Fourier transform of function What is Fourier transform of
$$f(x)=\frac{e^{-|x|}}{\sqrt{|x|}}?$$
I tried to calculate it using
$$F(e^{-|x|})=\sqrt{\frac{\pi}{2}}e^{-|a|}$$
and $$F(\frac{1}{\sqrt{|x|}})=\frac{1}{\sqrt{|a|}}$$
and convolution, but it seems to even more complicated.
 A: Your FT is
$$\begin{align}\int_{-\infty}^{\infty} dx \, |x|^{-1/2} \, e^{-|x|} e^{i k x} &= \int_{-\infty}^{0} dx \, (-x)^{-1/2} \,  e^{(1+i k) x} + \int_{0}^{\infty} dx \, x^{-1/2} \, e^{-(1-i k) x}\\ &= 2 \int_{0}^{\infty} du \, \left (e^{-(1-i k) u^2} + e^{-(1+i k) u^2}\right )\\ &= \sqrt{\pi} \left [(1-i k)^{-1/2}+(1+i k)^{-1/2} \right ] \\ &= 2 \sqrt{\pi} \, \Re{[(1+i k)^{-1/2}]}\\ &=2 \sqrt{\pi} (1+k^2)^{-1/4} \cos{\left(\frac12 \arctan{k}\right )}\\ &= \sqrt{2 \pi} \frac{\sqrt{1+\sqrt{1+k^2}}}{  \sqrt{1+k^2}}\end{align}$$
Note that I did not rely on the convolution theorem.  The lesson here is that sometimes it is easier just to evaluate the FT directly.
A: $\newcommand{\+}{^{\dagger}}
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 \newcommand{\equalby}[1]{{#1 \atop {= \atop \vphantom{\huge A}}}}
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 \newcommand{\fermi}{\,{\rm f}}
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\begin{align}
\fermi\pars{x} & \equiv {\expo{-\verts{x}} \over \root{\verts{x}}}
     =\int_{-\infty}^{\infty}
     \tilde{\fermi}\pars{k}\expo{-\ic kx}
\,{\dd k \over 2\pi}
\\[3mm] &
\imp\quad\tilde{\fermi}\pars{k} = \int_{-\infty}^{\infty}{\expo{-\verts{x}} \over \root{\verts{x}}}\,\expo{\ic k x}\,\dd x
\end{align}

\begin{align}
\color{#00f}{\large\tilde{\fermi}\pars{k}}&=
\int_{-\infty}^{\infty}{\expo{-\verts{x}} \over \root{\verts{x}}}\,
\expo{\ic k x}\,\dd x
=\int_{-\infty}^{\infty}{\expo{-\verts{x}} \over \root{\verts{x}}}\,\cos\pars{kx}
\,\dd x
\\[3mm] & =
2\Re\int_{0}^{\infty}{\expo{-x} \over \root{x}}\,
\expo{\ic kx}\,\dd x
\\[3mm]&=2\Re\int_{0}^{\infty}x^{-1/2}\expo{-\pars{1 - \ic k}x}\,\dd x
\\[3mm] & =
2\Re\bracks{\pars{1 - \ic k}^{-1/2}\
\overbrace{\int_{0}^{\infty}x^{-1/2}\expo{-x}\,\dd x}
^{\ds{\Gamma\pars{\half} = \root{\pi}}}}
\\[3mm]&=2\root{\pi}\Re\pars{1 - \ic k}^{-1/2}
=\bracks{\root{1 + k^{2}}\exp\pars{-\ic\arctan\pars{k}}}^{-1/2}
\\[3mm]&=2\root{\pi}\pars{1 + k^{2}}^{-1/4}\cos\pars{\arctan\pars{k} \over 2}
\\[3mm] & =2\root{\pi}\pars{1 + k^{2}}^{-1/4}
\root{1 + \cos\pars{\arctan\pars{k}} \over 2}
\\[3mm]&=\root{2\pi}\pars{1 + k^{2}}^{-1/4}
\root{1 + {1 \over \root{\tan^{2}\pars{\arctan\pars{k}} + 1}}}
\\[3mm]&=\root{2\pi}\pars{1 + k^{2}}^{-1/4}
\root{1 + {1 \over \root{k^{2} + 1}}}
\\ [3mm] & =\root{2\pi}
\root{{1 \over \root{1 + k^{2}}}\,{1 + \root{1 + k^{2}} \over \root{1 + k^{2}}}}
\\[3mm]&=\color{#00f}{\large\root{2\pi}\,\root{1 + \root{1 + k^{2}} \over 1 + k^{2}}}
\end{align}
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
\tilde{\on{f}}\pars{k} & \equiv
\bbox[5px,#ffd]{\int_{-\infty}^{\infty}{%
\expo{-\verts{x}} \over \root{\verts{x}}}\,\expo{\ic k x}\,\dd x} =
2\,\Re\int_{0}^{\infty}{%
\expo{-x} \over \root{x}}\,\expo{\ic k x}\,\dd x
\\[5mm] & =
2\,\Re\int_{0}^{\infty}x^{\color{red}{1/2} - 1}\,\,
\expo{-\pars{1 - \ic k}x}\,\,\,\dd x
\end{align}
Note that
$\ds{\expo{-\pars{1 - \ic k}x}\ = \sum_{0}^{\infty}{\bracks{-\pars{1 - \ic k}x}^{\,n} \over n!} =
\sum_{0}^{\infty}\color{red}{\pars{1 - \ic k}^{n}}
\,\,{\pars{-x}^{\,n} \over n!}}$.
Then,
\begin{align}
\tilde{\on{f}}\pars{k} & =
2\,\Re\bracks{\Gamma\pars{\color{red}{1 \over 2}}
\pars{1 - \ic k}^{-\color{red}{1/2}}}\
\pars{\substack{\ds{Ramanujan's} \\[0.5mm] \ds{Master}\\[0.5mm] \ds{Theorem}}}
\\[5mm] & = 
2\root{\pi}\Re\braces{\bracks{\root{1 + k^{2}}\expo{\ic\arctan\pars{-k}}\,\,}^{-1/2}}
\\[5mm] & =
2\root{\pi}\pars{1 + k^{2}}^{-1/4}\,\,
\cos\pars{\arctan\pars{k} \over 2}
\\[5mm] & =
2\root{\pi}\pars{1 + k^{2}}^{-1/4}\,\,
\root{1 + \cos\pars{\arctan\pars{k}} \over 2}
\\[5mm] & =
2\root{\pi}\pars{1 + k^{2}}^{-1/4}\,\,
\root{\sec\pars{\arctan\pars{k}} + 1 \over 2\sec\pars{\arctan\pars{k}}}
\\[5mm] & =
2\root{\pi}\pars{1 + k^{2}}^{-1/4}\,\,
\root{\root{k^{2} + 1} + 1 \over 2\root{k^{2} + 1}}
\\[5mm] & =
\bbx{\root{2\pi}
\root{\root{k^{2} + 1} + 1 \over k^{2} + 1}} \\ &
\end{align}
