Show that a function which is defined by integral is entire by using Morera’s theorem. 
Let $f : [-1,1] \rightarrow \mathbb{C}$ be a continuous function and we define $F$ on $\mathbb{C}$ by
  $$\displaystyle F(z) = \int_{-1}^{1} f(t) e^{itz} dt $$
  1) Show that $F$ is well-defined and continuous on $\mathbb{C}$.
  2) Show that $F$ is entire by using Morera’s theorem.

Since $f$ is a single-variable complex-valued function, I could show that $F$ is well-defined and continuous on $\mathbb{C}$ easily.
But I have difficulty with second part of the above problem,
What I tried is as follows:
Since the problem asks me to use Morera’s theorem,
I have to show that for every possible contour $C$,  $\int_{C}^{\ } F(z) dz=0$.
Then 

$$\begin{align}
\Bigl \vert \int_{C}^{\ }F(z)dz \Bigr \vert &= \Bigl \vert \int_{C}^{\ } \Bigl ( \int_{-1}^{1}\,f(t)e^{itz}dt \Bigr ) dz \Bigr \vert \\ &=  \Bigl \vert \int_{C}^{\ } e^{itz}\Bigl(\int_{-1}^{1} f(t) dt\Bigr)dz \Bigr \vert \qquad \cdots (1) \\ &= \Bigl \vert \int_{C}^{\ } e^{itz} \Bigl( const. \Bigr) dz \Bigr \vert \\ &= 0
\end{align}$$

I’m not sure at the equation (1), is it okay to pull $e^{itz}$ out from inner integral.
If it is allowed to do, then how exactly it is allowed?
And if not, then how do I proceed?
Edit:
I guess it is okay to change the order of integrals provided that the integral still gives you finite value even if you change the integrand by the absolute value of the integrand.
So, for any contour $C$,
$$\begin{align}
\int_{C}^{\ } \Bigl (\int_{-1}^{1} \vert \, f(t)e^{itz}\, \vert dt \Bigr ) dz &= \int_{C}^{\ } \Bigl (\int_{-1}^{1} \vert \, f(t) \, \vert dt \Bigr ) dz \\ &= (const.)
\end{align}$$
This shows that it is okay to interchange the order of integration, then
$$\begin{align}
\int_{C}^{\ } \Bigl (\int_{-1}^{1} f(t)e^{itz} dt \Bigr ) dz &= \int_{-1}^{1} \Bigl (\int_{C}^{\ } f(t)e^{itz} dz \Bigr ) dt \\ &= \int_{-1}^{1} 0 dt \\&=0
\end{align}$$
This shows that $F$ is entire by Morera’s theorem.
Is this proof right?

Your intuition on separating integrals is correct, but you have pulled out the wrong thing :)
Let's do it in this manner:
$$\int_{C}^{\ } \Bigl ( \int_{-1}^{1}\,f(t)e^{itz}dt \Bigr ) dz {=\int_{C}  \int_{-1}^{1}f(t)e^{itz}dtdz\\=\int_{-1}^{1}\int_{C}f(t)e^{itz}dzdt\\=\int_{-1}^{1}f(t)\int_{C}e^{itz}dzdt}$$and $\int_{C}e^{itz}dz=0$ since $e^{\imath tz}$ is entire.
