# What is $\int_{-\infty}^{\infty} \frac{e^{-\alpha t} \cos[t + y]}{1+\beta e^{-2\alpha t} } dt$?

I want to compute the following integral:

$\int_{-\infty}^{\infty} \frac{e^{-\alpha t} \cos[t + y]}{1+\beta e^{-2\alpha t} } dt$

with $\alpha, \beta, c$ real constants, and $\alpha>0,\beta=0$. Mathematica was unable to give a solution.

Attempt: $\int_{-\infty}^{\infty} \frac{e^{-\alpha t} \cos[t + y]}{1+\beta e^{-2\alpha t} } dt = \frac{1}{2i} \int_{-\infty}^{\infty} \frac{e^{-\alpha t} (e^{i(t+y)} - e^{-i(t+y)})}{1+\beta e^{-2\alpha t} } dt = \frac{e^{iy}}{2i} \int_{-\infty}^{\infty} \frac{(e^{it} - e^{-it})}{e^{\alpha t} +\beta e^{-\alpha t} } dt = \frac{e^{iy}}{2i} \int_{-\infty}^{\infty} \frac{e^{it}}{e^{\alpha t} +\beta e^{-\alpha t} } dt - \frac{e^{iy}}{2i} \int_{-\infty}^{\infty} \frac{e^{-it}}{e^{\alpha t} -\beta e^{-\alpha t} } dt$

-
The $c$ does nothing for the problem; it may be absorbed in the $\beta$ and another constant outside the integral. –  Ron Gordon May 15 '13 at 15:03
Ok, so I renamed $c$ into $y$, because in fact I want to get a result that is an expression in terms of $y$. –  Peter May 15 '13 at 15:09
Mathematica computes it. Also, you can set a context via "$\text{\$Assumptions}=\{\alpha>0,\alpha\in\text{Reals},\beta\in\text{Reals},y \in\text{Reals}\};$", just browse the help for "$\text{\$Assumptions}$". –  NikolajK May 15 '13 at 15:12
@Nick: Mathematica still gives the same expression (unevaluated integral) after I add the assumptions. –  Peter May 15 '13 at 15:15
Then your Mathematica doens't like you. If you don't expect people to care enough about your ugly question here, you can go to SE Mathematica and let them help you with your code/evaluation :) –  NikolajK May 15 '13 at 15:18