Existence of antiderivative without Cauchy-Goursat's theorem

I wonder if anybody has tried the following kind of direct proof for the existence of an antiderivative of an analytic function on a star-shaped domain.

Theorem: Let $f:D\to \mathbb{C}$ be an analytic function on a star-shaped domain D. Then f has an antiderivative F on D.

"proof": For simplicity, assume that every point in D is connected to $0\in \mathbb{C}$ by a line segment. Define

$F(z)={\int}_{0}^{1}zf(zt)dt.$

Then $\underset{h\to 0}{lim}\left(\frac{F(z+h)-F(z)}{h}\right)=\underset{h\to 0}{lim}{\int}_{0}^{1}\left(\frac{(z+h)f(zt+ht)-zf(t)}{h}\right)dt.$

It can be checked that as $h\to 0$, the integrand converges to $\frac{d}{dt}tf(zt).$.

Therefore, ${F}^{\prime}(z)=f(z)$, provided that the limit and integral are interchangeable. qed.

Of course, a limit and integral cannot always be interchanged. But I wonder if anybody seriously considered the above line of proof.

Thanks. As far as I can search from several textbooks in complex variables, the above theorem is proved by using Cauchy-Goursat's theorem. More concretely, they use the equality ${\int}_{{z}_{0}}^{z+h}f(z)dz\text{}-\text{}{\int}_{{z}_{0}}^{z}f(z)dz\text{}=\text{}{\int}_{z}^{z+h}f(z)dz,$, which can be justified by Cauchy-Goursat's theorem. The point of my question is: Is it possible to directly invoke to the computation as above, without using Cauchy-Goursat's theorem? If this is possible, we get another proof of Cauchy-Goursat's theorem, at least for star-convex domains.