Show that the identity $\int_0^\infty\sin(t)t^{z-1}\ dt = \Gamma(z)\sin\left(\pi{z\over 2}\right)$ holds on $-1<\operatorname{Re}(z)<1$

Question

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 $-1<\operatorname{Re}z<1$?

Answer 1

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.

Answer 2

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, $$\mathcal M(\sin)(s)=\Gamma(s)\varphi(-s) \\ \mathcal M(\sin)(s)=\Gamma(s)\sin(\pi s/2) \\ \implies \boxed{\int_0^\infty \sin(x)x^{s-1}\mathrm dx=\Gamma(s)\sin\left(\frac{\pi s}{2}\right)}$$