TITLE: Solving an improper integral contour integral, calculated via Wolfram but in need of analytic derivation possibly
QUESTION [1 upvotes]: In my studies of dynamical systems I have just encountered this supposedly tough looking improper integral, which is (not really relevant for my predicament) the Melnikov function, with the integral as follows:

$$ M(t_0) = \int_{-\infty}^{\infty} \frac{(1-e^{2At\pi})e^{At\pi}}{(1+e^{2At\pi})^2} \cos{(\omega(t+t_0))} dt $$

where $ \omega > 0 $ is a constant and A is a non zero real constant. In both cases of positive and negative A I plugged that integral (a function of $t_0$) into Wolfram Mathematica and got compact expressions for answers, but I am truly unable to solve this integral analytically, so I would certainly appreciate the help on it, I have no idea where to begin but seeing as how the computer could solve it then I assumed it is possible to solve it analytically. I thank all helpers.

REPLY [2 votes]: Hint. One may first assume that $A>0$. One may observe the expansion
$$
\frac{(1-e^{2At\pi})e^{At\pi}}{(1+e^{2At\pi})^2}=-\frac{(1-e^{-2At\pi})e^{-At\pi}}{(1+e^{-2At\pi})^2}=\sum_{n=0}^\infty(-1)^{n+1}(2n+1) e^{-A(2n+1)\pi t}.
$$ Then one gets
$$
\begin{align}
& \int_{-\infty}^{\infty} \frac{(1-e^{2At\pi})e^{At\pi}}{(1+e^{2At\pi})^2}  \cos{(\omega(t+t_0))} dt
\\\\&=\int_0^{\infty}\frac{(1-e^{-2At\pi})e^{-At\pi}}{(1+e^{-2At\pi})^2}\cos{(\omega(-t+t_0))} dt-\int_0^{\infty} \frac{(1-e^{-2At\pi})e^{-At\pi}}{(1+e^{-2At\pi})^2}\cos{(\omega(t+t_0))} dt
\\\\&=2\sin{(\omega t_0)}\int_0^{\infty}\frac{(1-e^{-2At\pi})e^{-At\pi}}{(1+e^{-2At\pi})^2}\sin{(\omega t)} dt
\\\\&=2\sin{(\omega t_0)}\sum_{n=0}^\infty (-1)^{n+1}(2n+1)\int_0^{\infty} e^{-A(2n+1)\pi t}\sin{(\omega t)}\:dt
\\\\&=2\omega\sin{(\omega t_0)}\sum_{n=0}^\infty \frac{ (-1)^{n+1}(2n+1)}{A^2(2n+1)^2\pi^2+\omega ^2},
\end{align}
$$ giving

$$
 \int_{-\infty}^{\infty} \frac{(1-e^{2At\pi})e^{At\pi}}{(1+e^{2At\pi})^2}  \cos{(\omega(t+t_0))} dt=\frac{\sin{(\omega t_0)}}{A\cosh \big(\frac{\omega}{2A} \big)}.
$$
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