Question: 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!

Answer: 
You might find it easier to rearrange the equation as $$z=\frac{2w}{w-1}$$ and then apply the condition $$|z|=1\Rightarrow 2|w|=|w-1|$$.

This leads to the equation of the circle $$3u^2+3v^2+2u-1=0$$

This has centre $(-\frac 13,0)$ and radius $\frac 23$, so there appears to be a misprint in the book.
If you do it the way you started, you have to push all the way for both $x$ and $y$:

$$
\begin{align}
u&=\Re\left(\frac{x+yi}{x+yi-2}\right)\\
v&=\Im\left(\frac{x+yi}{x+yi-2}\right)
\end{align}
$$

getting:

$$
\begin{align}
u&=\frac{x(x-2)+y^2}{(x-2)^2+y^2}\text{ (1)}\\
v&=\frac{y(x-2)-xy}{(x-2)^2+y^2}
\end{align}
$$

System (1) has the solution:

$$
\begin{align}
x&=\frac{2(u^2+v^2-u)}{u^2+v^2-2u+1}\text{ (2)}\\
y&=\frac{-2v}{u^2+v^2-2u+1}
\end{align}
$$

Now push $x$ and $y$ from (2) into equation $x^2+y^2=1$ to get:

$$\frac{4(u^2+v^2)}{u^2+v^2-2u+1}=1$$

which if you simplify, will give you the equation:

$$3u^2+2u+3v^2=1$$

which is the same as that in the answer of David.