Fourier Transform of $\frac{1}{\sqrt{|x|}}$ I want to find the fourier transform of $\frac{1}{\sqrt{|x|}}$. I checked the table of common fourier transforms in Wikipedia, and I know the answer should be $$\sqrt{\frac{2\pi}{|\omega|}}$$
What I can't find out, however, is why that is the answer.
I tried 
$$ \hat{f}(\omega) = \int_{-\infty}^{\infty} \frac{1}{\sqrt{|x|}} e^{-i\omega x} dx$$
$$ = \int_{0}^{\infty} \frac{1}{\sqrt{x}} e^{-i\omega x} dx + \int_{0}^{\infty} \frac{1}{\sqrt{x}} e^{i\omega x} dx$$
but that just gives me two unsolvable exponential integrals.
I also tried finding the answer through residue calculus, as the function has a single singularity at 0, which yields
$$ \hat{f}(\omega) = 2\pi i \ Res_{z = 0} \frac{e^{-i \omega z}}{\sqrt{|z|}} = 2\pi i \lim_{z \to 0} (e^{-i \omega z}) = 2\pi i$$
What am I doing wrong? Or am I thinking completely in the wrong direction?
Thanks in advance!
 A: Make the change of variables $x=t^2$ in both integrals in $\hat f(\omega)$ and use parity to extend the limits of integration to $-\infty$ and $\infty$. 
Things will then become much more clear.
A: Apart from potentially interesting and compelling heuristics, the literal integrals for Fourier transforms do not converge at all for such functions. But that's ok, because the Fourier transform extends to tempered distributions not by the literal integral description, but by an extension-by-continuity in the dual topology to Schwartz functions (which is quite weak, and does not refer at all to pointwise convergence of integrals).
Even better than just looking at weak-dual limits is to look at the properties of functions such as $|x|^s$. They are homogeneous. It is easy to show that Fourier transform converts homogeneous tempered distributions of degree $s$ to homogeneous tempered distributions of degree $1-s$ (up to normalizations and conventions...). So the only question is to determine the constant, which can be done by applying the functional(s) to things like Gaussians or Gaussians multiplied by $x$...
EDIT: as noted by @F.H. in a comment, the Fourier transform is homogeneous of degree $-1-s$, not $1-s$.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[#ffd,5px]{\int_{-\infty}^{\infty}{\expo{-\ic\omega x} \over \root{\verts{x}}}\,\dd x} =
2\int_{0}^{\infty}{\cos\pars{\verts{\omega}x} \over \root{x}}\,\dd x
\\[5mm] \stackrel{\verts{\omega}x\ \mapsto\ x}{=}\,\,\,&
{2 \over \root{\verts{\omega}}}
\int_{0}^{\infty}{\cos\pars{x} \over \root{x}}\,\dd x
\\[5mm]
\stackrel{\root{x}\ \mapsto\ x}{=}\,\,\,&
{4 \over \root{\verts{\omega}}}\
\underbrace{\int_{0}^{\infty}\cos\pars{x^{2}}\,\dd x}
_{\ds{{\root{2\pi} \over 4}}}
\\ = &\
\bbx{\root{2\pi} \over \root{\verts{\omega}}}
\\ &
\end{align}
The last integral is a
Fresnel Integral.
