Both examples are $L^1_{loc}$ functions which are bounded at infinity, thus are tempered distributions, that is continuous linear forms on the Schwartz space $\mathscr S$ of rapidly decreasing function. The definition of the Fourier transform on $\mathscr S'$ is
$$
\langle \hat u,\phi\rangle_{\mathscr S',\mathscr S}=
\langle u,\hat \phi\rangle_{\mathscr S',\mathscr S}.
$$
From this definition, you get for instance that the Fourier transform of $x^{-1/2}1_{\mathbb R_+}(x)$ is an homogeneous distribution of degree $-1/2$. Let's do the explicit computation. For $z\in \mathbb C, \Re z>0$,
$$
\int_0^{+\infty}x^{-1/2} e^{-2π z x} dx=(2π z)^{-1/2}\int_0^{+\infty}t^{-1/2} e^{-t} dt
=(2 z)^{-1/2}=2^{-1/2}e^{-\frac{1}{2}Log z}
$$
with
$Log z=\int_{[1,z]}\frac{d\zeta}{\zeta}$ for $z\in \mathbb C\backslash\mathbb R_-$.
We may as well extend that formula for $z=i\xi+0$ so that
$$
Log z=\ln \vert \xi\vert+\frac{i\pi}{2}sign \xi
$$
and the sought Fourier transform is
$$
2^{-1/2}\vert \xi\vert^{-1/2}e^{-\frac{i\pi}{4}sign \xi}.
$$
That formal computation is well justified by the weak definition given at the beginning of the answer.

Bazin.

haveto do these computations". So you haveto make a choice. Please make sure that you know how tojustifyyour choice. In other words, it's not enough justify the computation after the fact by saying "it provided the answer that I wanted to get". You need to understand why the particular choice that you made is the correct one to make. – André Henriques Jun 15 '11 at 12:49