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.

Solution 1:

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.

Solution 2:

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 $z=x+\sqrt{-1}y\in\mathbb{C}$.