Evaluating the integral $\int_0^{\infty}e^{-\alpha\cosh(u-\beta)}\,e^{-n u}\,du$

To calculate \begin{equation} I_n=\int_0^{\infty}e^{-\alpha\cosh(x)}\,e^{-n x}\,du \end{equation} we can decompose \begin{align} I_n&=\int_0^{\infty}e^{-\alpha\cosh(x)}\left( \cosh nx+\sinh nx \right)\,du\\ &=\int_0^{\infty}e^{-\alpha\cosh(x)}\cosh nx\,dx+\int_0^{\infty}e^{-\alpha\cosh(x)}\sinh nx \,dx\\ &=K_n\left( \alpha \right)-L_n\left( \alpha \right) \end{align} where the integral representation of the modified Bessel function is used and \begin{equation} L_n\left( \alpha \right)=\int_0^{\infty}e^{-\alpha\cosh(x)}\sinh nx \,dx \end{equation} We have directly \begin{equation} L_1(\alpha)=\frac{e^{-\alpha}}{\alpha} \end{equation} and \begin{align} L_{n+1}&=\int_0^{\infty}e^{-\alpha\cosh(x)}\cosh x\sinh nx \,dx+\int_0^{\infty}e^{-\alpha\cosh(x)}\cosh nx \sinh x\,dx\\ &=-\frac{dL_n}{d\alpha}+\frac{e^{-\alpha}}{\alpha}+\frac{n}{\alpha}\int_0^{\infty}e^{-\alpha\cosh(x)}\sinh nx \,dx \end{align} where we used an integration by parts to evaluate the second integral. Then, \begin{equation} L_{n+1}=\frac{e^{-\alpha}}{\alpha}+\frac{n}{\alpha}L_n-\frac{dL_n}{d\alpha} \end{equation} we obtain \begin{align} L_2&=\frac{2(\alpha+1)}{\alpha^2}e^{-\alpha}\\ L_3&=\frac{3\alpha^2+8\alpha+8}{\alpha^3}e^{-\alpha}\\ \ldots \end{align} Defining the polynomials \begin{align} P_{n+1}(\alpha)&=\left( 2n+\alpha \right)P_n(\alpha)-\alpha P'_n(\alpha)+\alpha^n\\ P_1(\alpha)&=1 \end{align} we find \begin{equation} L_n=\frac{P_{n}(\alpha)e^{-\alpha}}{\alpha^n} \end{equation} and finally \begin{equation} I_n=K_n\left( \alpha \right)-\frac{P_{n}(\alpha)e^{-\alpha}}{\alpha^n} \end{equation} You may also be interested in the papers by Jones which define an incomplete Bessel function as \begin{equation} K_\nu\left( z,w \right)=\int_w^\infty e^{-z\cosh} \cosh \nu t\,dt \end{equation} or in the articles of Harris on the "leaky aquifer function".