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 $s^*$?
Best Answer
If we assume that $f(s,\alpha)$ is analytic wrt $\alpha,s$, then $\frac{1}{f(s,\alpha)}$ is meromorphic wrt $\alpha,s$. Thus, for each $s$, there exists an integer $0<N(s)<\infty$, constants $\{a_n(s)\}_{n=1}^{N(s)}$, and a function $g(s,\alpha)$ analytic wrt $\alpha$ for which $$\dfrac{1}{f(s,\alpha)}=g(s,\alpha)+\sum_{n=1}^{N(s)}\dfrac{a_n(s)}{(\alpha-\alpha_r(s))^n}.$$ This essentially follows from the converse to Mittag-Leffler's theorem. Thus \begin{align} I(s)&=\int_{0}^{1}\dfrac{d\alpha}{f(s,\alpha)}=\int_{0}^{1}\left(g(s,\alpha)+\sum_{n=1}^{N(s)}\dfrac{a_n(s)}{(\alpha-\alpha_r(s))^n}\right)d\alpha\\ &=\int_{0}^{1}g(s,\alpha)d\alpha+\sum_{n=2}^{N(s)}a_n(s)\int_{0}^{1}\dfrac{d\alpha}{(\alpha-\alpha_r(s))^n}+a_1(s)\int_{0}^{1}\dfrac{d\alpha}{\alpha-\alpha_r(s)}, \end{align} We'll call the first part $G(s)$. For the second part, the integrals of the even terms are divergent, and the integrals for the odd terms exist: $$\int_{0}^{1}\dfrac{d\alpha}{(\alpha-\alpha_r(s))^{2n+1}}=\dfrac{1}{2n}\left(\dfrac{1}{\alpha_r(s)^{2n}}-\dfrac{1}{(1-\alpha_r(s))^{2n}}\right).$$ (where we're interpreting the integrals in terms of their Cauchy principal value). Now for the logarithmic term: \begin{align} \int_{0}^{1}\dfrac{d\alpha}{\alpha-\alpha_r(s)}&=\lim_{\epsilon\to0^{+}}\left(\int_{0}^{\alpha_r(s)-\epsilon}+\int_{\alpha_r(s)+\epsilon}^{1}\right)\dfrac{d\alpha}{\alpha-\alpha_r(s)}\\ &=\ln(1-\alpha_r(s))-\ln(-\alpha_r(s))-\lim_{\epsilon\to0^{+}}\int_{-\epsilon}^{\epsilon}\dfrac{d\alpha}{\alpha}\\ &=\ln\left(\dfrac{\alpha_r(s)-1}{\alpha_r(s)}\right). \end{align} (Where we're using the principal branch of the complex logarithm). So, when the integral exists, it admits a representation of the form \begin{align} I(s)&=G(s)+a_1(s)\ln\left(\dfrac{\alpha_r(s)-1}{\alpha_r(s)}\right)+\sum_{n=1}^{\left\lfloor\frac{N(s)-1}{2}\right\rfloor}\dfrac{a_{2n+1}(s)}{2n}\left(\dfrac{1}{\alpha_r(s)^{2n}}-\dfrac{1}{(1-\alpha_r(s))^{2n}}\right). \end{align} So yes, the branch cuts of $I(s)$ are precisely $\alpha_r^{-1}((0,1))=\{s:\alpha_r(s)\in(0,1)\}$ (assuming $a_1(s)\neq0$).
Some additional regularity assumptions may be needed for the analyticity of $G(s)$ and the $\{a_n(s)\}$ away from the locus. My multivariable complex analysis is a little rusty, so I will need to think about it.