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In the paper "Quantum Langevin Equation" by G. W. Ford, J. T. Lewis, and R. F. O’Connell I found the following statement \begin{equation} \frac{1}{\pi} \int_{0}^{\infty} \omega \cos[\omega t] \coth[\alpha \omega/2] \text{d} \omega = \frac{1}{\alpha} \frac{\text{d}}{\text{d}t} \coth[\pi t/\alpha]. \end{equation} It is Equation (2.11) in the paper I linked.

I would like to understand how to obtain this result. I have noticed that the left hand side can be written as \begin{equation} \frac{\text{d}}{\text{d}t}\frac{1}{\pi} \int_{0}^{\infty} \sin[\omega t] \coth[\alpha \omega/2] \text{d} \omega, \end{equation} and then tried using Wolframalpha to evaluate the integral \begin{equation} \int_{0}^{\infty} \sin[\omega] \coth[\omega] \text{d} \omega, \end{equation} but the result is that this integral diverges.

Furthermore I know that \begin{equation} \frac{1}{\pi} \int_{0}^{\infty} \cos[\omega t] \text{d} \omega = \delta(t), \end{equation} I thought maybe I could use this in the following way, starting at the l.h.s. \begin{align} \pi\frac{\text{d}}{\text{d}t} \coth[\pi t/ \alpha] &= \pi\frac{\text{d}}{\text{d}t} \int_{R} \text{d} \mu \delta(t-\mu) \coth[\pi \mu/\alpha] \\ &= \frac{\text{d}}{\text{d}t}\int_{R} \text{d}\mu \int_{0}^{\infty} \text{d} \omega \cos[\omega(t-\mu)] \coth[\pi \mu/\alpha] \\ &= \frac{\text{d}}{\text{d}t} \int_{R} \text{d} \mu \int_{0}^{\infty} \text{d} \omega \cos[\omega \mu] \coth[\pi(\mu + t) /\alpha] \\ &= \int_{0}^{\infty} \text{d} \omega \int_{R} \text{d} \mu \cos[\omega \mu] \frac{\text{d}}{\text{d}\mu} \coth[\pi(\mu+ t)/\alpha], \end{align} and from there I don't really know what to do... One could try partial integration, but I don't really see that working out either.

Any help to figure this out would be greatly appreciated!

Peter
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1 Answers1

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I suggest that either they use some other sense of integration (see edit below), or there is an error.

As far as I can see, the integral (forgetting different scaling constants) $$ \int_0^{+\infty}w\cos(w)\coth(w)\,dw $$ is divergent. One argument, that could be done more rigorous, is that $$ \coth(w)\approx 1 $$ where the $\approx$ is really exponentially fast as $w$ increases. Practically, $\coth w=1$ for $w>10$.

Since the integral $$ \int_0^{+\infty} w\cos (w)\,dw $$ is divergent, the integral $$ \int_0^{+\infty} w\cos(w)\coth(w)\,dw $$ is also divergent.

One way out could maybe be to consider complex constants $t$ and $\alpha$?

Edit

What they most likely do is that they do a Fourier cosine transform, considering $w\coth(w)$ as a distribution. Indeed, then $$ \mathcal F_{\cos}(w\coth w)(s)=-\frac{\pi^2}{2}\bigl(\text{csch}\,(\pi s/2)\bigr)^2. $$ Shamelessly, I suggest you to read this answer of mine to see how one can proceed in calculating such transforms.

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