= Cosine integral for a symmetric stable exponent
{title2=$\int_0^\infty(1-\cos u)u^{-1-\alpha}\,du=\pi/[2\Gamma(1+\alpha)\sin(\pi\alpha/2)]$}
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.
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