TITLE: Why does $f(z) = z^n$ have no antiderivative only for $n=-1$?
QUESTION [10 upvotes]: The complex valued function $f(z) = z^n$ has an analytic antiderivative on $\mathbb{C} \setminus \{0 \}$ for every $n$ except for $n=-1$. What is so special about $-1$? 
To show why this is such an anomaly, imagine if $z^n$ had an analytic antiderivative on $\mathbb{C} \setminus \{0 \}$ for every $n$ except for $n=3456$. People would demand to know what is so special about $3456$. However, it seems like no one feels the need to explain the anomaly at $n = -1$. What is going on at $-1$?

REPLY [10 votes]: What's special about $n=-1$ is that it's the only exponent such that $(az)^n\,d(az)=z^n\,dz$.  Here's why that's important:
Keep in mind that if $F(z)$ is analytic in a domain $\Omega$ (such as $\mathbb{C}\setminus\{0\}$), then $F(b)-F(a)=\int_a^bF'(z)\,dz$, where the $\int_a^b$ is understood as a contour integral along any path entirely within $\Omega$ connecting two points $a$ and $b$ in $\Omega$. Now suppose $f(z)=1/z$ had an antiderivative in the domain $\mathbb{C}\setminus\{0\}$, i.e., there were an analytic function $F(z)$ such that $F'(z)=1/z$. Then the function
$$L(z)=F(z)-F(1)=\int_1^z{d\omega\over\omega}$$
is also analytic in the domain $\mathbb{C}\setminus\{0\}$.  But we now see that, for any two nonzero complex numbers $a$ and $b$, we have
$$L(ab)=\int_1^{ab}{d\omega\over\omega}=\int_1^a{d\omega\over\omega}+\int_a^{ab}{d\omega\over\omega}=\int_1^a{d\omega\over\omega}+\int_1^b{d(a\omega)\over a\omega}=\int_1^a{d\omega\over\omega}+\int_1^b{d\omega\over\omega}=L(a)+L(b)$$
(i.e., we've just confirmed that the function $L$ behaves like a logarithm). In particular, if $\zeta$ is an $N$th root of unity, we have
$$NL(\zeta)=L(\zeta^N)=L(1)=\int_1^1{d\omega\over\omega}=0$$
Since the totality of all $N$th roots of unity (for all $N\in\mathbb{N}$) is dense on the unit circle, this shows that the analytic function $L(z)$ is identically $0$ on the unit circle, and that implies $L(z)$ is identically $0$ on its entire domain, which implies $L'(z)$ is also identically $0$, which is a contradiction, since $L'(z)=1/z$.