# 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 isn't a question of finding which theorem to apply. You are asked to show that the function $f$ is entire, i.e. differentiable at every point. So write down an expression that looks as though it might be the derivative (easy) and then prove by first principles that it equals $\lim_{h\to0} (f(z+h)-f(z))/h$.This amounts to no more than an exercise in Riemann integration. – Richard Martin Jan 28 at 12:57
• As a final point, you can also do it without differentiation if you use Morera's theorem. To do so, you just need to show that $f$ is continuous, and also that $\int_C f(z) \, dz=0$ for all closed contours $C$. The latter is clear because $e^{itz}$ is entire---again you have to justify the interchange of order of integration, but this is just an exercise in Riemann integration. – Richard Martin Jan 28 at 13:00