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

Solution 1:

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.

Solution 2:

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