Let $F(s)=\int_a^b f(t,s)\,dt$ where $f(t,s)$ is a continuous complex function in the rectangle $$R=\{(t,s): a\leq t\leq b,\, c\leq s \leq d\}$$ If the partial derivative $f_s(t,s)$ exists and is continuous on $R$, then the derivative $F’(s)$ exists for every $s\in[c,d]$ and is given by $$F’(s)=\int_a^b f_s(t,s)\,dt.$$
Is continuity of $f(s,t)$ and $ f_s(s,t )$ the only condition required for this theorem to be true? I am familiar with differentiation under the integral sign for the real case (Lebesgue integration) in which other conditions (other then continuity) must be satisfied.
Let $f(t,s)=g(t,s)+ih(t,s)$ for real $g,h$, then $g,g_{s},h,h_{s}$ are continuous. Let $G(s)=\displaystyle\int_{a}^{b}g(t,s)dt$, $H(s)=\displaystyle\int_{a}^{b}h(t,s)dt$, then $F(s)=G(s)+iH(s)$ for real $G,H$.
Assume that we have the result for the real case, then $G'(s)=\displaystyle\int_{a}^{b}g_{s}(t,s)dt$, $H'(s)=\displaystyle\int_{a}^{b}h_{s}(t,s)dt$, then $\displaystyle\int_{a}^{b}f_{s}(t,s)dt=\int_{a}^{b}[g_{s}(t,s)+ih_{s}(t,s)]dt=G'(s)+H'(s)$, then $F'(s)$ exists and the result follows.