\[\boldsymbol{Q}^{\mu}_{\nu}\left(z\right)=\frac{\pi^{1/2}\left(z^{2}-1\right)^{%
\mu/2}}{2^{\mu}\Gamma\left(\mu+\frac{1}{2}\right)\Gamma\left(\nu-\mu+1\right)}%
\*\int_{0}^{\infty}\frac{(\sinh t)^{2\mu}}{\left(z+(z^{2}-1)^{1/2}\cosh t%
\right)^{\nu+\mu+1}}\,\mathrm{d}t,\]