\[\frac{2\sin\left(\mu\pi\right)}{\pi}\boldsymbol{Q}^{\mu}_{\nu}\left(x\right)=%
\frac{(x+1)^{\mu/2}}{\Gamma\left(\nu+\mu+1\right)(x-1)^{\mu/2}}\mathbf{F}\left%
(\nu+1,-\nu;1-\mu;\tfrac{1}{2}-\tfrac{1}{2}x\right)-\frac{(x-1)^{\mu/2}}{%
\Gamma\left(\nu-\mu+1\right)(x+1)^{\mu/2}}\mathbf{F}\left(\nu+1,-\nu;\mu+1;%
\tfrac{1}{2}-\tfrac{1}{2}x\right).\]