Formula:3915

Digital Library of Mathematical Functions ID 10.19#Ex1

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

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