Integral from thermodynamics

Question

I've been trying to solve analytically an integral I stumbled into while working in a thermodynamics problem. I could solve it numerically, but it would be nicer if I can find an actual solution. I've been trying to find something but I'm stuck, so if anyone knows a technique I could use, it would be greatly appreciated.

Enthalpy can be defined as: $$h = \int c_p(T) dT$$ But: $$c_p(T) = c_{pp}\cdot \left\{ 1 + \frac{\gamma_p - 1}{\gamma_p}\left[ \left(\frac{\Theta}{T}\right)^2 \cdot \frac{e^{\Theta/T}}{(e^{\Theta/T}-1)^2} \right] \right\}$$ Where $c_{pp}, \gamma_p$ and $\Theta$ are constants. Thus: $$h = c_{pp} \cdot\left[ T + \frac{\gamma_p-1}{\gamma_p} \int \left(\frac{\Theta}{T}\right)^2 \cdot \frac{e^{\Theta/T}}{(e^{\Theta/T} - 1)^2}dT \right]$$

The integral in question would then be: $$ \int \left(\frac{\Theta}{T}\right)^2 \cdot \frac{e^{\Theta/T}}{(e^{\Theta/T} - 1)^2}dT$$

Here it is also in another form if it helps: $$ \Theta ^2 \int \left( \frac{1}{T^2} \cdot \frac{1}{(e^{\Theta/T} - 1)} + \frac{1}{T^2} \cdot \frac{1}{(e^{\Theta/T} - 1)^2} \right)dT$$

Thanks in advance!

Answer 1

Your integral is $\int\frac{\Theta^2}{4T^2}\operatorname{csch}^2\frac{\Theta}{2T}dT=\frac{\Theta}{2}\coth\frac{\Theta}{2T}+C$.

Answer 2

For a more pedestrian approach, if you don't immediately see the connection to the hyperbolic functions, you can substitute $u=\Theta/T$, which works out nicely since $\mathrm du=-\Theta/T^2\mathrm dT$ and you have $T^2$ in the denominator.