Growth order of entire functions This is an exercise 5 in chapter 5-Stein complex analysis. 
Let $\alpha\gt1$. Prove that $F_\alpha(z)=\int_{-\infty}^{\infty}e^{-|t|^\alpha}e^{2\pi izt}dt$ is of growth order $\alpha\over{\alpha-1}$. 
Actually, I found the following link. But, I can't understand some answer. 
how can i calculate growth order of this entire function?
Is there any good elementary method without "Laplace method" to exact computation about order of growth $\frac{\alpha}{\alpha-1}$ ? 
(I already proved the inequality case.)
I tried the substitution $t=|z|^{\frac{\alpha}{\alpha-1}} \mu$. 
Then,  $|z|^{\frac{\alpha}{\alpha-1}} $ could be extracted out of the integrals.
The integral is finite. Then, did I get what I want? I can't assure this.
 A: Another approach: 
You already proved the growth order is less or equal to $\frac{\alpha}{\alpha-1}.$   We prove that the order of growth is NOT less than $\frac{\alpha}{\alpha-1}$. 
For simplicity we consider $G_\alpha (z)=F_\alpha (z/(2\pi i))=\int_{-\infty}^\infty e^{-|t|^\alpha} e^{zt}dt.$  Of course $F_\alpha $ and $G_\alpha $ have the same order of growth.   
Suppose that the growth order, say $\rho $, is actually less than $\lambda =\frac{\alpha }{\alpha -1}$.
Then \begin{align}
|G_\alpha (z)|\le Ae^{B|z|^\rho},\quad (\rho <\lambda)    
\end{align}
holds for all $z\in \mathbb{C}$, where  $A,B$ are some positive constants.
We estimate the value of $G_\alpha (R)$. We have that 
\begin{align}
G_\alpha (R)&=\int_{-\infty}^\infty e^{-|t|^\alpha} e^{Rt}dt>\int_0^\infty e^{-t^\alpha} e^{Rt}dt\\
&> \int_0^{\frac{1}{2}R^\frac{1}{\alpha -1}} e^{-t^\alpha} e^{Rt}dt\\
&>e^{-\frac{1}{2^\alpha }R^\lambda }\int_0^{\frac{1}{2}R^\frac{1}{\alpha -1}}e^{Rt}dt,
\end{align}
since $e^{-t^\alpha} \ge e^{-\frac{1}{2^\alpha }}R^\lambda $ for $0\le t\le {\frac{1}{2}R^\frac{1}{\alpha -1}}$. 
Therefore we have $$
G_\alpha (R)>e^{-\frac{1}{2^\alpha }R^\lambda }\cdot\frac{1}{R}\left(e^{\frac{1}{2}R^\lambda }-1 \right)=\frac{1}{R}\left(e^{\left(\frac{1}{2}-\frac{1}{2^\alpha }  \right)R^\lambda } -1 \right).
$$
From $(1)$ we have $$
\frac{1}{R}\left(e^{\left(\frac{1}{2}-\frac{1}{2^\alpha }  \right)R^\lambda } -1 \right)<Ae^{BR^\rho}.
$$
This is a contradiction, since it does not hold for large $R$. Notice that $\rho <\lambda $.
A: The order $\lambda $ of an entire function $f(z)$ is defined by  $$\limsup_{r\to \infty}\frac{\log \log M(r)}{\log r},$$
where $ M(r)$ denotes $\max \{ |f(z)| : |z| = r \}$. If the Maclaurin expansion of $f(z)$ is $\sum_{n=0}^\infty c_nz^n,$ then $\lambda$ equals $$
\limsup_{n\to \infty}\frac{n \log n}{\log \frac{1}{|c_n|}}.$$ 
See here.
Now we have
\begin{align}
F_\alpha(z)&=\int_{-\infty}^{\infty}e^{-|t|^\alpha}e^{2\pi izt}dt=\int_{-\infty}^{\infty}e^{-|t|^\alpha}\left(\sum_{n=0}^\infty \frac{(2\pi izt)^n}{n!}\right)dt\\
&=\sum_{n=0}^\infty \left(\frac{(2\pi i)^n}{n!}\int_{-\infty}^{\infty}e^{-|t|^\alpha}t^ndt\right)z^n.
\end{align}
For odd $n$ the integral $\int_{-\infty}^{\infty}e^{-|t|^\alpha}t^ndt$ is $0$, and for even $n$ we have 
\begin{align}
\int_{-\infty}^{\infty}e^{-|t|^\alpha}t^ndt&=2\int_0^{\infty}e^{-t^\alpha}t^ndt=\frac{2}{\alpha }\int_0^{\infty}e^{-s}s^{\frac{n+1}{\alpha }-1}ds\\
&=\frac{2}{\alpha }\Gamma \left(\frac{n+1}{\alpha }\right).
\end{align}
Therefore \begin{align}
\lambda &=\limsup_{n\to \infty}\frac{n \log n}{\log \Big(\frac{\alpha n!}{2(2\pi)^n}\cdot\frac{1}{\Gamma\left(\frac{n+1}{\alpha }\right)}\Big)}\\
&=\limsup_{n\to \infty}\frac{n \log n}{O(n)+\log \Gamma(n+1)-\log \Gamma \left(\frac{n+1}{\alpha }\right)}.
\end{align}
Recall Stirling's formula$$
\log \Gamma(x)=\left(x-\frac{1}{2}\right)\log x-x +O(1)\quad (x\to \infty).
$$
Then we have \begin{align}
\lambda &=\limsup_{n\to \infty}\frac{n \log n}{O(n)+\left(1-\frac{1}{\alpha }\right)(n+1)\log (n+1)}\\
&=\frac{\alpha }{\alpha -1},
\end{align}
since $\frac{O(n)}{n\log n}\to 0$ and $\frac{(n+1)\log (n+1)}{n\log n}\to 1$ as $n \to \infty$.
