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!
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 $z_2 = e^{i\pi/2} = i$ $$ \int_\gamma \frac{\ln(1+z)}{1+z}dz = \frac{1}{2}(\ln^2(1+e^{i\pi/2}) - \ln^2(1+1)) = \frac{1}{2}(\ln^2(1+i) - \ln^2(2))\\ = \frac{1}{2}\left(\ln^2(\sqrt{2}e^{i\pi/2}) - \ln^2(2)\right) = -\frac{3\ln^2(2)+\pi^2}{8} + i\frac{\pi \ln(2)}{4} $$