Fourier transform of a function with exponential and powers How can I calculate this Fourier transform $F(y)$ ?
$$F(y)= \int_0^{\infty}(1+x)^{\frac{1}{2}} x^{-\frac{1}{2}-a} e^{-a x} \cos(2 \pi xy) dx$$
with $a$ complex ($0<Re(a)<\frac{1}{2}$)
This is in fact a Cosine transform.
 A: You can use a number of Fourier Transform theorems and tables of transforms to calculate this Fourier Transform.
Using this following form of the Fourier Transform:
$$F(s) = \int_{-\infty}^{\infty} f(x) e^{-2\pi i xs} ds$$
The convolution theorem tells you that a product of functions in the $x$ domain, is a convolution of Fourier Transforms in the $s$ domain:
$$\mathscr{F}\{f(x)g(x)...h(x)\} = F(s)*G(s)*...*H(s)$$
The shift theorem tells you a shift in the $x$ domain is multiplication by a complex exponential in the $s$ domain:
$$\mathscr{F}\{f(x-a)\} = e^{-2\pi i as} F(s)$$
The similarity theorem tells you how a scale (by a real scalar) in the $x$ domain, maps to the $s$ domain:
$$\mathscr{F}\{f(ax)\} = \dfrac{1}{|a|}F\left(\dfrac{s}{a}\right)$$
You mention you have a complex constant $a = c + id$, so the complex part can be factored our of the exponential and absorbed into the Fourier Kernel:
$$e^{-ax}e^{-2\pi i xs} = e^{-cx}e^{-idx}e^{-2\pi i xs}=e^{-cx}e^{-ix(2\pi s+d)}$$
which could then be integrated as a normal Fourier transform with a change of variable, $s' = s + \dfrac{d}{2 \pi}$.
And some transform pairs from a table that you may find useful:
$$\mathscr{F}\left\{|x|^{-\frac{1}{2}}\right\} = |s|^{-\frac{1}{2}}$$
$$\mathscr{F}\left\{e^{-|x|}\right\} = \dfrac{2}{1+(2\pi s)^2}$$
Given all of the above, I think you can calculate your desired transform.
I'm not quite sure about the $|x|^{-a}$ term.  That might be the hard part.
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

$\ds{\mrm{F}\pars{y} \equiv
\int_{0}^{\infty}\pars{1 + x}^{1/2}\,x^{-1/2 - a}\expo{-ax}\cos\pars{2\pi xy} \,\dd x:\ {\large ?}.\qquad 0 < \Re\pars{a} < 1/2}$.

With $\ds{z \equiv a + 2\pi y\ic}$:
\begin{align}
\mrm{F}\pars{y} & \equiv
\int_{0}^{\infty}\pars{1 + x}^{1/2}\,x^{-1/2 - a}\expo{-ax}\cos\pars{2\pi xy} \,\dd x
\\[5mm] & = \mrm{f}\pars{a + 2\pi y\ic} + \mrm{f}\pars{a - 2\pi y\ic}
\quad\mbox{where}\quad
\mrm{f}\pars{z} \equiv
{1 \over 2}\int_{0}^{\infty}\pars{1 + x}^{1/2}\,x^{-1/2 - a}\expo{-xz} \,\dd x
\end{align}

Lets $\ds{k \equiv {1 + a \over 2}}$ and $\ds{m \equiv {1 - a \over 2}}$. Then,
\begin{align}
\mrm{f}\pars{z} & \equiv
{1 \over 2}\int_{0}^{\infty}x^{-k - 1/2 + m}\pars{1 + x}^{k - 1/2 + m}\expo{-xz} \,\dd x
\\[5mm] & =
{1 \over 2}\,{1 \over z^{-k + 1/2 + m}}\int_{0}^{\infty}
x^{-k - 1/2 + m}\pars{1 + {x \over z}}^{k - 1/2 + m}\expo{-x} \,\dd x
\\[5mm] & =
{1 \over 2}\,{1 \over z^{-k + 1/2 + m}}\bracks{{\Gamma\pars{1/2 - k + m} \over \expo{-z/2}z^{k}}\,\mrm{W}_{k,m}\pars{z}}
\end{align}


where $\ds{\,\mrm{W}_{k,m}}$ is the
  Whittaker Function.

Then,
\begin{align}
\mrm{f}\pars{z} & =
{1 \over 2}\,\Gamma\pars{{1 \over 2} - a}z^{a/2 - 1}\expo{z/2}
\,\mrm{W}_{\pars{1 + a}/2,\pars{1 - a}/2}\pars{z}
\end{align}

$$
\begin{array}{|rcl|}\hline\\
\ds{\quad\mrm{F}\pars{y}} & \ds{=} & \ds{{1 \over 2}\,\Gamma\pars{{1 \over 2} - a}
\left[\vphantom{\LARGE A}\pars{a + 2\pi y\ic}^{a/2 - 1}\expo{a/2 + \pi y\ic}
\,\mrm{W}_{\pars{1 + a}/2,\pars{1 - a}/2}\pars{a + 2\pi y\ic}\
{\large +}\right.\quad}
\\&&
\ds{\phantom{{1 \over 2}\,\Gamma\pars{{1 \over 2} - a}\,\,\,}\left.
\vphantom{\LARGE A}
\pars{a - 2\pi y\ic}^{a/2 - 1}\expo{a/2 - \pi y\ic}
\,\mrm{W}_{\pars{1 + a}/2,\pars{1 - a}/2}\pars{a - 2\pi y\ic}\right]}
\\ && \mbox{}\\ \hline
\end{array}
$$
