I am getting stuck on some detail part of a practice problem.
I am trying to deduce an upper bound for $|\int_{-\infty}^{\infty} e^{-|t|^\alpha} e^{2\pi izt}dt|$ to get the order of growth in the form $Ae^{B|x|^\beta}$. This is the exercise 5.5 in Stein complex analysis. [I have looked through all other posts related to this question, but I don't the responses under those posts helped me understand how it works. ]
According to the hint, I have successfully deduced $-\frac{|t|^\alpha}{2}+2\pi |z||t| \leq c|z|^{\frac{\alpha}{\alpha -1}}$.
Therefore, I claimed $|\int_{-\infty}^{\infty} e^{-|t|^\alpha} e^{2\pi izt}dt|\leq \int_{-\infty}^{\infty} e^{-|t|^\alpha/2} e^{-|t|^\alpha / 2 + 2\pi i|z||t|}dt$.
I want to show $\int_{-\infty}^{\infty} e^{-|t|^\alpha/2} e^{-|t|^\alpha / 2 + 2\pi |z||t|}dt \leq ke^{c|z|^\frac{\alpha}{\alpha-1}}$, but I am not sure how I should take off the integral part.
I see $e^{-|t|^\alpha / 2 + 2\pi |z||t|}\leq e^{c|z|^\frac{\alpha}{\alpha-1}}$, but then how should I handle the other half integrand $e^{-|t|^\alpha/2} $?
- I don't think I can divide $e^{2\pi |k| |t|}$ because I do want to have $|t|$ on one side.
- I have taken the absolute value, so I can't see $e^{2\pi |k| |t|}$ as a change of orientation either.
Also, I thought in general $\int f(x)\leq \int g(x)$ when $f(x)\leq g(x)$, but here I am trying to get something like $\int f(x) \leq g(x)$. How can I achieve this?
Thank you in advance for your help!