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

Question

I'm trying to calculate the following integral

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

with $\alpha\geq 0$ , $\beta\in \mathbb{R}$ and $n=0,1,2,...$

It seems that may be related with Modified Bessel functions but I'm not able to see the relationship.

Thanks in advance.

Progress: Performing the change of variable $u-\beta=x$ we arrive

$$\int_0^{\infty}e^{-\alpha\cosh(u-\beta)}\,e^{-n u}du=e^{-n \beta}\,\int_{-\beta}^{\infty}e^{-\alpha\cosh(x)}\,e^{-n x}du=e^{-n \beta}\left(\,\int_{0}^{\infty}e^{-\alpha\cosh(x)}\,e^{-n x}dx+\int_{-\beta}^{0}e^{-\alpha\cosh(x)}\,e^{-n x}dx\right)$$

and the second integral may be evaluated (with some technic, using Taylor series for example). Then, I'm interested in calculate the first one $\displaystyle \int_{0}^{\infty}e^{-\alpha\cosh(x)}\,e^{-n x}dx$ . That is, putting $\beta=0$ at the original problem.

Answer 1

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".

Answer 2

$\int_0^\infty e^{-\alpha\cosh(u-\beta)}e^{-nu}~du$

$=\int_{-\beta}^\infty e^{-\alpha\cosh u}e^{-n(u+\beta)}~d(u+\beta)$

$=e^{-n\beta}\int_{-\beta}^\infty e^{-\alpha\cosh u}e^{-nu}~du$

$=e^{-n\beta}\int_{-\beta}^\infty e^{-\frac{\alpha e^u}{2}-\frac{\alpha}{2e^u}}e^{-nu}~du$

$=e^{-n\beta}\int_{e^{-\beta}}^\infty e^{-\frac{\alpha u}{2}-\frac{\alpha}{2u}}u^{-n}~d(\ln u)$

$=e^{-n\beta}\int_{e^{-\beta}}^\infty\dfrac{e^{-\frac{\alpha u}{2}-\frac{\alpha}{2u}}}{u^{n+1}}~du$

$=e^{-n\beta}\int_1^\infty\dfrac{e^{-\frac{\alpha e^{-\beta}u}{2}-\frac{\alpha}{2e^{-\beta}u}}}{(e^{-\beta}u)^{n+1}}~d(e^{-\beta}u)$

$=\int_1^\infty\dfrac{e^{-\frac{\alpha u}{2e^\beta}-\frac{\alpha e^\beta}{2u}}}{u^{n+1}}~du$

$=K_n\left(\dfrac{\alpha}{2e^\beta},\dfrac{\alpha e^\beta}{2}\right)$ (according to https://core.ac.uk/download/pdf/81935301.pdf)