How to prove this integral function is analytic?

Question

Given $$G(z) = \int_{1+i}^{z}\text{sin} (\theta^2) d\theta$$

Prove that $G(z)$ is an analytic function of $z$.

I read that integration preserves analyticity. But why is that true when $z$ is in the integral bounds? What if the function is something different instead of $\text{sin}(\theta^2)$, say $f(z)$ which is only analytic in a certain region $R$ and I integrate along a path that goes outside $R$?

Answer 1

You are asking about one of the standard theorems in complex analysis.

Let $f:\Omega \to \mathbb C$ be holomorphic function on open star-domain $\Omega \subseteq \mathbb C$. Let $p\in \Omega$ be point such that for any $z \in \Omega$, the line segment $<p, z>$ is contained in $\Omega$. Then function $F(z) = \int_{<p, z>} f(\theta)\operatorname d \! \theta$ is holomorphic on $\Omega$ and $F' = f$.

This theorem is step in proving Morera's theorem. For proof see Rudin's Real and Complex Analysis chapter 10: proof of 10.14 (what you are interested in), and statements of 10.6 and 10.16 (holomorphic functions are exactly the ones representable by power series).

Answer 2

If $f(z)$ is analytic on simply connected subset $D$ of complex plane $\mathbb C$ and $\gamma$ is any arbitrary curve inside $D$, then the function $F(z)=\int_\gamma f(t)dt$ is also analytic.

In your case $f(z)=\sin (z^2)$ is entire i.e. analytic on whole $\mathbb C$, so the antiderivative is also analytic regardless of the choice of $\gamma$.