2
$\begingroup$

Is it possible to calculate the following integral explicitly:

$$ I = \frac{2i}{\pi} \int_\gamma \ln(z) dz, $$ where $\gamma$ can be a disk at the origin of $\mathbb{C}$? Unfortunately $\ln$ isn't holomorphic at $0$ so we can't use Cauchy's integral theorem.

But it seems like it should be possible as if we identify $\mathbb{C}$ with $\mathbb{R^2}$ then this would be a considered a weakly singular integral in potential theory in $\mathbb{R}^2$ which I know is integrable.

So can this integral be evaluated explicitly if the contour is a disk at the origin?

$\endgroup$
3
  • $\begingroup$ How do even define the logarithm? There is no branch in $\mathbb{C}\setminus\{0\}$. It isn't a function with a singularity at the origin. It is a function which can be defined only on a specific branch. $\endgroup$
    – Mark
    Commented Jan 11, 2019 at 21:27
  • $\begingroup$ Yes there is a singularity at the origin..and it has a branch cut. I would like to be able to evaluate it explicitly if possible. $\endgroup$ Commented Jan 11, 2019 at 21:34
  • 1
    $\begingroup$ If $\gamma$ is a curve in $\mathbb{C}^*$ and the principal branch of $\log(z)$ is continued analytically along $\gamma$ then $\int_\gamma \log (z) dz= F(\gamma(1))-F(\gamma(0))$ where $F(z) = z\log z-z$ and $F(\gamma(t))$ is continued continuously in $t \in [0,1]$. With the unit circle $\gamma(t) = e^{2i \pi t}, t \in [0,1] $ then $\gamma(0) = \gamma(1)=1, F(\gamma(1) ) = 2i\pi - 1, F(\gamma(0)) = -1$. $\endgroup$
    – reuns
    Commented Jan 11, 2019 at 21:54

1 Answer 1

1
$\begingroup$

This is a difficult question because there is some monodromy.

You can integrate this directly. If $z=r\mathrm e^{\mathrm i\theta}$, then we can circle the origin many different ways, e.g. $0 < \theta < 2\pi$, or $-\pi < \theta < \pi$, or $5\pi < \theta < 7\pi$. In general, $t < \theta < t+2\pi$ would give a loop around the origin. As Mark Viola points out: we need the branch cut to pass through $|z|=r$ and $\theta = t$.

If $z=r\mathrm e^{\mathrm i \theta}$, then $\mathrm dz=\mathrm ir\mathrm e^{\mathrm i\theta}~\mathrm d\theta$, and so \begin{eqnarray*} \int_{\gamma} \ln z~\mathrm dz &=& \int_t^{t+2\pi}\ln\left(r\mathrm e^{\mathrm i\theta}\right)~\mathrm ir\mathrm e^{\mathrm i\theta}~\mathrm d\theta \\ \\ &=& \int_t^{t+2\pi}\left[\ln r + \mathrm i\theta\right]~\mathrm ir\mathrm e^{\mathrm i\theta}~\mathrm d\theta \\ \\ &=& \mathrm i r\ln r\int_t^{t+2\pi} \mathrm e^{\mathrm i \theta}~\mathrm d\theta \ \ - \ \ r\int_t^{t+2\pi}\theta\mathrm e^{\mathrm i \theta}~\mathrm d\theta \\ \\ &=& 0 \ \ - \ \ r\left[ (1-\mathrm i\theta)\mathrm e^{\mathrm i\theta}\right]_t^{t+2\pi} \\ \\ &=& 2\pi\mathrm ir\mathrm e^{\mathrm it} \end{eqnarray*}

$\endgroup$
2
  • 1
    $\begingroup$ I assume that the branch cut passes through $|z|=r$ at $\theta =t$. Is that what you have in mind here? $\endgroup$
    – Mark Viola
    Commented Jan 11, 2019 at 22:51
  • $\begingroup$ That's right. Thank you Mark :-) $\endgroup$ Commented Jan 11, 2019 at 22:54

You must log in to answer this question.

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