Is there a special function defined for this integral?

Question

Recently, I came across this integral: $$\int_0^\infty \frac{xe^{-sx}}{1-e^{-px}}dx$$ where $s, p$ are, in general, complex constants.

As far as I know, this integral does not have an elementary form. However, this integral looks so neat and simple that I guess a special function is defined for this- just like the Gamma function.

Is there such a special function? If so, have its properties been studied extensively?

p.s. Some might say the integral is simply the Laplace transform of $\frac{x}{1-e^{-px}}$. Well, if that’s the best answer, I would be a little bit disappointed.

Thanks for any help in advance.

EDIT: I realized I missed a $x$ in the numerator. Sorry for it.

Answer 1

For $\operatorname{Re}(s) > 0$ and $\operatorname{Re}(p) > 0$, we have

$$ \int_{0}^{\infty} \frac{x e^{-sx}}{1 - e^{-px}} \, dx = \sum_{n=0}^{\infty} \int_{0}^{\infty} x e^{-(s+np)x} \, dx = \sum_{n=0}^{\infty} \frac{1}{(s+np)^2} = \frac{1}{p^2}\psi'\left(\frac{s}{p}\right) $$

where $\psi$ is the digamma function (and hence $\psi'$ is the trigamma function). Alternatively,

$$ \int_{0}^{\infty} \frac{x e^{-sx}}{1 - e^{-px}} \, dx = \frac{1}{p^2}\zeta\left(2,\frac{s}{p}\right) $$

where $\zeta(s, z)$ is the Hurwitz zeta function.

Answer 2

Yeah, it's called "$\infty$": $$\int_{\epsilon}^{\infty}\frac{\mathrm{e}^{-sx}\mathrm{d}x}{1-\mathrm{e}^{-px}}\sim\frac{1}{p}\ln\frac{1}{\epsilon}$$ because the integrand diverges as $\frac{1}{px}$ as $x\to 0^+$.

On the other hand, if you subtract out this divergent part, you'll probably get something like the digamma function:

$$\psi\left(z\right)=\int_{0}^{\infty}\left(\frac{e^{-t}}{t}-\frac{e^{-zt}}{1-e^% {-t}}\right)\mathrm{d}t\quad (\Re z>0)$$

(DLMF 5.9.12).

Answer 3

HINT, the hypergeometric function : $$\int \frac{e^{-sx}}{1-e^{-px}}dx=\frac{1}{s-p}e^{x(p-s)}\:_2F_1\left(1\:,\:1-\frac{s}{p}\:;\:2-\frac{s}{p}\:;\:e^{px} \right)+\text{constant}$$