Evaluating Bessel-Schläfli-type Integral with Infinite contour

I am attempting to evaluate the integral

\begin{equation} \int_{\infty-i\pi}^{\infty+i\pi} dt \frac{e^{-\frac{l}{n}t}}{[(X'-X)^2+r^2+s^2- 2rs \cosh(t)]}, \end{equation} the only variable being $t$, all other terms are considered constant for this integral.

How can I define the contour in Mathematica?

The integral is from the modified Bessel function in the Schläfli representation

\begin{equation} I_\nu = \frac{1}{2\pi i}\int_{\infty-i\pi}^{\infty+i\pi} dt e^{\nu \cosh(t)-\nu \mu}, \end{equation} with the contour on the complex plane shown on the figure The integral is from the paper http://arxiv.org/abs/1405.5875v2, Pg 9 Eqn. 2.22

My Code is:

f[z_] := (E^(-a z))/(A + B (Cosh[z]))

Integrate[f[z], {z, Infinity - I * Pi, 0 - I * Pi, 0 + I * Pi, Infinity + I * Pi,Infinity - I*Pi}]
• Please include Mathematica code if it all possible. – Mr.Wizard Jun 21 '16 at 11:55
• And do you want to evaluate it symbolically or numerically? – xzczd Jun 21 '16 at 11:56
• I want to do it symbolically. – Kay Modikai Jun 21 '16 at 12:34
• Are those vectors in the denominator? Can you provide a reference on where you encountered this integral? – J. M. will be back soon Jun 21 '16 at 12:35
• I have added the reference @J.M. – Kay Modikai Jun 21 '16 at 12:59

Here is how to do the contour integral. Shown for some specific parameters.

A = 1; B = -1/2; a = 1;
f[z_] := (E^(-a z))/(A + B (Cosh[z]))
Integrate[f[rz - I Pi], {rz, Infinity, 0}] +
Integrate[f[z], {z, -I Pi, I Pi}] +
Integrate[f[rz + I Pi], {rz, 0, Infinity}]
// Simplify

4/3 I (-3 + 2 Sqrt) Pi

You will need to add appropriate assumptions to have any chance at a general solution. eg. Running again with:

A = 1; B = -1/2; Clear[a];
\$Assumptions = {a > 0}

yields after some time a large expression with several Hypergeometric functions.