I need to evaluate an integral of the form

$$\int_0^Q \frac{f(q)\cdot q^5 \mathop{dq}}{(1-\exp(-q/c))(\exp(q/c)+1)}.$$

If I got it right, then this looks a lot like the Debye integral of order 5 for which I haven't found a table yet..but I'm working on it. Anyway, there's another function in the numerator $f(q)$ I need to get rid of somehow.

Any ideas on how to approach this beast?

Edit: I need to perform numerical integration, which doesnt really work when $q$ approaches 0. I just noticed that $$ \frac{q^5}{(1-\exp(-q/c))(\exp(q/c)+1)}=\frac{q^5}{2(\exp(q)-1)}+\frac{q^5}{2(\exp(q)+1)}$$ The latter term is no problem in terms of numerical integration, while the first term is just a Debye function, I hope can find some series expansion for..


The integral in question can be seen in the form $$\int_0^Q \frac{f(q)\cdot q^5 \mathop{dq}}{(1-\exp(-q/c))(\exp(q/c)+1)} = \frac{1}{2} \, \int_{0}^{Q} f(t) \, t^{5} \, csch(t) \, dt$$ and depending on what form $f(t)$ can be placed into depends on how to further integrate. One method is to expand $csch(t)$ into a power series and then integrate as seen by \begin{align} \int_0^Q \frac{f(q)\cdot q^5 \mathop{dq}}{(1-\exp(-q/c))(\exp(q/c)+1)} &= \frac{1}{2} \, \int_{0}^{Q} f(t) \, t^{5} \, csch(t) \, dt \\ &= \int_{0}^{Q} f(t) \, t^4 \, dt + \sum_{n=1}^{\infty} \frac{(2 - 4^n) B_{n}}{(2n)!} \, \int_{0}^{Q} f(t) \, t^{2n+4} \, dt. \end{align}

  • $\begingroup$ Thanks for your effort. Unfortunately, from a computational perspective, that last part $\int_0^Q f(t) t^{2n+4}\mathop{dt}$ goes to infinity pretty fast, so It's not really useful in this case. But I appreciate your effort.good job $\endgroup$ – OD IUM Sep 15 '18 at 13:17

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