I am trying to integrate this function:
$\int_0^\infty\frac{1}{x}\coth(\frac{\hbar x}{2kT})\sin^2(\frac{xt}{2})dx$
which Wolframalpha (for me) returns nothing, just a blank screen. I thought that it may be because it has no solution, or diverges, since removing $t$ from the equation had the same result. But if I try a definite integral such as:
$\int_0^{1000}\frac{1}{x}\coth(x)\sin^2(\frac{x}{2})dx$
But Wolfram also doesn't answer:
$\int_0^{1000}\frac{1}{x}\coth(x)\sin^2(\frac{xt}{2})dx$
Is there a solution to this integral, or at least a way to estimate a definite integral of it numerically?
Hopefully this may help some more. I am attemping to perform simulation along the lines of this paper and it is Eq. 4 in the paper which I am attempting to resolve.
My attempt at figuring it out/simplifying it is as follows:
$\Gamma^x(t) = \frac{\hbar}{m_x}\int_0^\infty\frac{g^x(\omega)}{\omega}(\coth(\frac{\hbar\omega}{2k_bT})(1-\cos(\omega t)) - i\sin(\omega t)d\omega$
$ = \frac{\hbar g^x(\omega)}{m_x}\int_0^\infty\frac{2}{\omega}\coth(a\omega)\sin^2(\omega t/2)d\omega - \frac{i\hbar g^x(\omega)}{m_x}\int_0^\infty\frac{\sin(\omega t)}{\omega}d\omega$
$ = \frac{2g^x(\omega)\hbar}{m_x}\int_0^\infty\frac{\coth(a\omega)\sin^2(\omega t/2)}{\omega}d\omega - \frac{ig^x(\omega)\hbar}{m_x}\frac{\pi t}{2} $
The $g^x(\omega)$ is evaluated out to a constant in my simulation, that is why I took it out of the integral, so my problem comes down to evaluating the $\coth(a\omega)$