TITLE: Is $\int\limits_0^\infty\frac{\sin y}{y^{s+1}}dy=-\Gamma(-s)\sin(\frac{\pi s}{2})$ for $\operatorname{Re}(s)\in (-1,0)$ obvious?
QUESTION [13 upvotes]: This is a part of computation in Titchmash, Theories of Zeta Functions which I do not find obvious but there is no explanation. I did figure out the computation. 

$$\int_0^\infty\frac{\sin(y)}{y^{s+1}}dy=-\Gamma(-s)\sin\left(\frac{\pi s}{2}\right)$$

Q: There is no explanation in the book for this step. Why is this obvious without explanation? My recipe goes as the following. It suffices to restrict to real axis part with $s\in (-1,0)$ region. Now integral is real valued in this region. Here I need $\Gamma(-s)=\frac{\Gamma(-s+1)}{s}$ extension to obtain real valuedness. Consider the integral as the imaginary part of $\int_0^{i\infty} \frac{e^{z}}{i^s z^{s+1}}dz$ where I have already rotated axis by $i$ multiplication. Now to obtain $\Gamma$ function, close contour from $(+\infty,0)$ axis portion and connect to $(0,i\infty)$ portion. Then close the contour by arc. The arc contour contribution is $0$ via exponential suppresion. Then apply residue theorem easily as the whole thing is holomorphic by $s\in (-1,0)$ region. Hence equality follows. This is not $1-2$ line naive computation though not hard. However, it did take me a while to figure out.

REPLY [3 votes]: For $0<\Re(s)<1$, we have
$$
\int^{\infty}_{0}\sin(t)t^{s-1}dt=-\operatorname{Im}\left(\int^{\infty}_{0}e^{-it}t^{s-1}dt\right)=-\operatorname{Im}\left((-i)^{s}\int^{\infty i}_{0 i}e^{-z}z^{s-1}dz\right)=
$$
$$
-\operatorname{Im}\left(e^{-i\pi s/2}\Gamma(s)\right)=\sin\left(\frac{\pi s}{2}\right)\Gamma(s).
$$ 
About the question below, it have to be proved that
$$
\int^{i\infty}_{0}e^{-z}z^{s-1}dz=\Gamma(s),
$$ 
when $\Re(s)\in(0,1)$.