Finding the order of an entire function defined from an integral The following problem is posed in Greene and Krantz, page 297, Problem 11.
Let $g: \mathbb{R} \to \mathbb{C}$ be a continuous function, $\alpha \in \mathbb{R}$, and define $f: \mathbb{C} \to \mathbb{C}$ s.t. 
$$f(z) = \int^b_a e^{\alpha z t} g(t) dt .$$
Show that $f$ is entire and is of finite order. Determine whether or not the order of $f$ depends on $g$.
I am able to show that $f$ is entire and of finite order. Intuitively, it seems to me that assuming $\alpha \neq 0$ and $g$ not identically $0$, $f$ should always be of order 1, but I can't show this nor find a counterexample. Insights would be appreciated.
 A: The following generalized statement holds. To recover your original setting, we can simply define $\mu$ as $\mu(E):=\int_{E\cap[a,b]}g(t)dt$ and replace $f(z)$ by $f(\alpha z)$. 

Proposition: Let $\mu$ be a Borel complex measure on $\Bbb R$. Denote its support by ${\rm supp}(\mu)$ and its total variation by $\|\mu\|$. Suppose that 
  $$\sigma:=\sup \{|x|:x\in {\rm supp}(\mu)\}\in (0,+\infty). \tag{1}$$
  Then for 
  $$f:\Bbb C\to \Bbb C,\quad f(z):=\int_{\Bbb R}e^{zt}d\mu (t), \tag{2}$$
  $f$ is an entire function satisfying that for $x,y\in \Bbb R$, 
  $$|f(x+iy)|\le \|\mu\|\cdot e^{\sigma|x|}, \tag{3}$$
  and for $x\in \Bbb R$,
  $$\limsup_{|x|\to+\infty}\frac{\log |f(x)|}{|x|}=\sigma. \tag{4}$$
  As a corollary, the order and type of $f$ are $1$ and $\sigma$ respectively.


Proof:
Since $\mu$ has compact support, it is evident that $f$ is entire and $(3)$ holds. To prove $(4)$, without loss of generality, we may assume that $\sigma\in {\rm supp}(\mu)$(the case $-\sigma\in {\rm supp}(\mu)$ is similar), and it suffices to show that for every $\tau\in (0,\sigma)$, 
$$\limsup_{x\to+\infty}|f(x)e^{-\tau x}|=+\infty. \tag{5}$$
To prove $(5)$, let us define
$$F:\Bbb C\to \Bbb C,\quad F(z):=\int_\tau^\sigma e^{z(t-\tau)}d\mu (t). \tag{6}$$
By definition, when $x\ge 0$, $|f(x)e^{-\tau x}-F(x)|\le \|\mu\|$, so $(5)$ is equivalent to 
$$\limsup_{x\to+\infty}|F(x)|=+\infty. \tag{7}$$
To begin with the proof of $(7)$, note that $$|F(x+iy)|\le \|\mu\|\cdot e^{(\sigma-\tau)\cdot\max\{x,0\}}.\tag{8}$$ 
In particular, the order of $F$ is no more than $1$. By reduction to absurdity, suppose that $(7)$ fails. Then from $(8)$ we know that there exists $M\ge \|\mu\|$, such that $|F(x)|\le M$ when $x\ge 0$ and $|F(x+iy)|\le M$ when $x\le 0$. Now we can apply Phragmén–Lindelöf principle in the linked page to $F$ on the first and the forth quadrants respectively with $\lambda=2$ and $\rho =1$, which implies that $|F(x+iy)|\le M$ when $x\ge 0$. Therefore, $|F|\le M$ on $\Bbb C$, so by Liouville's theorem, $F$ must be constant. On the other hand, from $(6)$ we know that the function $y\mapsto F(-iy)e^{-i\tau y}$ is the Fourier transform of the complex measure $\nu:=\mu|_{[\tau,\sigma]}$, so $F$ being constant implies that  ${\rm supp}(\nu)=\{\tau\}$. However, by our assumption, $\sigma\in {\rm supp}(\nu)$. This contradiction completes the proof.  

Remark: On the one hand, $(4)$ asserts that as $x\to +\infty$, $\max\limits_{t\in [-x,x]}|f(t)|\to +\infty$ exponentially fast in $x$; on the other hand, even if $\mu$ is defined by $\mu(E):=\int_{E\cap[a,b]}g(t)dt$ for some smooth function $g$, it is possible that $f(x)$ has infinitely many zeros on $[0,+\infty)$.
