Question about the Fourier transform of $\ln(1+x^2)\exp(-x^2)$

Question

So given the function $f(x) = \ln(1+x^2)\exp(-x^2)$, is the Fourier transformation F infinitely differentiable? I tried to calculate the Fourier transformation but I couldn't find a solution so I was wondering if there is a quick way to verify this.

Answer 1

A simple lemma, presumably something that was covered in class before the exercise was assigned:

Suppose $f\in L^1(\Bbb R)$ and $\int|t||f(t)|\,dt<\infty$. Then the Fourier transform $\hat f$ is differentiable, with derivative equal to the Fourier transform of the function $-it f(t)$.

Proof: Dominated convergence shows that $$\frac{\hat f(\xi+h)-\hat f(\xi)}{h}=\int\frac{e^{-i(\xi+h)t}-e^{-i\xi t}}{h}f(t)\,dt\to\int e^{-i\xi t}(-it)f(t)\,dt$$as $h\to0$. (For the "dominated" part, recall that $$\left|e^{is}-e^{it}\right|\le|s-t|,\quad s,t\in\Bbb R.)$$

Now if $f$ is as in the question it's clear that $$\int|t|^k|f(t)|\,dt<\infty$$for $k=1,2\dots$; hence it follows by induction that the derivative $F^{(k)}$ exists.

An example showing that the converse of the lemma is false (here I assume the reader is familiar with basic Fourier analysis):

Say $\phi\in C^\infty_c (\Bbb R)$, with support in $(-1/2,1/2)$. There exists $g\in L^1$ with $$\hat g=\phi.$$

Now let $$\phi_n(\xi)=\frac1{n^2}\phi(2^n(\xi-n)).$$Then $$\phi_n=\hat g_n,$$where $$||g_n||_1=\frac1{n^2}||g||_1.$$Let $$f=\sum_{n=1}^\infty g_n.$$Then $f\in L^1$ and $\hat f=\sum\phi_n$ is certainly differentiable, but the Riemann-Lebesgue lemma shows that $\int|t||f(t)|\,dt=\infty$, since the derivative of $\hat f$ does not tend to $0$ at infinity.