Find $$\oint_{|z|=1}\log\left|\frac{1}{1+z^p}\right|\frac{dz}{z} $$ where $0<p<\frac{1}{2}$
Attempt $$I=\oint_{|z|=1}\log\left|\frac{1}{1+z^p}\right|\frac{dz}{z} $$ Define, $$f(z)= \frac{1}{1+z^p} $$ Then since $0<p<\frac{1}{2}$, so $f(z)$ has a branch point at $z=0.$ so $f(z)$ is not analytic at $z=0$. How to solve this integral using branch cuts or otherwise?.
As was correctly noted in the comment the integral is not unambiguously defined without specifying the branch cut. For an arbitrary cut intersecting the circle at the argument $\phi-\pi$ one obtains using substitution $z=e^{it}$: $$\begin{align} -i\int\limits_{\phi-\pi}^{\phi+\pi}\log|1+e^{ipt}|dt &=-i\int\limits_{\phi-\pi}^{\phi+\pi}\log\sqrt{(1+e^{ipt})(1+e^{-ipt})}dt\\ &=-\frac i2\int\limits_{\phi-\pi}^{\phi+\pi}\left[\log\left(1+e^{ipt}\right)+\log\left(1+e^{-ipt}\right)\right]dt\\ &=\frac i2\int\limits_{\phi-\pi}^{\phi+\pi}\left[\sum_{k=1}^\infty(-1)^k\frac{e^{iptk}+e^{-iptk}}k\right]dt\\ &=\frac i2\sum_{k=1}^\infty\frac{(-1)^k}k \int\limits_{\phi-\pi}^{\phi+\pi}\left[{e^{iptk}+e^{-iptk}}\right]dt\\ &=\frac i2\sum_{k=1}^\infty\frac{(-1)^k}k\left[\frac{e^{iptk}-e^{-iptk}}{ipk}\right]_{\phi-\pi}^{\phi+\pi}\\ &=\frac i{p}\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\Im\left[e^{iptk}\right]_{\phi-\pi}^{\phi+\pi}\\ &=\frac ip\Im\sum_{k=1}^\infty\frac{(-1)^k}{k^2}\left[e^{ip(\phi+\pi)k}-e^{ip(\phi-\pi)k}\right]\\ &=\frac ip{\Im\left[ \operatorname{Li}_2\left(-e^{ip(\phi+\pi)}\right)- \operatorname{Li}_2\left(-e^{ip(\phi-\pi)}\right)\right]}.\tag1 \end{align}$$
Particularlry for the principal branch of the logarithm with the cut along the negative real semi-axis $\phi=0$ and one obtains from $(1)$: $$ \frac{2i}p{\Im\operatorname{Li}_2\left(-e^{ip\pi }\right)}.\tag2 $$