Mellin transform of $\sin x$ aka $\int^{\infty}_0 x^{s-1}\sin x dx $ I am trying to find the Mellin transform of $\sin x $, put in other words to solve:
$$\int^{\infty}_0  x^{s-1}\sin x \mathrm{d} x $$
And I know that the answer is:
$$\Gamma(s) \sin \left(\frac{\pi s}{2}\right)$$
From several tables on the internet but I couldn't find any justification.
How can this identity be proven?
 A: Alternatively, the Mellin transform for $\sin x$ can be found by employing the following useful property for the Laplace transform:
$$\int_0^\infty f(x) g(x) \, dx = \int_0^\infty \mathcal{L} \{f(x)\} (t) \cdot \mathcal{L}^{-1} \{g(x)\} (t) \, dt.$$
Noting that
$$\mathcal{L} \{\sin x\}(t) = \frac{1}{1 + t^2},$$
and
$$\mathcal{L}^{-1} \left \{\frac{1}{x^{1-s}} \right \} (t)= \frac{1}{\Gamma (1 - s)} \mathcal{L}^{-1} \left \{\frac{\Gamma (1 - s)}{x^{1-s}} \right \} (t) = \frac{t^{-s}}{\Gamma (1 - s)},$$
then
\begin{align}
\mathcal{M} \{\sin x\} &= \int_0^\infty \sin x \cdot \frac{1}{x^{1 - s}} \, dx\\
&= \int_0^\infty \mathcal{L} \{\sin x\} (t) \cdot \mathcal{L}^{-1} \left \{\frac{1}{x^{1 - s}} \right \} (t) \, dt\\
&= \frac{1}{\Gamma (1 - s)} \int_0^\infty \frac{t^{-s}}{1 + t^2} \, dt.
\end{align}
Setting $u = t^2$, one has
\begin{align}
\mathcal{M} \{\sin x\} &= \frac{1}{2 \Gamma (1 - s)} \int_0^\infty \frac{u^{-\frac{s}{2} - \frac{1}{2}}}{1 + u} \, du\\
&= \frac{1}{2 \Gamma (1 - s)} \operatorname{B} \left (\frac{1}{2} - \frac{s}{2}, \frac{1}{2} + \frac{s}{2} \right ) \tag1\\
&= \frac{1}{2 \Gamma (1 - s)} \Gamma \left (\frac{1}{2} - \frac{s}{2} \right ) \Gamma \left (\frac{1}{2} + \frac{s}{2} \right ) \tag2\\
&= \frac{1}{2 \Gamma (1 - s)} \Gamma \left [1 - \left (\frac{1}{2} + \frac{s}{2} \right ) \right ] \Gamma \left (\frac{1}{2} + \frac{s}{2} \right )\\
&= \frac{1}{2 \Gamma (1 - s)} \frac{\pi}{\sin \left (\frac{\pi}{2} + \frac{\pi s}{2} \right )} \tag3\\
&= \frac{1}{2 \Gamma (1 - s)} \frac{\pi}{\cos \left (\frac{\pi s}{2} \right )}\\
&= \frac{\Gamma (s) \sin (\pi s)}{2 \pi} \cdot \frac{\pi}{\cos \left (\frac{\pi s}{2} \right )} \tag4\\
&=\frac{\Gamma (s) \sin \left (\frac{\pi s}{2} \right ) \cos \left (\frac{\pi s}{2} \right )}{\cos \left (\frac{\pi s}{2} \right )}\\
&= \Gamma (s) \sin \left (\frac{\pi s}{2} \right )
\end{align}
This is valid for $-1 < s < 1$.
Explanation
(1) Using $\operatorname{B} (x,y) = \displaystyle{\int_0^\infty \frac{t^{x - 1}}{(1 + t)^{x + y}} \, dt}$.
(2) Using $\operatorname{B}(x,y) = \dfrac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$.
(3) Using the reflection formula for the gamma function: $\Gamma (1 - z) \Gamma (z) = \dfrac{\pi}{\sin (\pi z)}$.
(4) Again using the reflection formula for the gamma function.
A: The basic idea is to use Euler's formula: $\sin x = \dfrac{e^{ix}-e^{-ix}}{2i}$.
Let's look at the Mellin transform of $e^{ix}$:
$$\int_0^\infty e^{ix} x^{s-1}\,dx = \begin{bmatrix} x=it \\ dx = i\,dt\end{bmatrix} = \int_{0}^{-i\cdot\infty} e^{-t} (it)^{s-1} i\,dt = -i^s \int_{-i\cdot \infty}^0 e^{-t}t^{s-1}\,dt.$$
For certain values of $s$ (I will leave it to you to work out the details), you can deform the integral over the negative imaginary axis to an integral over the positive real axis:
$$\int_{-i\cdot \infty}^0 e^{-t}t^{s-1}\,dt = -\int_0^\infty e^{-t}t^{s-1}\,dt = -\Gamma(s).$$
(Add a large quarter circle in the fourth quadrant, use Cauchys integral theorem and estimate $f(z) = e^{-z}z^{s-1}$ on the new quarter circle.)
Summing up, we get that Mellin transform of $e^{ix}$ is
$$i^s\Gamma(s) = \exp(i\pi s/2) \Gamma(s).$$
Similarly, the Mellin transform of $e^{-ix}$ turns out to be $$i^{-s}\Gamma(s) = \exp(-i\pi s/2) \Gamma(s).$$
Forming the appropriate linear combination, the Mellin transform of $\sin x$ ends up as
$$ \frac{\exp(i\pi s/2) + \exp(-i\pi s/2)}{2i} \Gamma(s)  = \sin\frac{\pi s}2 \Gamma(s).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\large\mbox{Ramanujan Master Theorem}:$
With
$\ds{\quad{\sin\pars{\root{x}} \over \root{x}} =
\sum_{k = 0}^{\infty}\color{red}{\Gamma\pars{k + 1} \over \Gamma\pars{2k + 2}}{\pars{-x}^{k} \over k!}}$:
\begin{align}
&\bbox[5px,#ffd]{\int_{0}^{\infty}x^{s - 1}\sin\pars{x}
\,\dd x} =
{1 \over 2}\int_{0}^{\infty}
x^{\pars{\color{red}{s/2 + 1/2}}\ -\ 1}\,\,\,{\sin\pars{\root{x}} \over \root{x}}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\Gamma\pars{{s \over 2} + {1 \over 2}}\,
{\Gamma\pars{-\bracks{s/2 + 1/2} + 1} \over \Gamma\pars{-2\bracks{s/2 + 1/2} + 2}}
\\[5mm] = &\
{1 \over 2}\,\,
{\Gamma\pars{1/2 + s/2}\Gamma\pars{1/2 - s/2} \over \Gamma\pars{1 - s}} =
{1 \over 2}\,\,
{\pi/\sin\pars{\pi\bracks{1/2 + s/2}} \over
\pi/\bracks{\Gamma\pars{s}\sin\pars{\pi s}}}
\\[5mm] = &\
{1 \over 2}\,\Gamma\pars{s}\,{\sin\pars{\pi s} \over \cos\pars{\pi s/2}} =
\bbx{\Gamma\pars{s}\sin\pars{\pi s \over 2}} \\ &
\end{align}
