 Circle integral of a logarithm with a radial branch cut

Take a complex logarithm whose radial branch cut has angle $\beta$. On a positively oriented circle, use the one-sided branch values on either side of its crossing with the cut. Parametrize from $\theta =\beta$ to $\beta +2\pi$; adding an integer multiple of $2\pi i$ to the branch does not change the integral because the integral of $dz$ is zero. Along the open arc, a primitive is $z\mathrm{Log}z-z$. Its endpoint difference is $2\pi i$ times the cut-crossing point $R{e}^{i\beta }$, proving the displayed formula. The circle crosses the cut, so this is a piecewise contour integral of boundary values, not the integral of a single holomorphic logarithm on a neighborhood of a closed circle. The Cauchy integral theorem therefore does not make it zero.