TITLE: Proof of the identity $\int_0^{+\infty}\frac{\sin(x)}{x^\alpha}dx=\frac{\Gamma(\alpha/2)\Gamma(1-\alpha/2)}{2\Gamma(\alpha)}$ for $\alpha\in (0,2)$.
QUESTION [5 upvotes]: Let $0<\alpha<2.$ Looking for a proof for the following: $$\int_0^{+\infty}\frac{\sin(x)}{x^\alpha}dx=\frac{\Gamma(\alpha/2)\Gamma(1-\alpha/2)}{2\Gamma(\alpha)}.$$
Any ideas?

REPLY [5 votes]: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\sin\pars{x} \over x^{\alpha}}\,\dd x}
\,\,\,\stackrel{x\ \mapsto\ \root{x}}{=}\,\,\,
\int_{0}^{\infty}{\sin\pars{\root{x}} \over x^{\alpha/2}}
\pars{{1 \over 2}\,x^{-1/2}}\dd x
\\[5mm] = &\
{1 \over 2}\int_{0}^{\infty}x^{\pars{\color{red}{1 - \alpha/2}} - 1}\,
{\sin\pars{\root{x}} \over \root{x}}\,\dd x
\end{align}
Note that
$\ds{{\sin\pars{\root{x}} \over \root{x}} =
\sum_{k = 0}^{\infty}\pars{-1}^{k}\,{x^{k} \over \pars{2k + 1}!} =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{k + 1} \over
\Gamma\pars{2k + 2}}\,{\pars{-x}^{k} \over k!}}$.
With Ramanujan-MT:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}{\sin\pars{x} \over x^{\alpha}}\,\dd x} =
{1 \over 2}\,\Gamma\pars{1 - {\alpha \over 2}}
\color{red}{\Gamma\pars{\color{black}{-\bracks{1 - \alpha/2}} + 1} \over
\Gamma\pars{2\color{black}{\braces{-\bracks{1 - \alpha/2}}} + 2}}
\\[5mm] = &\
\bbx{{1 \over 2}\,\Gamma\pars{1 - {\alpha \over 2}}\,
{\Gamma\pars{\alpha/2} \over \Gamma\pars{\alpha}}} =
{\pi \over 2}{\csc\pars{\pi\alpha/2} \over \Gamma\pars{\alpha}} \\ &
\end{align}