1
$\begingroup$

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 enter image description here

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}] 
$\endgroup$
5
  • 1
    $\begingroup$ Please include Mathematica code if it all possible. $\endgroup$
    – Mr.Wizard
    Jun 21, 2016 at 11:55
  • 1
    $\begingroup$ And do you want to evaluate it symbolically or numerically? $\endgroup$
    – xzczd
    Jun 21, 2016 at 11:56
  • $\begingroup$ I want to do it symbolically. $\endgroup$ Jun 21, 2016 at 12:34
  • $\begingroup$ Are those vectors in the denominator? Can you provide a reference on where you encountered this integral? $\endgroup$ Jun 21, 2016 at 12:35
  • $\begingroup$ I have added the reference @J.M. $\endgroup$ Jun 21, 2016 at 12:59

1 Answer 1

1
$\begingroup$

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[3]) 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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.