\[J_{\nu}\left(\nu\operatorname{sech}\alpha\right)\sim\frac{e^{\nu(\tanh\alpha-%
\alpha)}}{(2\pi\nu\tanh\alpha)^{\frac{1}{2}}}\sum_{k=0}^{\infty}\frac{U_{k}(%
\coth\alpha)}{\nu^{k}},\]