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Compute $$\int_0^{2\pi } \ln \vert z - \tfrac{\pi}{2} \vert d\theta$$ for $z=\pi e^{i\theta}$. (Here the $\ln$ is natural logarithm.)

Since $dz = \pi i e^{i\theta} d\theta$. Therefore, $d\theta = \frac{dz}{zi}$.

Therefore, $$\int_0^{2\pi } \ln \vert z - \tfrac{\pi}{2} \vert d\theta = \int _{\vert z\vert =\pi} \frac {\ln\vert z - \frac{\pi}{2} \vert }{zi} dz $$

I can't proceed for next step anymore. What should I do next to find that value?

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  • $\begingroup$ Surely the contour is $|z|=\pi$. $\endgroup$
    – J.G.
    Jun 23 at 9:07
  • $\begingroup$ You can use the fact that your integral is $ {\mathop{\rm Re}\nolimits} \int_0^{2\pi } {\log \left( {\pi e^{i\theta } - \frac{\pi }{2}} \right)d\theta } . $ $\endgroup$
    – Gary
    Jun 23 at 9:07
  • $\begingroup$ @J.G., Yes that was the typo. I edited. $\endgroup$
    – kechang lee
    Jun 23 at 9:12
  • $\begingroup$ How did you get the last equality? $\endgroup$
    – Gary
    Jun 23 at 9:14
  • $\begingroup$ @Gary. In fact I've got some that from math.stackexchange.com/questions/238272/…. But I'll deleted it if that does not hold. $\endgroup$
    – kechang lee
    Jun 23 at 9:26

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