Why does $f(z) = z^n$ have no antiderivative only for $n=-1$?

Question

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 $-1$?

Answer 1

The reason is that the power rule for derivatives won't work: we get division by zero. Of course, as we know from calculus, this is the birth of the natural log. Well, the complex logarithm is analogous.

It is rather amazing that when defining, in the real case, $\ln x=\int_1^x1/t\operatorname dt$, it turns out that we get the inverse of the function $e^x$, where $e=\lim_{x\to\infty}(1+1/x)^x$.

So, in other words, $\ln x=\log_ex$.

Of course, in the complex case we need a branch of $\log$. But branches of $\ln$ are defined in terms of the real $\ln$ via $\ln z=\ln|z|+2\pi k+i \operatorname{arg}z$.

Answer 2

When you integrate around the circle, the little bits $x^k dx$ rotate through $k+1$ multiples of $2\pi$, so they cancel each other out unless $k=-1$.
We want the integral from $x$ to $y$ along one path to be the same as the integral along another path. Then the integral is just a function of $x$ and $y$. Imagine putting a path from $y$ to $w$ on the end of that, then the integral from $x$ to $w$ equals the integral from $x$ to $y$ plus the integral from $y$ to $w$. So the integral is now a function $F(y)-F(x)$.
If two different paths from $x$ to $y$ have the same integral, then the loop from $x$ to $y$ along the first path, and back along the second path, has to be $F(y)-F(x)+F(x)-F(y)=0$. To repeat, the integral along a closed loop has to be zero.
Suppose a loop can be shrunk to zero, and f(x) is bounded inside the loop. Break up the area into $N^2$ tiny areas of width $O(1/N)$. The integral around the large loop equals the sum of the integrals around all the small areas because the inner paths cancel out. Now we want a condition that makes the integral around a small area to be $O(N^{-3})$. From memory, that condition is the 'Cauchy Riemann' equations, which is the condition the function f is a function of $z$ and not its conjugate $\overline{z}$.
Now the integral is the sum of $N^2$ integrals, each $O(N^{-3})$, so it is $O(1/N)$. Let $N$ be very large, and so the integral is zero. So: The integral around a loop is zero, when $f(z)$ is bounded within the loop, and $f(z)$ does not involve the conjugate.
The only thing left in this case is an integral around zero. This boils down to a single integral along the unit circle. So $z=e^{i\theta}$,$dz=e^{i\theta}id\theta$, $z^kdz = ie^{(k+1)\theta}d\theta$, and the result follows.

Answer 3

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 $L'(z)=1/z$.