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This innocent-looking problem came to me some years ago.

These two most basic integration formulas are, of course, disturbingly different, in the eyes of any good mathematicians [ just joking ;-) ].

The challenge is to put them together in a nice manner.

Smells like analytical continuation and the zeta(1) pole is playing a role here, so I don't expect the answer to be trivial. Bonus for anyone using representation theory or p-adic methods [yes, i am thinking about Tate's thesis], or any appearances of cohomology.

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    $\begingroup$ I'm puzzled by the question. Obviously you don't want someone to just say "$\log(x)$ is the limit as $a \to -1$ of $x^{a+1}/(a + 1)$". The word "challenge", and the last sentence of your question, suggest you know something that you're not telling us. It's not really in the spirit of this site to do that... share whatever you know! $\endgroup$
    – Tom Leinster
    Commented Mar 30, 2010 at 7:37
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    $\begingroup$ I'm not sure I get the question either, but I find your remark puzzling as well, Tom. Obviously we don't want someone to say "log(x) is the limit as $a \rightarrow -1$ of $\frac{x^{a+1}}{a+1}$" because it isn't true: that limit diverges! $\endgroup$
    – Pete L. Clark
    Commented Mar 30, 2010 at 8:36
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    $\begingroup$ The issue can be resolved by starting the integral from 1 to define the antiderivative. Then one takes the limit as a approaches -1 of (x^{a+1} - 1)/(a+1) and this really does equal log x. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 30, 2010 at 8:48
  • $\begingroup$ No, I don't know the answer at the moment... one day I shall sit down and try hard at it. The challenge is, $\log(x)$ is not the limit as $a \to -1$ of $x^{a+1}/(a+1)$, at least not in any naive way. Let me clarify the question a little bit. The simplest solution will be to embed $\log(x)$ and $x^n$ and possibly other functions in a nice space (probably Hilbert, etc., and the space shall not be very artificial), such that we can find a smooth functional "A" always sending $x^n$ to the correct anti-derivative [forget the constant factor]. $\endgroup$
    – Bo Peng
    Commented Mar 30, 2010 at 9:04
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    $\begingroup$ @Qiaochu: your comment is a perfectly good answer to the question; you might even want to leave it as an answer. The OP seems to want a fancier answer, but I don't yet understand why. $\endgroup$
    – Pete L. Clark
    Commented Mar 30, 2010 at 9:27

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First of all, I'll repeat what I said in the comments: writing $\frac{x^h - 1}{h} = \frac{e^{h \log x} - 1}{h}$ makes it fairly clear that as $h \to 0$, this expression tends to $\log x$, and setting $h = a + 1$ this is precisely the desired result.

I should mention that this result is implicit in a certain fact well-known to people who do competition math, which is as follows. Given non-negative real numbers $x_1, ... x_n$, let $A_p(x_1, ... x_n) = \sqrt[p]{ \frac{x_1^p + ... + x_n^p}{n} }$ denote the $p$-power mean for $p \neq 0$. For $p = 0$, define $A_0(x_1, ... x_n) = \sqrt[n]{x_1 ... x_n}$ (the geometric mean, and also the limit as $p \to 0$ of the above).

Theorem (Power Mean Inequality): If $p \le q$, then $A_p \le A_q$.

If you like fancy keywords, then I will bring to your attention that as $p \to \infty$ the $p$-power mean approaches $\text{max}(x_1, ... x_n)$, which one can think of as the "low-temperature limit" of ordinary addition becoming tropical addition. Then $p \to 0$ can be thought of as the "high-temperature limit," in which ordinary addition becomes multiplication instead. Somebody who knows more statistical mechanics than I do (that is, any) can probably tell you the physical significance of this.

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  • $\begingroup$ Nice $e^{h \log x}$ trick, although I cannot see why the well-known power mean inequality will have anything to do with it... $\endgroup$
    – Bo Peng
    Commented Mar 30, 2010 at 10:06
  • $\begingroup$ Essentially the same limit is involved in both computations. You should pretend that all those p-power terms turn into logarithms and the pth root turns into an exponential to see the connection. $\endgroup$
    – Qiaochu Yuan
    Commented Mar 30, 2010 at 10:28
  • $\begingroup$ As $p \to -\infty$, the $p$-mean approaches the minimum. $\endgroup$
    – Michael Lugo
    Commented Mar 30, 2010 at 12:53
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    $\begingroup$ So, rather than "tropical geometry", we should be speaking of "arctic geometry"? $\endgroup$
    – Pete L. Clark
    Commented Mar 30, 2010 at 13:52
  • $\begingroup$ Yes. Allen Knutson made a push a while back to convince tropical people that "tropicalization" should refer to deforming away from the low temperature limit, not towards it. But I think most people were nervous about changing terminology in a field that already has too many definitions. $\endgroup$
    – David E Speyer
    Commented Mar 30, 2010 at 14:16

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