the analyticity of integrals depending on a parameter
Many important functions are not given by a formula but by an integral with a parameter, such as F(z) = the integral over some path or set of g(z, w) dw, where z is a complex parameter and w is integrated out. A natural and useful question: if the integrand depends holomorphically on z, is the resulting function F holomorphic in z too? Cauchy's machinery answers yes, under mild conditions — analyticity passes through the integral sign.
The clean statement: if g(z, w) is holomorphic in z for each fixed w, continuous in both variables, and the integral over a fixed bounded set converges nicely (for instance uniformly on compact z-sets), then F(z) = the integral of g(z, w) dw is holomorphic in z, and you may even differentiate under the integral, F'(z) = the integral of the z-derivative of g. The proof is a Morera argument: integrate F around any triangle, swap the order of integration (justified by the convergence), use that g is holomorphic in z so its triangle integral is zero, conclude F's triangle integral is zero, and Morera makes F holomorphic.
This is the quiet workhorse that legitimizes a huge family of special functions. The gamma function defined by an integral, the Laplace and Fourier transforms, and the Cauchy integral formula itself (an integral whose parameter is z_0) are all holomorphic precisely because of this principle. The caveat that earns its keep: you must control the convergence — uniform convergence on compact sets is the usual passport, and without it interchanging limit, derivative, and integral can fail.
The gamma function Gamma(z) = the integral from 0 to infinity of t^(z-1) e^(-t) dt is holomorphic for Re z greater than 0, because the integrand is holomorphic in z and the integral converges uniformly on compact subsets of that half-plane.
An integral defines a holomorphic function whenever the convergence is uniform on compacta.
Holomorphy in the parameter is not automatic — it relies on adequate convergence (typically uniform on compact sets); for integrals over infinite ranges you must check the tails, or differentiation under the integral can give a wrong answer.