I'm not very versed in complex analysis and I'm trying to understand some concepts on branch cuts and contour integration. Consider a function
$$ I(s)=\int_0^1 d\alpha\ \frac{1}{f(s,\alpha)}, $$
such that $\alpha_r(s)$ are the roots of $f(s,\alpha)$, i.e. $f(s,\alpha_r(s))=0$. Let's say now that we take some path in the complex $s$-plane, for which the root $\alpha_r$ draws some curve in the $\alpha$-plane (depicted in red in the figure that I attach). At the point $s=s^*$, the root crosses the contour of integration, which we simply choose to stay on the real axis $\alpha\in[0,1]$ (drawn in blue in the figure). Thus, for $s_1<s^*$ (in our chosen path), the root $\alpha_r$ lies at one side of the contour, while for $s_2>s^*$, it lies on the opposite side. My first question is: would the integral $I(s)$ jump discontinuously at $s=s^*$ and, if so, how to justify it? If it does, then does that mean that there is a branch cut in $I(s)$ which we cross at $s^*$?
