Fourier transform of $\frac{1}{\sqrt{1 + x^2}}$ I have a question about Fourier transforms:
What does the following question means?
" Does the Fourier transform of the function $f(x)= \frac{1}{\sqrt{1 + x^2}}$ belongs to $L_2(\mathbb{R})$?"
We know that $L_2(\mathbb{R})= \{ f: \mathbb{R} \rightarrow \mathbb{C} \mid \int_{- \infty}^{+ \infty} |f(t)|^2 dt < \infty \} $ and we know that $\hat{f}(t) = \int_{- \infty}^{+ \infty} e^{-itx} f(x) dx$.
Now for to show that $\hat{f}(t) \in L_2(\mathbb{R})$, what do we need to prove?
$\textbf{(1)}$ $~$ Is it enough to show that $f(x) \in L_1(\mathbb{R}) \cap L_2(\mathbb{R})$, then we have some results that shows $\hat{f}(t) \in L_2(\mathbb{R})!$
$\textbf{(2)}$ $~$ Or we have to prove $\int_{- \infty}^{+ \infty} |\hat{f}(t)|^2 dt < \infty$?
If $\textbf{(2)}$ is true? Can you please help me to show that?
Here linked-to result someone has found its Fourier transform, but I cannot understand it! I mostly prefer to find it by usual integration ways such as variable changes or substitution or even by inverse transform theorem!
Please let me know if I am wrong about using $\textbf{(1)}$ for to prove that $\hat{f}(t) \in L_2(\mathbb{R})$?
Thanks! 
 A: As already stated it depends upon how you really define the Fourier transform. 
If you define it as an improper integral the question becomes non-trivial (otherwise it is just Plancherel). For $n\geq 1$ let
  $$ f_n(x) = f(x)\; {\rm \bf 1}_{[-n,n]}(x) = \frac{1}{\sqrt{1+x^2}} {\rm \bf 1}_{[-n,n]}(x) $$
where ${\rm \bf 1}_{[-n,n]}$ is the indicator function for the interval $[-n,n]$. Then $f_n$ is in both $L^1\cap L^2$ and the Fourier transform
$$ \hat{f}_n(t) = \int_{-n}^n \frac{e^{itx}}{\sqrt{1+x^2}} dx $$
converges for every $t\neq 0$. This follows from partial integration:
 $$ \hat{f}_n(t) = \left[ \frac{e^{itx}}{it\sqrt{1+x^2}} \right]_{-n}^n +   \int_{-n}^n \frac{\sin(tx)}{t} \frac{x}{\sqrt{1+x^2}^3} dx . $$
As $n\rightarrow \infty$, the first term goes to zero and the latter converges absolutely.
Thus, you may define $\hat{f}(t)=\lim_n \hat{f}_n(t)$ for every $t\neq 0$ and the question is then if $\hat{f}\in L^2$?
The answer is yes: First note, that $f_n \rightarrow f$ in $L^2$ so $(f_n)_{n\geq 1}$ is Cauchy in $L^2$.
By Plancherel $\|\hat{f}_n-\hat{f}_m\|_2 = \|f_n-f_m\|_2$, so $(\hat{f}_n)_{n\geq 1}$ is also Cauchy in $L^2$, whence converges in $L^2$ to some function $g$. 
Any convergent sequence in $L^2$ has, however, a subsequence that converges a.e. and we already know that for $t\neq 0$ (i.e. a.e.) the limit is $\hat{f}(t)$, so  $\hat{f}=g$ (mod 0)  is in $L^2$ 
A: Answering confusion in the comments about the Plancherel theorem:
A rough version of the theorem is


$\hat f\in L^2(\mathbb R)$ if and only if $f\in L^2(\mathbb R)$.


That's true, but it really needs to be stated more carefully. Because for example  it's not clear what $\hat f$ even means given just $f\in L^2$. A more careful version starts here:


If $f\in L^1\cap L^2$ then $\hat f\in L^2$ and $\lVert\hat f\rVert_2=\lVert f\rVert_2$.


There there's no problem with the definition of $\hat f$, since $f\in L^1$.
$\newcommand{\ft}{\mathcal F}$
At this point it's convenient to introduce the notation
$$\ft f=\hat f.$$
Now $\ft$ is defined on a dense subspace of $L^2$ and $\ft$ is an isometry (in the $L^2$ norm) on that dense subspace. Hence $\ft$ extends to an isometry  $\ft:L^2\to L^2$. When people talk about $\hat f$ for $f\in L^2$ they're referring to that extension. (And now we're done with the problem in the question: Since $f\in L^2$ it follows that $\hat f\in L^2$.)
It follows easily that $\ft$ is its own inverse, almost:


If $f\in L^2$ then $\ft^2f(t)=f(-t)$.


Proof: The $L^1$ inversion theorem shows that this holds for, say, $f$ in the Schwartz space. Since the Schwartz space is dense and $\ft$ is continuous on $L^2$ (as is the operator $f(t)\mapsto f(-t)$) it holds for all $f\in L^2$.
In particular $\ft$ maps $L^2$ onto $L^2$: that is, $\ft$ is an isometric isomorphism, which is to say $\ft$ is unitary. (And in fact $\ft^4=I$, which leads easily to a decomposition of $L^2$ into four eigenspaces...)
