmark Thank you. Ok, I did not see this post, but still, it is not exactly the answer I was looking for. I mean, of course, I am aware of the content of the answer in the post you have referenced. But I was looking for a more general answer in some sense, that would explain the phenomenon, and not just the specific example. Do you see what I mean? – Malkoun Mar 29 '17 at 19:14
  • If you're aware that you get the correct limit using the definite integral, then this whole question is a bit strange. The point is not “ignoring an infinite term”, it's choosing the right constant of integration so that the limit exists. (That is, consider the antiderivative $x^{n+1}/(n+1) + C$, with the suitable choice $C=-1/(n+1)$.) – Hans Lundmark Mar 29 '17 at 19:20
  • @HansLundmark I guess what I would like, is some general theorem, where a certain limit is obtained like the above, by throwing away some term that would give infinity, or at least some satisfactory general explanation. – Malkoun Mar 29 '17 at 19:20
  • @HansLundmark I understand what you are writing. To make my concern a bit more precise, the way I wrote things, the constant of integration is in some sense infinite, so it feels like there is a Riemann sphere appearing somewhere. – Malkoun Mar 29 '17 at 19:24
  • Well, there is really no way of saying what the constant of integration is. If you take $(x^{n+1}-1)/(n+1)+C$ as your starting point, you would say that $C=0$ is the natural thing, and that $x^{n+1}/(n+1)$ feels “off” by something which approaches infinity as $n \to -1$. – Hans Lundmark Mar 29 '17 at 19:32

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