The Fourier transform of a power of the absolute value function (and a related integral) What (Fourier-analytic?) methods would I use to compute the following two integrals?

$\displaystyle\int_{\mathbb{R}} e^{2 \pi i t} |t|^a dt \:\:\:\:\:\:\: \:\:\:\:\:\:\: \text{ and } \:\:\:\:\:\:\: \:\:\:\:\:\:\: 
\displaystyle\int_{\mathbb{C}} e^{2 \pi i (z + \overline z) } |z|^a dz  \:\:\:\:\:\:\: \:\:\:\:\:\:\:
 (0 < a < 1)$

(These do converge, don't they?) Just hints please. 
 A: $\newcommand{\+}{^{\dagger}}
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$\ds{\int_{\mathbb{R}}\expo{2\pi\ic t}\verts{t}^{a}\,\dd t:\ {\large ?}
\quad\mbox{and}\quad
\int_{\mathbb{C}}\expo{2\pi\ic\pars{z + \ol{z}}}\verts{z}^{a}\,\dd z:\
{\large ?}.\qquad 0 < a < 1}$

\begin{align}
&\color{#c00000}{\int_{\mathbb{R}}\expo{2\pi\ic t}\verts{t}^{a}\,\dd t}
=2\int_{0}^{\infty}\cos\pars{2\pi t}t^{a}\,\dd t
=2\Re\int_{0}^{\infty}\expo{2\pi t\,\ic}t^{a}\,\dd t
\\[3mm]&=
2\Re\int_{0}^{\infty}\expo{-t}\pars{\ic t \over 2\pi}^{a}
\,\pars{\ic\,{\dd t \over 2\pi}}
=-\,{2 \over \pars{2\pi}^{a + 1}}\,\Im\pars{\expo{\ic\pi a/2}
\int_{0}^{-\infty\ic}\expo{-t}t^{a}\,\dd t}\tag{1}
\\[3mm]&=-\,{2 \over \pars{2\pi}^{a + 1}}\,\Im\bracks{\expo{\ic\pi a/2}\pars{%
-\int_{0}^{\infty}\expo{-t}t^{a}\,\dd t
-\lim_{R \to \infty}\int_{z = R\expo{\ic\theta} \atop
{\vphantom{\Huge A}-\pi/2 < \theta < 0}}
\expo{-z}z^{a}\,\dd z}}
\\[3mm]&={2^{-a} \over \pi^{a + 1}}\,\sin\pars{\pi a \over 2}\Gamma\pars{a + 1}
+{2^{-a} \over \pi^{a + 1}}\,
\lim_{R \to \infty}\Im\pars{\expo{\ic\pi a/2}\int_{z = R\expo{\ic\theta} \atop
{\vphantom{\Huge A}-\pi/2 < \theta < 0}}
\expo{-z}z^{a}\,\dd z}
\end{align}
  $\ds{\Gamma\pars{z}}$ is the
  Gamma Function. The first integral converges when $\ds{a > -1}$. The second integral impose additional constraints on $\ds{a}$ as we will see below. 

The $R \to \infty$-limit vanishes out since
\begin{align}
&\verts{\Im\int_{z = R\expo{\ic\theta} \atop {\vphantom{\Huge A}-\pi/2 < \theta < 0}}
\expo{-z + \ic\pi a/2}z^{a}\,\dd z}
<\int_{-\pi/2}^{0}\exp\pars{-R\cos\pars{\theta}}R^{a + 1}\,\dd\theta
\\[3mm]&=\int_{0}^{\pi/2}\exp\pars{-R\sin\pars{\theta}}R^{a + 1}\,\dd\theta
<\int_{0}^{\pi/2}\exp\pars{-R\bracks{2\theta/\pi}}R^{a + 1}\,\dd\theta
\\[3mm]&=\half\,\pi R^{a}\pars{1 - \expo{-R}}
\color{#c00000}{\large \to 0}\ \mbox{when}\ R \to \infty
\quad\pars{~\mbox{whenever}\ -1 < a < 0~}
\end{align}

$$
\color{#00f}{\large\int_{\mathbb{R}}\expo{2\pi\ic t}\verts{t}^{a}\,\dd t
={2^{-a} \over \pi^{a + 1}}\,\sin\pars{\pi a \over 2}\Gamma\pars{a + 1}}\,,
\qquad -1 < a < 0
$$

\begin{align}
&\color{#c00000}{\int_{\mathbb{C}}\expo{2\pi\ic\pars{z + \ol{z}}}\verts{z}^{a}
\,\dd z}
=\int_{r = 0}^{r \to \infty}\int_{\theta = 0}^{\theta = 2\pi}
\expo{4\pi\ic r\cos\pars{\theta}}r^{a}\,
\pars{r\,\dd r\,\dd\theta}
\\[3mm]&=
\int_{0}^{\infty}\dd r\,r^{a + 1}
\int_{0}^{2\pi}\expo{4\pi r\cos\pars{\theta}\ic}\,\dd\theta
=2\pi\color{#00f}{\int_{0}^{\infty}r^{a + 1}{\rm J}_{0}\pars{4\pi r}\,\dd r}
\end{align}
where $\ds{{\rm J}_{\nu}\pars{z}}$ is the
Bessel Function of the First Kind.
When $\ds{0 < a < 1}$ the $\ds{\color{#00f}{\mbox{blue integral}}}$ diverges. It converges whenever $\ds{\Large -2 < \Re\pars{a} < \half}$.
In that case the result is
$$\color{#00f}{\large%
\int_{\mathbb{C}}\expo{2\pi\ic\pars{z + \ol{z}}}\verts{z}^{a}\,\dd z}=
-\frac{(2 \pi )^{1-2 (a+1)} \left(a (2 \pi )^a+2^{a+1} \pi ^a\right) \Gamma \left(\frac{a}{2}+1\right)}{(a+2)^2 \Gamma \left(-\frac{a}{2}-1\right)}
$$

$\large\tt\mbox{It is clear that both integrals converges simultaneously}$
  $\tt\large\mbox{whenever}\ -1 < a < 0$.

A: Ah - so here's how we can compute the first integral:
$\int_{\mathbb{R}} e^{2 \pi i t} |t|^{a-1}  dt  = \int_{0}^{\infty} (e^{2 \pi i t} + e^{- 2 \pi i t}) |t|^{a-1} dt = 2 \int_0^{\infty} \cos(2 \pi t)  t^{a-1} dt $
which is the Mellin transformation of $\cos(2 \pi t)$, giving $2 (2 \pi)^{-a} \cos(\pi a/2) \Gamma(a)$. This doesn't seem to generalise to the second integral.
