Mellin Transform of $\sin$ Can one show that the following integral converges on $-1<\Re s < 1$ 
and define holomorphic function of $s$?
$$\int_0^\infty \sin(y) y^{s-1} dy$$
I've googled for a while, but I could not find any good reference.
 A: I assume you know that for every $0<a<b<\infty$ the integral $\int_a^b \sin(y)y^{s-1}\,dy$ is a holomorphic function of $s$. We need to pass to the limit $a\to 0$ and $b\to \infty$, and show that convergence is locally uniform with respect to $s$; this is enough to preserve holomorphicity. 
Let us write $s=\sigma+it$ for no particular reason. Notice that every $s$ has a neighborhood in which $\epsilon-1<\sigma<1-\epsilon$.
The limit $a\to 0$ is easy because $|\sin (y)y^{s-1}|\le y^{\sigma}$. In any neighborhood of $s$ we have a bound of the kind $y^{\sigma}\le y^{\epsilon-1}$, which ensures uniformity of convergence. Thus, the integral $\int_0^1 \sin(y)y^{s-1}\,dy$ is a holomorphic function of $s$. 
The other tail needs to be integrated by parts: 
$$\int_1^b \sin(y)y^{s-1}\,dy = -\cos(y)y^{s-1}\bigg|_1^b+ \frac{1}{s-2}\int_1^b \cos(y)y^{s-2}\,dy$$
Here $|\cos(y)y^{s-1}|\le y^{\sigma-1}\le y^{-\epsilon}$, which gives uniform convergence of the boundary term. Also, $|\cos(y)y^{s-2}|\le y^{\sigma-2}\le y^{-\epsilon-1}$, hence the integral tail is uniformly convergent.
A: by analytic continuation i think we have
$$ \int_{0}^{\infty}dtsin(t)t^{s-1} = \Gamma (s) sin( \frac{\pi s}{2})$$
the sine part comes from the imaginarhy part of $ i^{s}=e^{i\pi s} $
