TITLE: What is so special about $\alpha=-1$ in the integral of $x^\alpha$?
QUESTION [54 upvotes]: Of course, it is easy to see, that the integral (or the
antiderivative) of $f(x) = 1/x$ is $\log(|x|)$ and of course for
$\alpha\neq - 1$ the antiderivative of $f(x) = x^\alpha$ is
$x^{\alpha+1}/(\alpha+1)$.
I was wondering if there is an intuitive (probably geometric)
explanation why the case $\alpha=-1$ is so different and why the
logarithm appears?
Some answers which I thought of but which are not convincing:

Taking the limit $\alpha=-1$ either from above or below lead to diverging functions.
Some speciality of the case $\alpha=-1$ are that both asymptotes are non-integrable. However, the antidrivative is a local thing, and hence, shouldn't care about the behavior at infinity.

REPLY [2 votes]: OK, suppose you know some calculus but you fall asleep as your professor talks about the exponential function. Your true love is algebra and you've noticed the following:
$x \mapsto x^{-1}$ is a homomorphism of multiplication. You look at different forms of the function on algebraic objects,
$\Bbb R^* \to \Bbb R^*$
$(0, +\infty) \to (0, +\infty)$
$(0, 1] \to [1, +\infty)$
taking the inverse of a product to the product of the inverses.
As you study the mapping $(0, +\infty) \to (0, +\infty)$ you find it interesting to look at how the set
$\{\dots,6^{-1},5^{-1},4^{-1},3^{-1},2^{-1},1^0,2^1,3^1,4^1,5^1,6^1,\dots\}$
is mapped under inversion. It looks like multiplying by $-1$ on $\Bbb Z$, and as you explore it further, you wonder if you can create a homomorphism that takes multiplication to addition,
$(0, +\infty) \to \Bbb R$
that somehow uses the inversion function $1/x$. You are looking for a function that dilates and is negative when operating  on arugments less than $1$, is equal to $0$ at $1$, and is a contraction and positive when operating on arguments greater than $1$. 
But $1/x$ is greater than $1$ and explodes near 0 when $x \lt 1$, and is less than $1$ and shrinks to zero near $+\infty$ when $x \gt 1$. With a flash of insight you think, just maybe, the function
$F(x) =\int_1^x u^{-1}du$
will do the trick. It feels like it will stretch $(0, +\infty) \to \Bbb R$ in the right way and it takes $1$ to $0$ with $F^{'}(1) = 1$. 
OK, time to see if integrating a multiplicative homomorphism with the lower limit the identity $1$ will give you a new homomorphism that takes multiplication to addition.