Let $P(z, t)$ be a cubic with the parameter $t$, and consider $$\mathcal{I} = \int_{\gamma(t)} \frac{dz}{\sqrt{P(z, t)}}.$$ Here, $\gamma(t)$ is a contour in the complex plane that encloses any two of the three roots of $P(z, t)$ and does not go through the third. Since the roots are given in terms of $t$, the path $\gamma(t)$ is a function of $t$.
I would like to differentiate $\mathcal{I}$ under the integral sign with respect to $t$. Is it permissible? Could someone point me to a theorem I could use or help me prove it directly?
Fix a $t_0$. Then $ P(z,t_0)=c(z-z_1)(z-z_2)(z-z_3)$ for three complex numbers $z_1$, $z_2$, $z_3$. Assume that $\gamma_0:=\gamma(t_0)$ satisfies $$n(\gamma_0,z_1)=n(\gamma_0,z_2)=1, \quad n(\gamma_0,z_3)=0\ .$$ This includes that all three roots have a distance at least $\delta>0$ from the set $\gamma_0$. Furthermore $z\mapsto \sqrt{P(z,t_0)}$ picks up a factor $-1$ each from enclosing $z_1$ and $z_2$, so that your integrand $$f(z,t_0):={1\over\sqrt{P(z,t_0)}}$$ can be uniquely defined along $\gamma_0$.
Since the zeros $z_i(t)$ of $P(z,t)$ are continuous functions of the coeffficients, i.e., of $t$, there is an $\epsilon>0$ such for all $t$ with $|t-t_0|<\epsilon$ we have $$n(\gamma_0,z_1(t))=n(\gamma_0,z_2(t))=1, \quad n(\gamma_0,z_3(t))=0\ .$$ It follows that this same $\gamma_0$ may be used to compute ${\cal I}$ for all these $t$: $${\cal I}(t)=\int_{\gamma_0}{dz\over\sqrt{P(z,t)}}\qquad\bigl(|t-t_0|<\epsilon\bigr)\ .$$ It is now obvious that you can obtain ${\cal I}'(t_0)$ using differentiation under the integral sign.