# Help evaluating the integral $\int_{0}^{\infty} \omega \cos(\omega t) \coth(\alpha \omega) \text{d} \omega$.

In the paper “Quantum Langevin Equation” by G. W. Ford, J. T. Lewis, and R. F. O’Connell I found the following statement

\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].

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

\frac{\text{d}}{\text{d}t}\frac{1}{\pi} \int_{0}^{\infty} \sin[\omega t] \coth[\alpha \omega/2] \text{d} \omega,

and then tried using Wolframalpha to evaluate the integral

\int_{0}^{\infty} \sin[\omega] \coth[\omega] \text{d} \omega,

but the result is that this integral diverges.

Furthermore I know that

\frac{1}{\pi} \int_{0}^{\infty} \cos[\omega t] \text{d} \omega = \delta(t),

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!

#### Solutions Collecting From Web of "Help evaluating the integral $\int_{0}^{\infty} \omega \cos(\omega t) \coth(\alpha \omega) \text{d} \omega$."

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.