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 $\alpha\in (0,2)$. 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?
 A: The Mellin transform of $\sin{t}$ (as proven here) yields:
$$\mathcal{I}(z)=\int_0^{\infty} t^{z-1} \sin{t} \; \mathrm{d}t =\Gamma\left(z\right)\sin{\left(\frac{\pi}{2}z\right)}, \; -1 < \Re \left(z\right) < 1$$
And your integral is:
\begin{align*}
\mathcal{I}(1-\alpha) &= \Gamma\left(1-\alpha\right)\sin{\left(\frac{\pi}{2}\left(1-\alpha\right)\right)} \\
&= \Gamma\left(1-\alpha\right) \left( \frac{\pi }{\Gamma\left(\frac{1-\alpha}{2}\right)\Gamma\left(\frac{1+\alpha}{2}\right)} \right) \\
&=\Gamma\left(1-\alpha\right) \left(\frac{\pi \sin{\left(\pi \alpha\right)} \Gamma\left(\frac{\alpha}{2}\right) \Gamma\left(1-\frac{\alpha}{2}\right) }{2 \pi^2} \right) \\
&=\Gamma\left(1-\alpha\right) \left(\frac{\Gamma\left(\frac{\alpha}{2}\right) \Gamma\left(1-\frac{\alpha}{2}\right) }{2 \Gamma\left(\alpha\right) \Gamma \left(1-\alpha\right)} \right) \\
&= \boxed{\int_0^{+\infty} \frac{\sin{(x)}}{x^{\alpha}} \; \mathrm{d}x =\frac{\Gamma\left(\frac{\alpha}{2}\right) \Gamma\left(1-\frac{\alpha}{2}\right) }{2 \Gamma\left(\alpha\right)}, \; 0< \Re\left(\alpha\right)<2} 
\end{align*}
Where Euler's reflection formula and the Legendre relation were utilized to get the desired form of the answer: $$\Gamma\left(\alpha\right)\Gamma\left(1-\alpha\right)=\frac{\pi}{\sin{\left(\pi \alpha\right)}}$$
$$\pi^2=\Gamma\left(\frac{\alpha}{2}\right)\Gamma\left(1-\frac{\alpha}{2}\right)\sin{\left(\frac{\pi \alpha}{2}\right)} \cos{\left(\frac{\pi \alpha}{2}\right)} \color{blue}{\Gamma\left(\frac{\alpha+1}{2}\right)\Gamma\left(\frac{1-\alpha}{2}\right)} $$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \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}\,}\,}
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\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}
A: 
I thought that it might be instructive to present an approach that relies on a useful property of the Laplace Transform (See Here) to evaluate integrals over the positive reals.  To that end we proceed.


Let $F(s)=s^{-\alpha}$, $0<\alpha<2$ and $f(t)=\sin(t)$.  Then, the inverse Laplace Transform of $F(s)$ is
$$\mathscr{L}^{-1}\{F\}(x)=\frac{x^{\alpha-1}}{\Gamma(\alpha)}\tag1$$
and the Laplace Transform of $f(t)$ is given by
$$\mathscr{L}\{f\}(x)=\frac1{x^2+1}\tag2$$

Then, using $(1)$ and $(2)$ along with This Property of the Laplace Transform, we assert that
$$\int_0^\infty \frac{\sin(x)}{x^\alpha}\,dx=\frac1{\Gamma(\alpha)}\int_0^\infty \frac{x^{\alpha-1}}{x^2+1}\,dx\tag3$$

The integral on the right-hand side of $(3)$ can be evaluated using a host of methodologies See This, and is given by
$$\int_0^\infty \frac{x^{\alpha-1}}{x^2+1}\,dx =\frac\pi{2\sin(\pi\alpha/2)} \tag4$$

Substituting $(4)$ in $(3)$, we find that
$$\int_0^\infty \frac{\sin(x)}{x^\alpha}\,dx=\frac{\pi}{2\Gamma(\alpha)\sin(\pi \alpha/2)}\tag5$$

Finally, using the reflection formula for the Gamma Function (See this answer) as given by $\Gamma(x)\Gamma(1-x)=\frac{\pi}{\sin(\pi x)}$, we arrive at the expected result
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty \frac{\sin(x)}{x^\alpha}\,dx=\frac{\Gamma(\alpha/2)\Gamma(1-\alpha/2)}{2\Gamma(\alpha)}}$$
as was to be shown!

