nt.number theory - How to put "x^a => 1/(a+1) x^(a+1)" and "x^-1 => log(x)" together

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.