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?

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: $$\Gamma\left(\alpha\right)\Gamma\left(1-\alpha\right)=\frac{\pi}{\sin{\left(\pi \alpha\right)}}$$ $$\pi^2=\Gamma\left(\frac{\alpha}{2}\right)\Gamma\left(1-\frac{\alpha}{2}\right)\sin{\left(\frac{\pi \alpha}{2}\right)} \cos{\left(\frac{\pi \alpha}{2}\right)} \color{blue}{\Gamma\left(\frac{\alpha+1}{2}\right)\Gamma\left(\frac{1-\alpha}{2}\right)} $$

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

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!