Let $f$ : $[a, b] \rightarrow\mathbb C$ be continuous.
Show that the function $F : \mathbb C \rightarrow \mathbb C $ defined by $$ F(z) = \int_a^b f(t)\exp(tz) dt$$ is holomorphic.
I am unsure of how to start, I think I should get some property of holomorphic functions like composition or integral of holomorphic functions is holomorphic but I'm not sure if any such thing exists.
On
We can use Morera's theorem.
Let $\gamma: [0,1] \to \mathbb C$ be a closed piecewise $C^1$ path. Then
$\displaystyle \int_\gamma F(z)\,dz =\int_0^1 F(\gamma(s))\gamma'(s)\,ds $
$\displaystyle =\int_0^1 \int_a^b f(t)\exp(t\gamma(s))\,dt\, \gamma'(s) \,ds $
$\displaystyle =\int_a^b \int_0^1 f(t)\exp(t\gamma(s))\gamma'(s) \,ds\, \,dt $
$\displaystyle =\int_a^b 0 \,dt = 0 $
The switch in the order of integration is ok because the integrand is continuous and the domain is compact.
The last inner integral is $\displaystyle \int_\gamma f(t) \exp(tz)\,dz $ and so is zero by Cauchy's theorem, since the integrand is a holomorphic function for fixed $t$.
You just have to observe that$$\int_a^bf(t)\exp(tz)\,\mathrm dt=\int_a^bf(t)\sum_{n=0}^\infty\frac{t^nz^n}{n!}\,\mathrm dt=\sum_{n=0}^\infty\left(\int_a^b\frac{f(t)t^n}{n!}\,\mathrm dt\right)z^n.$$