Integrating fraction of exponentials

Question

I need to integrate \begin{align} \int_0^\infty{\frac{e^{t(4\omega-1)}}{1-e^{t2\omega}\kappa}dt} \end{align} with $\omega<0$ and $\kappa \in [-1,0]$. I tried substitution, but got stuck. Here is what I did so far. \begin{align} [1-\kappa,1] \ni s := 1-e^{t2\omega}\kappa \quad &\Longleftrightarrow \quad t = \frac{1}{2\omega}\ln\left(\frac{1-s}{\kappa}\right)\\ \frac{ds}{dt} = -2\omega e^{t2\omega}\kappa \quad &\Longleftrightarrow \quad dt = \frac{ds}{-2\omega \kappa e^{t2\omega}} \end{align} and hence consider \begin{align} \int_0^\infty{\frac{e^{t(4\omega-1)}}{1-e^{t2\omega}\kappa}dt} & = -\int_{1-\kappa}^1{\frac{e^{t(4\omega-1)}}{2s\omega \kappa e^{t2\omega}}ds}\\[4mm] & = -\int_{1-\kappa}^1{\frac{e^{t(2\omega-1)}}{2s\omega \kappa}ds}\\[4mm] & = -\int_{1-\kappa}^1{\frac{e^{\frac{1}{2\omega}\ln\left(\frac{1-s}{\kappa}\right)(2\omega-1)}}{2s\omega \kappa}ds}\\[4mm] & = -\int_{1-\kappa}^1{\frac{\left(\frac{1-s}{\kappa}\right)^\frac{2\omega-1}{2\omega}}{2s\omega \kappa}ds}\\[4mm] & = -\int_{1-\kappa}^1{\frac{(1-s)^\frac{2\omega-1}{2\omega}}{2s\omega\kappa^\frac{4\omega-1}{2\omega}}ds} \end{align}

Answer 1

The integral does not seem to be elementary. By taking a series in powers of $\kappa$ I get (at least for $|\kappa|<1$, though I think it works for all $\omega,\kappa < 0$ by analytic continuation)

$$\frac{- \Phi \left( \kappa,1,{\frac {4\,\omega-1}{2\omega}} \right) }{2\omega} $$ where $\Phi$ is the Lerch Phi function.

EDIT: Lerch Phi has the integral representation

$$ \Phi(\kappa, 1,a) = \int_0^\infty \frac{e^{-as}}{1-\kappa e^{-s}}\; ds$$ where $\text{Re}(a) > 0$ and $\kappa < 1$. Here we want $a = (4\omega - 1)/(2\omega) = 2 - 1/(2\omega)$; note that $\omega < 0$ implies $a > 2$. Change variables to $t = -s/(2\omega)$ and (with $\omega < 0$) this becomes $-2 \omega$ times your integral.