6
$\begingroup$

I tried three times, but in the end I am concluding that it equals infinity, after parametrizing, making a substitution and integrating directly (since the Residue Theorem is not applicable, because the unit circle encloses an infinite amount of non-isolated singularities.)

Any ideas are welcome.

Thanks,

$\endgroup$
5
  • 2
    $\begingroup$ do u know what a branch cut is? $\endgroup$
    – tired
    Commented Jan 6, 2016 at 21:57
  • 3
    $\begingroup$ You'd have to define the function on the unit circle, first. $\endgroup$ Commented Jan 6, 2016 at 21:59
  • 3
    $\begingroup$ How could a function bounded on the unit circle give infinity for its integral? $\endgroup$
    – zhw.
    Commented Jan 6, 2016 at 22:03
  • $\begingroup$ Hi @thomasandrews, hmmm ... I parametrized and tried to integrate explicitly ... of course the lower limit of 0 is problematic .... $\endgroup$
    – User001
    Commented Jan 6, 2016 at 22:12
  • $\begingroup$ Hi @ThomasAndrews, after parameterizing, there is no more log, as ln|z| = 0 on the unit circle, as the answer below shows. I missed this one :-( Thanks, $\endgroup$
    – User001
    Commented Jan 6, 2016 at 22:33

2 Answers 2

12
$\begingroup$

If $\log z$ is interpreted as principal value $${\rm Log}z:=\log|z|+i{\rm Arg} z\ ,$$ where ${\rm Arg}$ denotes the polar angle in the interval $\ ]{-\pi},\pi[\ $, then the integral in question is well defined, and comes out to $-2\pi i$. (This is the case $\alpha:=-\pi$ in the following computations).

But in reality the logarithm $\log z$ of a $z\in{\mathbb C}^*$ is, as we all know, not a complex number, but only an equivalence class modulo $2\pi i$. Of course it could be that due to miraculous cancellations the integral in question has a unique value nevertheless. For this to be the case we should expect that for any $\alpha\in{\mathbb R}$ and any choice of the branch of the $\log$ along $$\gamma:\quad t\mapsto z(t):=e^{it}\qquad(\alpha\leq t\leq\alpha+2\pi)$$ we obtain the same value of the integral. This boils down to computing $$\int_\alpha^{\alpha+2\pi}(it+2k\pi i)\>ie^{it}\>dt=-\int_\alpha^{\alpha+2\pi}t\>e^{it}\>dt=2\pi i\>e^{i\alpha}\ .$$ During the computation several things have cancelled, but the factor $e^{i\alpha}$ remains. This shows that the integral in question cannot be assigned a definite value without making some arbitrary choices.

$\endgroup$
2
  • $\begingroup$ Hi Christian, I'm wondering how you get $2k\pi i$ in the integrand from $i{\rm Arg} z$? $\endgroup$ Commented Jun 6, 2017 at 3:38
  • $\begingroup$ @MathStudent1324: I had said "for any $\alpha\in{\mathbb R}$ and any choice of the branch of the $\log$". Note that the capital A in ${\rm Arg}\,z$ refers to the principal branch and the cut along the negative real axis. $\endgroup$ Commented Jun 6, 2017 at 14:12
5
$\begingroup$

I would do it likes this:

Let $z = e^{i\theta}$. Then $\mathrm{d} z = i e^{i\theta} \mathrm{d} \theta$. Depending on you branch cut, $\ln(z) = i(\theta + 2\pi n)$ for whole $n$. Therefore the integral becomes $$ \int_0^{2 \pi} - (\theta + 2 \pi n) e^{i \theta} \mathrm{d} \theta \, . $$ Integrating by parts, I get $$ i [\theta e^{i \theta}]_0^{2 \pi} - [e^{i \theta}]_0^{2 \pi} + 2\pi n i [e^{i \theta}]_0^{2 \pi} = 2 \pi i \, . $$ Therefore $$ \oint_{|z| = 1} \ln z \, \mathrm{d}z = 2 \pi i $$

$\endgroup$
3
  • $\begingroup$ Apart from the conceptual issues, your integral should come out to $2\pi i$. $\endgroup$ Commented Jan 7, 2016 at 12:09
  • $\begingroup$ You are right -- my bad. $\endgroup$
    – SSF
    Commented Jan 7, 2016 at 12:10
  • $\begingroup$ I corrected it. $\endgroup$
    – SSF
    Commented Jan 7, 2016 at 12:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .