Integral $ \int_{-\infty}^\infty \frac{e^{ikx}}{x^{3/2}}dx$ Hi I'm trying to solve this integral Fourier Transform
$$
\int_{-\infty}^\infty \frac{e^{ikx}}{x^{3/2}}dx=\sqrt{2\pi|k|}(1+i)  (-1+\text{sgn}(k))
$$
where sgn(k)$=1$ for k>1 and $-1$ for k<1.
I am trying to use residues.  Thanks there is a singularity at $x=0$.  We can try and write
$$
e^{ikx}=\cos x+i\sin x
$$
but I don't think it will help.  IT will be nice to use a contour with $e^{ikx}$ instead.  Thanks
 A: $\newcommand{\+}{^{\dagger}}
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With $\ds{s \equiv \ic x\quad\imp\quad x = -\ic s = \expo{3\ic\pi/2}s}$:
\begin{align}
\int_{-\infty + \ic 0^{+}}^{\infty + \ic 0^{+}}
{\expo{\ic kx} \over x^{3/2}}\,\dd x&=
\Theta\pars{-k}\int_{-0^{+} - \ic\infty}^{-0^{+} + \ic\infty}
{\expo{\ic k\pars{-\ic s}} \over \pars{\expo{3\ic\pi/2}s}^{3/2}}\,\pars{-\ic\,\dd s}
\\[3mm]&=
\Theta\pars{-k}\pars{-\ic\expo{-9\ic\pi/4}}
\int_{0^{+} -\ic\infty}^{0^{+} + \ic\infty}{\expo{\verts{k}s} \over \expo{-3\ic\pi/2}s^{3/2}}\,\dd s
\\[3mm]&=\Theta\pars{-k}{\root{2} \over 2}\pars{-1 + i}
\int_{0^{+} -\ic\infty}^{0^{+} + \ic\infty}s^{-3/2}\expo{\verts{k}s}\,\dd s\tag{1}
\end{align}

$$
\int_{0^{+} - \ic\infty}^{0^{+} + \ic\infty}s^{-3/2}\expo{\verts{k}s}\,\dd s
=\lim_{\epsilon \to 0^{+}}\pars{%
{\cal I}_{+} + {\cal I}_{-}}\tag{2}
$$

\begin{align}
{\cal I}_{+}&=-\int_{-\infty + \ic\epsilon}^{\ic\epsilon}
\pars{-x}^{-3/2}\expo{-3\ic\pi/2}\expo{\verts{k}x}\,\dd x=
-\ic\int_{-\ic\epsilon}^{\infty - \ic\epsilon}x^{-3/2}\expo{-\verts{k}x}\,\dd x
\\[3mm]&=-2\ic\pars{-\ic\epsilon}^{-1/2}
+ 2\ic\verts{k}\int_{-\ic\epsilon}^{\infty - \ic\epsilon}x^{-1/2}\expo{-\verts{k}x}\,\dd x
\\[3mm]&=\root{2}\pars{-1 + i}\epsilon^{-1/2}
+ 2\ic\verts{k}\int_{-\ic\epsilon}^{\infty - \ic\epsilon}x^{-1/2}\expo{-\verts{k}x}\,\dd x\tag{3}
\end{align}

\begin{align}
{\cal I}_{-}&=-\int_{-\ic\epsilon}^{-\infty - \ic\epsilon}
\pars{-x}^{-3/2}\expo{3\ic\pi/2}\expo{\verts{k}x}\,\dd x=
-\ic\int_{\ic\epsilon}^{\infty + \ic\epsilon}x^{-3/2}\expo{-\verts{k}x}\,\dd x
\\[3mm]&=-2\ic\pars{\ic\epsilon}^{-1/2}
+2\ic\verts{k}\int_{\ic\epsilon}^{\infty + \ic\epsilon}x^{-1/2}\expo{-\verts{k}kx}\,\dd x
\\[3mm]&=\root{2}\pars{1 - i}\epsilon^{-1/2}
+ 2\ic\verts{k}\int_{\ic\epsilon}^{\infty + \ic\epsilon}x^{-1/2}\expo{-\verts{k}x}\,\dd x\tag{4}
\end{align}

We replace $\pars{3}$ and $\pars{4}$ in $\pars{2}$. In the limit
$\epsilon \to 0^{+}$:
\begin{align}
\Theta\pars{-k}\int_{0^{+} - \ic\infty}^{0^{+} + \ic\infty}s^{3/2}\expo{\verts{k}s}\,\dd s
&=\Theta\pars{-k}\bracks{4\ic\verts{k}\int_{0}^{\infty}x^{-1/2}\expo{-\verts{k}x}\,\dd x}
\\[3mm]&=4\ic\root{\verts{k}}\Theta\pars{-k}\
\underbrace{\int_{0}^{\infty}x^{-1/2}\expo{-x}\,\dd x}
_{\ds{\Gamma\pars{1/2} = \root{\pi}}}
\\[3mm]&=\Theta\pars{-k}\bracks{4\ic\root{\pi\verts{k}}}
=-2\ic\bracks{-1 + \sgn\pars{k}}\,\root{\pi\verts{k}}\tag{5}
\end{align}

We'll replace $\pars{5}$ in $\pars{1}$:
  \begin{align}
\color{#00f}{\large\int_{-\infty}^{\infty}{\expo{\ic kx} \over x^{3/2}}\,\dd x}&=
{\root{2} \over 2}\pars{-1 + i}\braces{%
-2\ic\bracks{-1 + \sgn\pars{k}}\,\root{\pi\verts{k}}}
\\[3mm]&=\color{#00f}{\large\root{2\pi\verts{k}}\pars{1 + i}\bracks{-1 + \sgn\pars{k}}}
\end{align}

