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.
Best Answer
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 $.