Evaluating $\operatorname{Re} \oint_{\mathcal{C}} \mathrm{d} z \frac{\arg z}{z}$ over a given contour

Question

I encountered a problem asking me to calculating the contour integral as shown in the picture.

I am asked to compute the energy using the following equation. $$E=-\frac{J}{2} n^{2} \operatorname{Re} \oint_{\mathcal{C}} \mathrm{d} z \frac{\arg z}{z}$$ I think I just need to compute the bigger loop integral minus the smaller loop integral.

So I let $z_{1}=Re^{i\omega t}$, $z_{2}=r_{0}e^{i\omega t}$, then $\oint\mathrm{d} z \frac{\arg z}{z}$ can be converted to $\oint i\omega^{2}t dt$, which yield the result $2\pi^{2}i$, and taking the real part I simply get nothing, which does not lead to the result.

Anyone know where did this go wrong?

Answer 1

We can parameterize the contour as follows:

\begin{align*} \gamma_{1}(t) = t \quad &t \in \left[r_{0},R\right]\\ \gamma_{2}(\theta) = Re^{i\theta} \quad & \theta\in(0,2\pi)\\ \gamma_{3}(t) = -t \quad& t\in \left[-R,-r_{0}\right]\\ \gamma_{4}(\theta) = r_{0}e^{-i\theta} \quad &\theta \in (-2\pi, 0) \end{align*} Hence

\begin{align*} \oint_{\gamma_{1}+...+\gamma_{4}} \frac{\arg{z}}{z} dz =& \int_{r_{0}}^{R} \frac{\arg(\gamma_{1}(t))}{\gamma_{1}(t)}\gamma_{1}'(t) dt + \int_{0}^{2\pi} \frac{\arg(\gamma_{2}(\theta))}{\gamma_{2}(\theta)}\gamma_{2}'(\theta) d\theta+ \int_{-R}^{-r_{0}} \frac{\arg(\gamma_{3}(t))}{\gamma_{3}(t)}\gamma_{3}'(t) dt + \int_{-2\pi}^{0} \frac{\arg(\gamma_{4}(\theta))}{\gamma_{4}(\theta)}\gamma_{4}'(\theta) d\theta\\ =& \int_{r_{0}}^{R} \frac{0}{t} dt + i\int_{0}^{2\pi} \theta d\theta + \int_{-R}^{-r_{0}} \frac{2\pi}{t} dt + i\int_{-2\pi}^{0} \theta d\theta \\ =& i\int_{0}^{2\pi} \theta d\theta - \int_{r_{0}}^{R} \frac{2\pi}{t} dt - i\int_{0}^{2\pi} \theta d\theta \\ =&- 2\pi \int_{r_{0}}^{R}\frac{1}{t}dt\\ =& 2\pi \ln\left(\frac{r_{0}}{R}\right) \end{align*}

Answer 2

I don't follow the computation that yields $\oint i \omega^2 t \,dt$, but here's another method that avoids fussing too much with the branch cut.

First, as the contour suggests, we take the branch cut of the logarithm along the positive real axis. Rearranging the decomposition $\log z = \log |z| + i \operatorname{arg} z$ and splitting the integral gives $$\oint_C \frac{\operatorname{arg} z \, dz}{z} = \frac{1}{i} \left(\oint_C \frac{\log z \, dz}{z} - \oint_C \frac{\log |z| \, dz}{z} \right).$$ Now, $\frac{\log z}{z}$ is analytic along and inside the contour, so $\oint_C \frac{\log z \, dz}{z} = 0$. The other integrand, $\frac{\log |z|}{z}$, does not depend on the choice of branch cut, so in the limit the integrals along the line segments cancel, and the original integral is just equal to the sum of the (appropriately signed) integrals along the inner and outer circles; we denote the anticlockwise contours parameterizing them by $C_{r_0}$ and $C_R$, respectively. We thus have: \begin{align*} \oint_C \frac{\operatorname{arg} z \, dz}{z} &= -\frac{1}{i} \oint_{-C_{r_0} \cup C_R} \frac{\log |z| \,dz}{z} \\ &= -\frac{1}{i} \left(-\oint_{C_{r_0}} \frac{\log |z| \,dz}{z} + \oint_{C_R} \frac{\log |z| \,dz}{z}\right) \\ &= \frac{1}{i} \left(\log r_0 \oint_{C_{r_0}} \frac{dz}{z} - \log r \oint_{C_{r_0}} \frac{dz}{z}\right) \\ &= \frac{1}{i} (2 \pi i \log r_0 - 2 \pi i \log R) \\ &= \boxed{2 \pi \log \frac{r_0}{R}} . \end{align*}