Proof to $\int_{0}^{\infty}\sin(t)t^{z-1}\,\mathrm{d}t= \sin\left ( \frac{\pi z}{2} \right )\Gamma(z)$ I tried to check the source of the proof to the equation 
$$\int_{0}^{\infty}\sin(t)t^{z-1}\,\mathrm{d}t= \sin\left ( \frac{\pi z}{2} \right )\Gamma(z),\qquad -1<\Re z < 1$$
but it only has a sketch of proof, since it was meant to be an exercise. In this exercise seems a bit unclear to me what to do. I am looking for a proof to this equation, if there is one. Another source says the domain should be $0<\Re z < 1$ (in which it has no proof). I do not know what this equation is called so searching on Google is kind of difficult. 
 A: For $\text{Re}(z)\in(-1,1)$, we may use the Laplace transform to get:
$$ \int_{0}^{+\infty}\sin(t)\,t^{z-1}\,dt = \frac{1}{\Gamma(1-z)}\int_{0}^{+\infty}\frac{ds}{s^z(1+s^2)}\tag{1}$$
since $\mathcal{L}(\sin t)=\frac{1}{1+s^2}$ and $\mathcal{L}^{-1}\left(t^{z-1}\right)=\frac{1}{s^z \Gamma(1-z)}.$ 
With the substitution $\frac{1}{1+s^2}=u$, the RHS of $(1)$ becomes:
$$ \frac{1}{2\,\Gamma(1-z)}\int_{0}^{1}u^{\frac{z-1}{2}}(1-u)^{\frac{-z-1}{2}}\,du=\frac{\Gamma\left(\frac{z+1}{2}\right)\,\Gamma\left(\frac{1-z}{2}\right)}{2\,\Gamma(1-z)} \tag{2}$$
by Euler's beta function, then the claim follows from the $\Gamma$ reflection formula:
$$ \Gamma(z)\,\Gamma(1-z) = \frac{\pi}{\sin(\pi z)}.\tag{3}$$
A: Let $I(s)$ be the integral given by
$$I(s)=\int_0^\infty t^{s-1}\sin(t)\,dt \tag 1$$
for $\text{Re}(s)<1$.  
We can use Euler's Formula to write $(1)$ as 
$$I(s)=\frac1{2i}\left(\int_0^\infty t^{s-1}e^{it}\,dt-\int_0^\infty t^{s-1}e^{-it}\,dt\right) \tag 2$$
Now, moving to the complex plane, we analyze the closed contour integral(s)
$$J(s)=\oint_{C^\pm} z^{s-1}e^{\pm iz}\,dz \tag 3$$
where $C^\pm$ is comprised of (i) the segment on the real line from $\epsilon$ to $R$, (ii) the quarter circle with radius $R$, centered at the origin, from $R$ to $\pm iR$, (iii) the line segment along the imaginary axis from $\pm iR$ to $\pm i\epsilon$, and (iv) the quarter circle with radius $\epsilon$, centered at the origin, from $ \pm i\epsilon$ to $\epsilon$.
With the branch cut chosen as the line along the non-positive real axis, from $0$ to $-\infty$, the integrands in $3$ are analytic in and on $C^\pm $.  Then, Cauchy's Integral Theorem guarantees that $J(s)=0$.
In the limit as $R\to \infty$ and $\epsilon \to 0$, the contributions to $J(s)$ from the integrations along the quarter circles vanish.  Hence, we find that
$$\begin{align}
\int_0^\infty t^{s-1}e^{\pm i t}\,dt&=(\pm i)^{s}\int_0^\infty t^{s-1}e^{-t}\,dy\\\\
&=e^{\pm i\pi s/2}\Gamma(s)\tag 4
\end{align}$$
Using $(4)$ in $(2)$, we find 
$$\bbox[5px,border:2px solid #C0A000]{\int_0^\infty t^{s-1}\sin(t)\,dt =\sin(\pi s/2)\Gamma(s)}$$
as was to be shown!