ALTERNATIVE METHODOLOGY:  CONTOUR INTEGRATION
We begin by analyzing the closed-contour integral $I(\alpha)$, $\alpha\in (0,2)$ given by
$$\begin{align}
I(\alpha)&=\int_\varepsilon^R \frac{e^{ix}}{x^\alpha}\,dx+\int_0^{\pi/2}\frac{e^{iRe^{i\phi}}}{(Re^{i\phi})^\alpha}\,iRe^{i\phi}\,d\phi\\\\
&-\int_\varepsilon^R \frac{e^{-x}}{(ix)^\alpha}\,i\,dx-\int_0^{\pi/2}\frac{e^{i\varepsilon e^{i\phi}}}{(\varepsilon 
 e^{i\phi})^\alpha}\,i\varepsilon e^{i\phi}\,d\phi\tag6
\end{align}$$
If we choose the branch cut of the natural logarithm to originate at $0$ and extend to the point at infinity along the real axis, Cauchy's Integral Theorem guarantees that $I(\alpha)=0$.  Furthermore, it is straightforward to show that as $R\to\infty$, the second integral on the right-hand side of $(6)$ vanishes.
So far, after letting $R\to\infty$ and then taking the imaginary parts of all terms in $(6)$ we have
$$\begin{align}
\int_\varepsilon^\infty \frac{\sin(x)}{x^\alpha}\,dx&=\sin\left(\frac{\pi (1-\alpha)}2\right)\int_\varepsilon^\infty\frac{e^{-x}}{x^\alpha}\,dx\\\\
&+\varepsilon^{1-\alpha}\int_0^{\pi/2} e^{-\varepsilon \sin(\phi)}\cos\left((1-\alpha)\phi+\varepsilon \cos(\phi)\right)\,d\phi\tag7
\end{align}$$
The last term on the right-hand side can be written as
$$\varepsilon^{1-\alpha}\int_0^{\pi/2} e^{-\varepsilon \sin(\phi)}\cos\left((1-\alpha)\phi+\varepsilon \cos(\phi)\right)\,d\phi=\varepsilon^{1-\alpha}\frac{\sin(\pi (1-\alpha)/2)}{1-\alpha}+O(\varepsilon^{2-\alpha})\tag8$$
Using $(8)$ in $(7)$, integrating by parts the first integral on the right-hand side of $(7)$ with $u=e^{-x}$ and $v=\frac{1}{(1-\alpha)x^{\alpha-1}}$, letting $\varepsilon\to0^+$, and exploiting the aforementioned reflection formula $\Gamma(x)\Gamma(1-x)=\frac\pi{\sin(\pi x)}$ yields
$$\begin{align}
\int_\varepsilon^\infty \frac{\sin(x)}{x^\alpha}\,dx&=\frac{\sin\left(\pi (1-\alpha)/2\right)}{1-\alpha}\int_\varepsilon^\infty\frac{e^{-x}}{x^{\alpha-1}}\,dx\\\\
&=\sin\left(\frac{\pi(1-\alpha)}2\right)\Gamma(1-\alpha)\\\\
&=\frac{\pi \sin\left(\frac{\pi (1-\alpha)}2\right)}{\sin(\pi \alpha)\Gamma(\alpha)}\\\\
&=\frac{\pi}{2\Gamma(\alpha)\sin(\pi \alpha/2)}\\\\
&=\frac{\Gamma(\alpha/2)\Gamma(1-\alpha/2)}{2\Gamma(\alpha)}
\end{align}$$
which agrees with the result obtained in the previous section!
