antiderivative of $\frac{1}{z(z-1)}$, complex logarithm

Question

I have the domain $\mathbb{C} \backslash [0,1]$ and want to show that $$\int_\gamma \frac{1}{z(z-1)}dz = 0$$ for all closed curves $\gamma$. I want to accomplish this by explicitly finding an antiderivative.

I did a partial fraction decomposition and got $$f(z) = \frac{1}{z(z-1)} = \frac{1}{z-1} - \frac{1}{z}$$

Now $$ F(z) =\text{log}(z-1) - \text{log}(z)$$ would be an antiderivative but I don't feel comfortable with the complex logarithm yet. Is my $F(z)$ well-defined on my whole domain? If not, how do I find a well-defined antiderivative?

Answer 1

The expression $$ F(z)=\log(z−1)−\log(z) $$ is not well-defined on your domain, because neither $\log(z−1)$ nor $\log(z)$ are holomorphic in $\mathbb{C} \backslash [0,1]$. One can probably solve that by choosing the arguments of the logarithms carefully, but it becomes much easier if the right-hand side is rewritten as $\log \frac{z-1}{z}$.

The Möbius transformation $T(z) = \dfrac{z-1}{z}$ maps $\mathbb{\hat C} \backslash [0,1]$ conformally onto $\mathbb{C} \backslash (-\infty,0]$, so you can define $$ F(z) = \log \frac{z-1}{z} $$ where $$\log w = \log|w| + i \arg w \quad (-\pi < \arg w < \pi) $$ is a holomorphic branch of the logarithm on $\mathbb{C} \backslash (-\infty,0]$.

Then $F$ is an "explicit" antiderivate of $$ \frac{1}{z-1} - \frac{1}{z} $$ This can either be verified directly, or you argue as follows: In any disc $D \subset \mathbb{C} \backslash [0,1]$ $$ F(z) = \log(z-1) - \log z + C $$ for some holomorphic branch of $\log(z-1)$ and $ \log z$ in $D$, and the result follows by differentiation.

Answer 2

The solution depends on your closed curve, which poles it does contain in the interior of γ. Since you excluded [0,1] any closed curve will not contain both the poles, or it will contain both and can't have $γ(t)=0$ or $γ(t)=1$. (Curve must have a finite length)

Lets consider:\begin{align}\oint_\gamma f(z)\, dz\end{align} where $f(z)=\frac{1}{z(z-1)}$

Theorem: \begin{align}\oint_\gamma f(z)\, dz = 2\pi i \sum_{k=1}^n \operatorname{I}(\gamma, a_k) \operatorname{Res}( f, a_k )\end{align}

In this specific case $n=2$ and $a_1=0,a_2=1$

Since a close curve will contain both of them, or none of them, then:

\begin{align}\operatorname{I}(\gamma, a_1)=\operatorname{I}(\gamma, a_2)\end{align}

Therefore: \begin{align}\oint_\gamma f(z)\, dz = 2\pi i \operatorname{I}(\gamma, a_1) \sum_{k=1}^2 \operatorname{Res}( f, a_k )\end{align}

Theorem 2: Pole order = 1 $\Rightarrow$ \begin{align}Res(f,z_i) = \lim_{z\to z_i} \frac{z-z_i}{f(z)}\end{align} At $z=0$,\begin{align}Res(f,0) = \lim_{z\to 0} \frac{z}{z(z-1)} = \lim_{z\to 0} \frac{1}{z-1}=-1\end{align} At $z=1$ \begin{align}Res(f,1) = \lim_{z\to 1} \frac{z-1}{z(z-1)}=1 \end{align}

Therefore:

\begin{align}\int_\gamma \frac{1}{z(z-1)}dz = 2\pi i \operatorname{I}(\gamma, a_1) \sum_{k=1}^2 \operatorname{Res}( f, a_k )=2\pi i \operatorname{I}(\gamma, a_1) (1-1)=0\end{align}

Answer 3

Another possible argument (in response to your above comment): With the substitution $$ z = \frac 1w \, ,\quad dz = -\frac{dw}{w^2} $$ you get $$ \int_\gamma \frac{1}{z(z-1)} \,dz = \int_{\gamma'} \frac{-1}{\frac 1w(\frac 1w-1)} \,\frac{dw}{w^2} = \int_{\gamma'} \frac{1}{w-1} \, dw $$ where $\gamma'$ is a closed curve in $D = \Bbb C \setminus [1, \infty) $. $D$ is simply-connected and $1/(w-1)$ holomorphic in $D$. It follows from Cauchy's integral theorem that the integral is zero.