I'm stuck on this particular integration. Here is my problem:

$$\int_{\mathcal{C}} Log (z)\,\mathrm{d}z,$$

where Log (z) is the branch cut of the complex logarithm and C(t) = $\ e^{it}$ with $t$ from -$\pi$ to $\pi$.

Because of the Branch cut of Log (z), I know that it is impossible to use the Couchy-Gorsat Theorem for this. If not, how can I solve this problem?

  • $\begingroup$ Instead of integrating from $-\pi$ to $\pi$, slightly change the contour so that the new limits are $-\pi + \epsilon$ and $\pi - \epsilon$. Add another contour parallel to the branch and wrapping around it starting from $e^{-i(\pi-\epsilon)}$, going toward zero, then a half a circle, then similarly toward $e^{i(\pi-\epsilon)}$. Since inside this new closed contour the function is holomorphic this intergral is zero. So the integration you wanted is equal to the integration for the new contour. The rest should be straightforward. $\endgroup$ – Hamed Oct 9 '15 at 1:03
  • $\begingroup$ Why not just do it directly? ${\rm Log} ( e^{it} )= it$ for $ -\pi < t\le \pi$, so the integral equals $$ \int_{-\pi}^\pi it e^{it}i \,dt,$$ which you can do by parts - seems right? $\endgroup$ – peter a g Oct 9 '15 at 1:21
  • $\begingroup$ To be clear, $dz = d( e^{it}) = e^{it}i\,dt$, although, of course, with integration by parts the $d (e^{it})$ version is more useful... $\endgroup$ – peter a g Oct 9 '15 at 1:33
  • $\begingroup$ I see. That makes sense, Thanks, Peter! $\endgroup$ – baek Oct 9 '15 at 2:43

@Peter is correct. Make the substitution $z=e^{it}$; then $dz=ie^{it}dt$ and we can write $$\int_{\mathcal{C}}\log z dz=\int_{-\pi}^\pi \log(e^{it})(ie^{it}dt)=\int_{-\pi}^\pi -te^{it}dt$$ By parts, we let $u=-t;\ dv=e^{it}ht$ and the integral becomes $$-t\frac{e^{it}}{i}\bigg|_{-\pi}^\pi-\int_{-\pi}^\pi\frac{e^{it}}{i}(-dt)$$ Simplifying and evaluating, $$=\left(-\pi\frac{-1}{i}-\pi\frac{-1}{i}\right)-(-1-(-1))=\boxed{-2\pi i}$$


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