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

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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. – Yiteng 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
Ah, solved the problem in Mathematica. The answer is indeed in terms of Hypergeometric functions... – Yiteng May 15 '13 at 15:22
For $\beta=1$ we have the beautiful result $~I=\dfrac\pi{2|a|}\cdot\text{sech}\dfrac\pi{2|a|}\cdot\cos y$. – Lucian Jul 4 '14 at 13:09

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