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Computer $$\displaystyle\int\limits_{\gamma }\frac{\log (1+z)}{1+z}dz$$ Where :

$$\gamma =\{ |z|=1~ ; ~\Re z≥0,\Im z≥0 \}$$

I try :

$z=e^{it}$ then $dz=ie^{it}dt$ And $t\in [0,\frac{π}{2}$ then integration become :

$$\displaystyle\int\limits_{0}^{\frac{π}{2}}\frac{\log (1+e^{it})}{1+e^{it}}dz$$ Using :

$$\log (1+e^{it})=\ln |1+e^{it}|+i \arg (1+e^{it})$$

but I don't know how I complete ? Can you assist ?


Thanks!

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The intgrand has "problems" at $x=-1$ but the curve $\gamma$ is far from it $$ \int \frac{\ln(1+z)}{1+z}dz =\int \frac{\ln(u)}{u}du =\int \ln(u)d(\ln(u)) =\frac{1}{2}\ln^2(u) +c =\frac{1}{2}\ln^2(1+z) +c $$ The limits of the integral are $z_1 = e^{i0} = 1$ and $z_2 = e^{i\pi/2} = i$ $$ \int_\gamma \frac{\ln(1+z)}{1+z}dz = \frac{1}{2}(\ln^2(1+e^{i\pi/2}) - \ln^2(1+1)) = \frac{1}{2}(\ln^2(1+i) - \ln^2(2))\\ = \frac{1}{2}\left(\ln^2(\sqrt{2}e^{i\pi/2}) - \ln^2(2)\right) = -\frac{3\ln^2(2)+\pi^2}{8} + i\frac{\pi \ln(2)}{4} $$

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  • $\begingroup$ Thank you much sir $\endgroup$ – Ellen Ellen Sep 22 at 5:04
  • $\begingroup$ you're welcome! $\endgroup$ – Physor Sep 22 at 7:55
  • $\begingroup$ @physor how you get the limits of integral? I use $z=e^{it}$ also but I think limits $0\to \frac{π}{2}$ $\endgroup$ – Roze flowers Sep 22 at 13:21

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