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

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?

• 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.
– Mark
Commented Jan 11, 2019 at 21:27
• 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. Commented Jan 11, 2019 at 21:34
• 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 $\gamma(0) = \gamma(1)=1, F(\gamma(1) ) = 2i\pi - 1, F(\gamma(0)) = -1$. Commented Jan 11, 2019 at 21:54

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 $$\begin{eqnarray*} \int_{\gamma} \ln z~\mathrm dz &=& \int_t^{t+2\pi}\ln\left(r\mathrm e^{\mathrm i\theta}\right)~\mathrm ir\mathrm e^{\mathrm i\theta}~\mathrm d\theta \\ \\ &=& \int_t^{t+2\pi}\left[\ln r + \mathrm i\theta\right]~\mathrm ir\mathrm e^{\mathrm i\theta}~\mathrm d\theta \\ \\ &=& \mathrm i r\ln r\int_t^{t+2\pi} \mathrm e^{\mathrm i \theta}~\mathrm d\theta \ \ - \ \ r\int_t^{t+2\pi}\theta\mathrm e^{\mathrm i \theta}~\mathrm d\theta \\ \\ &=& 0 \ \ - \ \ r\left[ (1-\mathrm i\theta)\mathrm e^{\mathrm i\theta}\right]_t^{t+2\pi} \\ \\ &=& 2\pi\mathrm ir\mathrm e^{\mathrm it} \end{eqnarray*}$$
• I assume that the branch cut passes through $|z|=r$ at $\theta =t$. Is that what you have in mind here? Commented Jan 11, 2019 at 22:51