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For context, I'm doing research in open quantum systems and we have an autocorrelation function which has the following form. $$ C(τ) = \int_0^\infty dω\, J(ω)\left(\coth\frac{βω}{2}\cos(ωτ)-i\sin(ωτ)\right) $$ where $J(ω)$ is a real-valued function that generally goes to zero as $ω$ gets large. For $$ J(ω) = αγω_0^2\frac{ω}{(ω_0^2-ω^2)^2+γ^2ω^2} $$ a Matsubara sum can be calculated and is given in this paper and can be proven with a contour integral where the contour is extended over the full real axis by noticing that the integrand is even.

I have been trying to do a similar calculation for when the integrand is not even but odd. Specifically, $$ J(ω) = αγω_0\frac{ω^2}{(ω_0^2-ω^2)^2+γ^2ω^2} $$ but without much luck. I've considered the approach in this answer but I don't see how I can get it to work.

We have poles in all four quadrants and a series of poles along the imaginary axis. What contour would work for functions like this? Or is there an alternative method handling this integral?

  • Was $\alpha \gamma \omega_0^2$ versus $\alpha \gamma \omega_0$ in your odd and even definitions of $J(\omega)$ a typo? – Steven Clark Jan 24 '25 at 16:25
  • Not a typo. I need the units to be consistent. But presumably that doesn't have any impact as it's independent of the integration variable? – Oliver Dudgeon Jan 26 '25 at 10:41

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