I'm trying to tackle this for several days now, and no matter what I do I never get it quite right. If some1 gets is right I will quid pro quo solve an open problem on this board.

Show that

$\displaystyle \int_0^{a}d\tau\int_0^{a}d\tau'\frac{q(\tau)q(\tau ')\cosh{\Bigl(\omega|\tau-\tau'|-\frac{a\omega}{2}\Bigr)}}{\sinh{\Bigl(\frac{a\omeg a}{2}\Bigr)}}$

reduces to $\displaystyle \int_{-\infty}^{\infty}d\tau'\int_0^{a}d\tau q(\tau)q(\tau')e^{-\omega|\tau-\tau'|}$

when $\displaystyle q(\tau+a)=q(\tau)$ outside $\displaystyle 0\leq\tau<a.$

You might want to use something like $\displaystyle \int_0^{\infty}dtq(t)e^{-st}=\int_0^{T}dt\frac{q(t)e^{-st}}{1-e^{-sT}}$ when

$\displaystyle q(t)$ has period $\displaystyle T.$