Fourier transform of a function of compact support My professor occasionally assigns optional difficult problems which we do not turn in from Stein and Shakarchi's Complex Analysis.  I am currently studying for a test in that class and try to get all of these optional problems answered.  One problem he gave us is Problem 2 from Chapter 4 on page 132 which you can find here 
http://carlossicoli.free.fr/S/Stein_E.M.,_Shakarchi_R.-Complex_Analysis-Princeton_univ_press(2003).pdf I am currently working on part (a)
Suppose f has bounded support and is of class $C^2$.  For $z \in \mathbb{C}$, let $\hat{f}(z)=\int_{-\infty}^{\infty} f(t)e^{-2\pi izt} dt$.  I am supposed to observe that $\hat{f}$ is an entire function, and using integration by parts show that for fixed $a\ge 0$ then $|y|\le a$ implies that for some constant $C_a$, $|\hat{f}(x+iy)|\le \frac{C_a}{1+x^2}$.  It says observe $\hat{f}$ is an entire function so I assume it is something simple but I don't see it.  Maybe I will have to evaluate the integral first. Which leads me to the integration by parts.  I am struggling with that without knowing the function f specifically.  I tried using $f$ as $u$ and the exponential function as $dv$ but got nowhere.  Thus, I am here asking for your help.  Thanks!
 A: A straightforward way you can show that $\hat{f}$ is an entire function is write
$$
         \hat{f}(z) = \int_{-R}^{R}f(t)e^{2\pi izt}\,dt 
                    = \int_{-R}^{R}f(t)\sum_{n=0}^{\infty}\frac{(2\pi izt)^{n}}{n!}\,dt
                    = \sum_{n=0}^{\infty}\frac{(2\pi iz)^{n}}{n!}\int_{-R}^{R}t^{n}f(t)\,dt,
$$
where $f(t)=0$ outside $[-R,R]$. Interchanging integration and summation is allowed because $f$ is supported on a finite interval $[-R,R]$ and because the series for the exponential converges absolutely and uniformly on $|tz| \le r$ for any fixed $r > 0$. So the series on the right converges for each fixed $z$, which means that the series must be an entire function of $z$ with an infinite radius of convergence.
Assume that $f$ vanishes for $|t| \ge R$, and let $M$ be a bound for $f$ on $[-R,R]$. Then, for $|\Im z| \le \alpha$,
$$
      \left|\int_{-R}^{R}f(t)e^{2\pi itz}\,dt\right| \le \int_{-R}^{R}|f(t)|e^{2\pi\alpha|t|}\,dt \le 2RMe^{2\pi R\alpha}.
$$
Let $N$ be a bound for $f''$ on $[-R,R]$. Then the far field, the bound is better. For $|\Im z| \le \alpha$ and $z\ne 0$, integration by parts twice gives
$$
   \left|\int_{-R}^{R}f(t)e^{2\pi izt}\,dt\right| = \left|\frac{1}{(2\pi i z)^{2}}\int_{-R}^{R}e^{2\pi itz}f''(t)\,dt\right| \le \frac{2RNe^{2\pi R\alpha}}{4\pi^{2}|z|^{2}}.
$$
It's not hard to combine these into a single estimate
$$
                      |\hat{f}(x+iy)| \le \frac{C_{\alpha}}{1+x^{2}},\;\;\; |y| \le \alpha.
$$
A: Yes, as you probably figured out you can do the differentiation under the integral sign; the reason would be that you are differentiating with respect to z, and all the stuff under the sign is wrt t.
I can get started here. You could write $e^{-2\pi it}$ as $e^{-2 \pi ixt}e^{2 \pi yt}$.    You've got a bound on $e^{2 \pi yt} \le e^{2 \pi at}$ and the other exponential is naturally bounded. I think the integration by parts should be u(t) = f(t)$ e^{2 \pi at}$ and v'(t) = $e^{-2\pi ixt}$.  It may be that you may want to integrate by parts twice and compare with the original integral.  You can do that because f is in $C^2$.  
See if you can get farther with this.
