Proof that $g(z) = \int_0^1 f(t)e^{tz} \, dt$ is entire Let $f(z)$ be a complex and continuous function on $[0,1]$. We'll define $g(z) = \int_0^1 f(t)e^{tz} \, dt $. Prove that $g(z)$ is an entire function. 
 A: If you write $f=f_1+\sqrt{-1}f_2$ where $f_1, f_2$ are continuous function on $[0,1]$ and $z=x+\sqrt{-1}y$, then 
$$g(z)=g(x,y)=\int_0^1\big(f_1(t)+\sqrt{-1}f_2(t)\big)e^{tx}\big(\cos(ty)+\sqrt{-1}\sin(ty)\big)dt.$$ Therefore the real part $u(x,y)$ and the imaginary part $v(x,y)$ of $g(x,y)$ are given by
$$u(x,y)=\int_0^1e^{tx}\big(f_1(t)\cos(ty)-f_2(t)\sin(ty)\big)dt$$
and
$$v(x,y)=\int_0^1e^{tx}\big(f_1(t)\sin(ty)+f_2(t)\cos(ty)\big)dt.$$
Then you can check easily that $u$ and $v$ satisfy the Cauchy-Riemann condition: $u_x=v_y$ and $u_y=-v_x$ for all $z=x+\sqrt{-1}y\in\mathbb{C}$. 
A: Morera's theorem implies that if for every circle $C$ in the complex plane,
$$
\int_C g(z) \; dz = 0
$$
then $g$ is entire.  To find that integral, first write
$$\int_C g(z) \; dz = \int_C\int_0^1 f(t)e^{tz} \; dt\; dz.$$
Now we're integrating continuous functions over compact sets, so everything in sight is Lebesgue-integrable and Fubini's theorem can be used to get
$$
\int_C\int_0^1 f(t)e^{tz} \; dt\; dz=\int_0^1 \int_C f(t) e^{tz} \; dz\;dt = \int_0^1\left(f(t)\int_C e^{tz} \; dz \right) \; dt.
$$
The inner integral is $0$ because it is the integral of an entire function around a circle.  So the whole thing is $0$ and the conclusion from Morera's theorem is just what is needed.
