I'm asked to calculate $\int_{|z| = 1} z^{n} \log z dz$ in two ways:
(1) if $\log 1 = 0$;
(2) if $\log (-1) = i \pi$.
I understand it means that in case (1) I have to work with the principal branch of the logarithm, that is, with $-\pi < \arg(z) < \pi$, and in case (2), with the branch $0 < \arg(z) < 2\pi$.
So I start my calculations. In case (1), I have that:
$\log (z) = \ln |z| + i \theta$
where $\theta$ is the (unique) number between $-\pi$ and $\pi$ s.t. $z = |z| e^{i \theta}$. Now i need a parametrization for the unit circle, so I choose:
$\gamma(t) = e^{it}$
with $- \pi \leq t \leq \pi$. This way, $\log (e^{it}) = it$. Then, it should be a matter of computing the integral:
$\int_{-\pi}^{\pi} – t e^{(n+1)t} dt$
splitting in the cases $n \neq -1$ and $n = -1$. But in the case $n \neq -1$, the right answers are given as:
(1) $\frac{2\pi i}{n+1}$, (2) $(-1)^{n+1}\frac{2\pi i}{n+1}$
and I get to the results with the opposite sign. What am I doing wrong?
NOTE: If I parametrize with $0 \leq t \leq 2\pi$ in case (I) I get the right answer, but, then, how can I make sure that $\log$ is defined for every $t$?
Best Answer
Let's do the general case, since the particular assumptions are unclear. Designate a point $e^{i\theta}$ of the unit circle $\mathbb T$ as the branch cut of $\log$. On the complementary arc $\mathbb T\setminus\{e^{i\theta}\}$ the logarithm has single-valued branches which differ by a multiple of $2\pi i$. When $n\ne -1$, it does not matter which branch we use, because $\int z^n (2\pi i)\,dz =0$. The integral is
$$\begin{split}\int_{\theta}^{\theta+2\pi} \exp(int)\, it\, i\exp(it)\,dt &= - \int_{\theta}^{\theta+2\pi} t \,\exp(i(n+1)t) \,dt \\ &= \frac{1}{(n+1)^2} \exp(i(n+1)t) (i(n+1)t-1) \bigg|_\theta^{\theta+2\pi} \\ &= \frac{2\pi i}{n+1} \exp(i(n+1)\theta) \end{split} $$
So, the two given answers correspond to cutting the circle at $\theta=0$ and at $\theta=\pi$. (Which gives a way to re-interpret the botched problem statement.)
Now take $n=-1$. Then $$\begin{split}\int_{\theta}^{\theta+2\pi} \exp(int)\, it\, i\exp(it)\,dt &= - \int_{\theta}^{\theta+2\pi} t\,dt \\ &= -\frac{(\theta+2\pi)^2-\theta^2}{2} =-2\pi (\theta +\pi) \end{split} $$ Here the choice of branch is encoded in the value of $\theta$: we get different results from $\theta$, $\theta+2\pi$, etc., although the cut is in the same place.