Is it possible to evaulate $I = \frac{2i}{\pi} \int_\gamma \ln(z) dz$ explicitly even though $\ln(z)$ isn't holomorphic at $z=0$?

Question

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?

Answer 1

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*}