Show that the identity $\int_0^\infty\sin(t)t^{z-1}\ dt = \Gamma(z)\sin\left(\pi{z\over 2}\right)$ holds on $-1<\operatorname{Re}(z)<1$ The question is from Stein complex analysis 6.10(b)

(b) Show that the following identity
$$\int_0^\infty\sin(t)t^{z-1}\ dt = \Gamma(z)\sin\left(\pi{z\over 2}\right)\quad 0<\operatorname{Re}z<1$$
is valid in the larger strip $-1<\operatorname{Re}z<1$.

Since $\Gamma$ has a meromorphic continuation on $\Bbb C$ with simple poles at $-\Bbb N\cup\{0\}$ and $\sin$ has a simple zero at $0$, I can conclude that the integral
$$\int_0^\infty\sin(t)t^{z-1}\ dt$$
has an analytic continuation on $-1<\operatorname{Re}z<1$ since the RHS of the above identity does. But this does not show the identity holds on $-1<\operatorname{Re}z<1$ right? I think I need to show the LHS is also holomorphic on $-1<\operatorname{Re}z<1$ then by identity theorem I can say the identity holds. Why the integral holomorphic on $-1<\operatorname{Re}z<1$?
 A: Break up the integral as
$$
\begin{align}
\int_0^\infty\sin(t)\,t^{z-1}\,\mathrm{d}t
&=\overbrace{\int_0^1\sin(t)\,t^{z-1}\,\mathrm{d}t}^{\substack{\text{holomorphic on}\\\text{$\mathrm{Re}(z)\gt-1$}}}
+\overbrace{\int_1^{\infty\vphantom{1}}\sin(t)\,t^{z-1}\,\mathrm{d}t}^{\substack{\text{holomorphic on}\\\text{$\mathrm{Re}(z)\lt1$}}}\tag1\\
&=\int_0^1\frac{\sin(t)}t\,t^z\,\mathrm{d}t
+\cos(1)+(z-1)\int_1^\infty\cos(t)\,t^{z-2}\,\mathrm{d}t\tag2
\end{align}
$$
where $(2)$ follows by integrating the right-hand integral by parts
The Identity Theorem should handle the rest.
A: Here's another approach. Formally you can write,
$$\int_0^\infty \sin(x)x^{s-1}\mathrm dx=\mathcal{M}(\sin)(s)$$
Where I have taken a formal Mellin transform. Now write
$$\sin(x)=\sum_{k=0}^\infty \frac{1}{k!}\overbrace{\big(-\sin(\pi k/2)\big)}^{:=\varphi(k)}(-x)^k$$
Then apply Ramunajan's Master Theorem,
$$\mathcal M(\sin)(s)=\Gamma(s)\varphi(-s) \\ \mathcal M(\sin)(s)=\Gamma(s)\sin(\pi s/2) \\ \implies \boxed{\int_0^\infty \sin(x)x^{s-1}\mathrm dx=\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}$$
