<s>[INST] Proving $\mathcal{M}\left(\sin(x)\right)(s) = \Gamma(s)\sin\left(\frac{\pi}{2}s \right)$ using Real Analysis recently I've been investigating Mellin Transforms and this morning solved for case of $\sin(x)$ using Ramunajan's Master Theorem. I was curious if there were any Real based methods to evaluate this integral as in searching around online all proofs seem to rely on Contour Integration. 
My approach:
\begin{equation}
\mathcal{M}\left(\sin(x)\right)(s) = \int_0^\infty x^{s - 1}\sin(x)\:dx = \Gamma(s)\sin\left(\frac{\pi}{2}s \right)
\end{equation}
By definiton:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \int_0^\infty x^{s - 1}\sin(x)\:dx = \int_0^\infty x^{s - 1}\left[\sum_{m = 0}^{\infty} (-1)^m \frac{x^{2m + 1}}{(2m + 1)!} \right]\:dx \\&= \int_0^\infty x^{s} \sum_{m = 0}^{\infty} (-1)^m \frac{\left(x^2\right)^m}{(2m + 1)!}\:dx
\end{align}
Here we make the substitution $u = x^2$:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \int_0^\infty \left(\sqrt{u}\right)^{s} \sum_{m = 0}^{\infty} (-1)^m \frac{\left(u\right)^m}{(2m + 1)!}\frac{du}{2\sqrt{u}} = \frac{1}{2}\int_0^\infty u^{\frac{s - 1}{2}}\sum_{m = 0}^{\infty} (-1)^m \frac{u^m}{(2m + 1)!}\:du  \\
&= \frac{1}{2}\int_0^\infty u^{\frac{s + 1}{2} - 1}\sum_{m = 0}^{\infty} \frac{(-u)^m}{(2m + 1)!}\:du = \frac{1}{2}\mathcal{M}\left(g(u)\right)\left(\frac{s + 1}{2}\right)
\end{align}
Where 
\begin{equation}
g(u) = \sum_{m = 0}^{\infty} \frac{(-u)^m}{(2m + 1)!} = \sum_{m = 0}^{\infty} \frac{m!}{(2m + 1)!}\frac{(-u)^m}{m!} = \sum_{m = 0}^{\infty} \frac{\Gamma(m + 1)}{\Gamma(2m + 2)}\frac{(-u)^m}{m!}
\end{equation}
We observe that this form enables the use of Ramanujan's Master Theorem. Thus:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \frac{1}{2}\mathcal{M}\left(g(u)\right)\left(\frac{s + 1}{2}\right) = \frac{1}{2} \Gamma\left( \frac{s + 1}{2}\right) \cdot \frac{\Gamma\left(- \frac{s + 1}{2} + 1\right)}{\Gamma\left(2\cdot -\frac{s + 1}{2} + 2\right)} \\
&= \frac{1}{2} \frac{\Gamma\left( \frac{s + 1}{2}\right)\Gamma\left(1 - \frac{s + 1}{2}\right)}{\Gamma(1 - s)}
\end{align}
Employing Euler's Reflection Formula on the terms in the numerator we arrive at:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \frac{1}{2} \frac{\Gamma\left( \frac{s + 1}{2}\right)\Gamma\left(1 - \frac{s + 1}{2}\right)}{\Gamma(1 - s)} = \frac{1}{2} \cdot \frac{1}{\Gamma(1 - s)}\cdot \frac{\pi}{\sin\left(\pi  \cdot \frac{s + 1}{2}\right)} = \frac{\pi}{2} \frac{1}{\Gamma\left(1 - s\right)\sin\left(\frac{\pi}{2} (s + 1)\right)}
\end{align}
Now Euler's Reflection Formula:
\begin{equation}
 \Gamma(s)\Gamma(1 - s) = \frac{\pi}{\sin(\pi s)} \rightarrow \Gamma(1 - s) =  \frac{\pi}{\Gamma(s)\sin(\pi s)}
\end{equation} 
Returning to our integral and observing that $\sin\left(\frac{\pi}{2} (s + 1)\right)= \cos\left(\frac{\pi}{2}s\right)$ we arrive at:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \frac{\pi}{2} \frac{1}{\Gamma\left(1 - s\right)\sin\left(\frac{\pi}{2} (s + 1)\right)} = \frac{\pi}{2} \cdot \frac{1}{\frac{\pi}{\Gamma(s)\sin(\pi s)} \cdot \cos\left(\frac{\pi}{2}s\right)} = \frac{\Gamma(s)\sin(\pi s)}{2 \cos\left(\frac{\pi}{2} s\right)}
\end{align}
We now use the double angle formula $\sin\left(\pi s\right) = 2\sin\left(\frac{\pi}{2} s\right)\cos\left(\frac{\pi}{2} s\right)$ to yield:
\begin{align}
\mathcal{M}\left(\sin(x)\right)(s) &= \frac{\Gamma(s)\sin\left(\pi s\right)}{2 \cos\left(\frac{\pi}{2} s\right)} = \frac{\Gamma(s)\cdot  2\sin\left(\frac{\pi}{2} s\right)\cos\left(\frac{\pi}{2} s\right)}{2 \cos\left(\frac{\pi}{2} s\right)} = \Gamma(s)\sin\left(\frac{\pi}{2} s\right)
\end{align}
As required. 
 [/INST] For $b > 0, \Re(s) > 0$ $$b^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-bx}dx $$
By induction if for some $ b,\Re(b) > 0 $  we have  $\forall s, \Re(s) >0,b^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-bx}dx$ then for every $|a/b|< 1, \Re(a+b) > 0$, using the binomial series   $$\int_0^\infty x^{s-1} e^{-(a+b)x}dx = \sum_{k=0}^\infty \int_0^\infty x^{s-1}\frac{(-ax)^k}{k!} e^{-bx}dx=\sum_{k=0}^\infty \frac{(-a)^k}{k!} b^{-s-k} \Gamma(s+k)\\ =b^{-s} \Gamma(s)\sum_{k=0}^\infty {-s \choose k} (a/b)^k = (a+b)^{-s} \Gamma(s)$$
From which $b^{-s} \Gamma(s) = \int_0^\infty x^{s-1} e^{-bx}dx $ is true for every $b, \Re(b) > 0$ 
And hence for $\Re(s) \in (0,1)$
$$\int_0^\infty x^{s-1}  \sin(x)dx = \lim_{b \to 0^+} \int_0^\infty x^{s-1} \frac{e^{-(b-i)x}-e^{-(b+i)x}}{2i}dx \\ = \lim_{b \to 0^+}  \frac{(b-i)^{-s}-(b+i)^{-s}}{2i} \Gamma(s)= \sin(\pi s/2) \Gamma(s)$$
</s>