\[\int_{-1}^{1}\left(\mathsf{P}_{\nu}\left(x\right)\right)^{2}\,\mathrm{d}x=%
\frac{\pi^{2}-2{\sin}^{2}\left(\nu\pi\right)\psi'\left(\nu+1\right)}{\pi^{2}%
\left(\nu+\frac{1}{2}\right)},\]