\[\mathbf{A}_{-\nu}\left(\lambda\nu\right)\sim\left(\frac{2}{\pi\nu}\right)^{1/2%
}e^{-\nu\mu}\sum_{k=0}^{\infty}\frac{{\left(\tfrac{1}{2}\right)_{k}}b_{k}(%
\lambda)}{\nu^{k}},\]