Line integral depending on a parameter is entire Suppose you have a continuous function:

$$\phi:[0,1]\rightarrow \mathbb{C}$$

define the complex function:

$$f(z)=\int_0^1\phi(t)e^{itz}dt$$

prove that it is entire and calculate it's Taylor expansion centered at $z=0$. Honestly I don't know where to start, I think I have to apply the theorem of holomorphy of a parametric integral but I don't understand how.

Also, how can I apply those results to the sequence of functions:

$$f_n(z)=\int_0^n \sqrt{t}e^{-tz}dt$$
It's obviously entire as the derivative is $$ \int_0^1 ite^{izt}  \phi(t) \, dt$$ which exists (is convergent) for all $z$  as the integrand is bounded.
The Taylor series is obtained by differentiating under the integral sign: the $n$th derivative is $$ \int_0^1 (it)^n \phi(t)\, dt.$$
It's obviously entire as the derivative is $$ \int_0^1 ite^{izt}  \phi(t) \, dt$$ which exists (is convergent) for all $z$  as the integrand is bounded.
The Taylor series is obtained by differentiating under the integral sign: the $n$th derivative is $$ \int_0^1 (it)^n \phi(t)\, dt.$$