\[\tan\left(\phi_{\nu}\left(x\right)-\theta_{\nu}\left(x\right)\right)=\frac{M_{%
\nu}\left(x\right)\theta_{\nu}'\left(x\right)}{M_{\nu}'\left(x\right)}=\frac{2%
}{\pi xM_{\nu}\left(x\right)M_{\nu}'\left(x\right)},\]