# What is the integral of log(z) over the unit circle?

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,

• do u know what a branch cut is? – tired Jan 6 '16 at 21:57
• You'd have to define the function on the unit circle, first. – Thomas Andrews Jan 6 '16 at 21:59
• How could a function bounded on the unit circle give infinity for its integral? – zhw. Jan 6 '16 at 22:03
• Hi @thomasandrews, hmmm ... I parametrized and tried to integrate explicitly ... of course the lower limit of 0 is problematic .... – User001 Jan 6 '16 at 22:12
• 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, – User001 Jan 6 '16 at 22:33

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.

• Hi Christian, I'm wondering how you get $2k\pi i$ in the integrand from $i{\rm Arg} z$? – MathStudent1324 Jun 6 '17 at 3:38
• @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. – Christian Blatter Jun 6 '17 at 14:12

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 $$\oint_{|z| = 1} \ln z \, \mathrm{d}z = 2 \pi i$$

• Apart from the conceptual issues, your integral should come out to $2\pi i$. – Christian Blatter Jan 7 '16 at 12:09
• You are right -- my bad. – SSF Jan 7 '16 at 12:10
• I corrected it. – SSF Jan 7 '16 at 12:27