 Solution

Parametrize the continued branch by $t=r{e}^{i\theta }$, $0\le \theta \le 2\pi$. Then $t^{z-1}dt=i{r}^z{e}^{iz\theta }d\theta$, givingThe apparent singularity at zero is removable, so $f$ is an entire function. The factor $r^z=e^{z\log r}$ shows its dependence on the contour radius; for instance $f(1/2)=-4\sqrt{r}$. The integrand is continued along the contour and generally returns on a different branch, so this is not an ordinary integral of a single-valued holomorphic function around a closed loop.