Cosine integral for a symmetric stable exponent (${\int }_0^{\infty }(1-\cos u)u^{-1-\alpha }du=\pi /[2\Gamma (1+\alpha )\sin (\pi \alpha /2)]$)

For $0<\alpha <2$, represent $u^{-1-\alpha }$ as $\Gamma (1+\alpha {)}^{-1}{\int }_0^{\infty }{v}^{\alpha }{e}^{-vu}dv$. The Tonelli theorem applies to the nonnegative integrand. The inner integral in $u$ is $1/[v(1+v^2)]$, leaving $\Gamma (1+\alpha {)}^{-1}{\int }_0^{\infty }{v}^{\alpha -1}/(1+v^2)dv$. Substitution $z=v^2$ and the Euler beta function together with the Gamma reflection formula give the result. The bounds at zero and infinity explain exactly the index range.