Trigonometric wizardry with the Gamma function I'm trying to understand the first proof of the functional equation in Titchmarsh's famous book on the Zeta function (as edited by Heath-Brown), and I'm stuck.
This is because the author claims that
$$
\int_0^\infty \frac{\sin(y)}{y^{1+s}} dy = -Γ(-s) \cdot \sin\left( \frac{πy}{2} \right).
$$
For some reason, I'm too stupid to prove this identity. Any ideas?
 A: Start with
$$I(y)=\int\frac{e^{iy}}{y^{1+s}}\,dy=-y^{-s} E_{s+1}(-i y)$$ Assuming $s>0$, $I(\infty)=0$ and $I(0)=-(-i)^s \Gamma (-s,0)$. So
$$\int y^{-(1+s)} \sin (y)\,dy=\Im\left(-i)^s\right)\Gamma (-s,0)=-\sin \left(\frac{\pi  s}{2}\right)\,\Gamma (-s,0)=-\sin \left(\frac{\pi  s}{2}\right)\,\Gamma (-s)$$
A: The integral $\int_{0}^{+\infty}\frac{\sin(y)}{y^{1+s}}\,dy$ is convergent for any $s> -1$ by Dirichlet's criterion. The Laplace transform of $\sin(y)$ is $\frac{1}{1+t^2}$ and the inverse Laplace transform of $\frac{1}{y^{1+s}}$ is $\frac{t^s}{\Gamma(s+1)}$, so
$$ \int_{0}^{+\infty}\frac{\sin(y)}{y^{1+s}}\,dy = \frac{1}{\Gamma(s+1)}\int_{0}^{+\infty}\frac{t^s}{t^2+1}\,dt \stackrel{\frac{1}{1+t^2}\mapsto u}{=}\frac{1}{2\,\Gamma(s+1)}\int_{0}^{1}(1-u)^{\frac{s-1}{2}}u^{-\frac{s+1}{2}}\,du$$
and by Euler's Beta function and the reflection formula for the $\Gamma$ function the RHS equals
$$ \frac{1}{2\,\Gamma(s+1)}\,\Gamma\left(\frac{1+s}{2}\right)\Gamma\left(\frac{1-s}{2}\right)=\frac{\frac{\pi}{2}}{\Gamma(s+1)\cos\left(\frac{\pi s}{2}\right)}=-\Gamma(-s)\sin\left(\frac{\pi s}{2}\right). $$
