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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 $t$?

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    $\begingroup$ I think you misunderstood how (1) and (2) were meant (not that I can blame you, I can only guess an intention from the results). In (1), you should apparently use $0 \leqslant \arg z < 2\pi$, and in (2) $-\pi < \arg z \leqslant \pi$ (or possibly $\pi \leqslant \arg z < 3\pi$). Well, that way, at least the given results come out. $\endgroup$ – Daniel Fischer Jul 23 '13 at 21:47
  • $\begingroup$ I'm afraid you're right, reverse engineering seems to be the only way to understand what's going on here. Also, after getting lots of sign-changed answers, including for problems available on the web, I convinced myself that my teacher considers the principal branch to be the one obtained by removing the nonnegative real axis, which is usually what physicists and engineers do. $\endgroup$ – ulilaka Jul 23 '13 at 23:32
  • $\begingroup$ I have three words for your teacher, then: What the fourletterword? $\endgroup$ – Daniel Fischer Jul 23 '13 at 23:38
  • $\begingroup$ I agree with you. But it seems, for example, that they do it in this MIT OCW too without further comments: ocw.mit.edu/courses/mathematics/… $\endgroup$ – ulilaka Jul 24 '13 at 4:52
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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.

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