# 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.

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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.

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Thank you for your precise answer! I understand perfectly! –  jsquare Sep 17 '12 at 1:41

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 $i^{s}=e^{i\pi s}$

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