The Fourier transform of a power of the absolute value function (and a related integral)

Question

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.

Answer 1

$\newcommand{\+}{^{\dagger}} \newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\down}{\downarrow} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\isdiv}{\,\left.\right\vert\,} \newcommand{\ket}[1]{\left\vert #1\right\rangle} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert} \newcommand{\wt}[1]{\widetilde{#1}}$ $\ds{\int_{\mathbb{R}}\expo{2\pi\ic t}\verts{t}^{a}\,\dd t:\ {\large ?} \quad\mbox{and}\quad \int_{\mathbb{C}}\expo{2\pi\ic\pars{z + \ol{z}}}\verts{z}^{a}\,\dd z:\ {\large ?}.\qquad 0 < a < 1}$

\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 $$\color{#00f}{\large% \int_{\mathbb{C}}\expo{2\pi\ic\pars{z + \ol{z}}}\verts{z}^{a}\,\dd z}= -\frac{(2 \pi )^{1-2 (a+1)} \left(a (2 \pi )^a+2^{a+1} \pi ^a\right) \Gamma \left(\frac{a}{2}+1\right)}{(a+2)^2 \Gamma \left(-\frac{a}{2}-1\right)} $$

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

Answer 2

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.