An explanation or hints to evaluate $\text{PV}\int_0^\infty \frac{e^{-x}(x+\log (x))}{1-e^{-x}}dx$

Question

After I was thinking how evaluate $\int_0^\infty \frac{e^{-x}}{1-e^{-x}}x\log (x)dx$ (the idea is to differentiate an integral representation for $\Gamma(s)\zeta(s)$ that holds for $\Re s>1$, and evaluating after at $s=2$), I was interested in the integral $$\int_0^\infty \frac{e^{-x}(x+\log (x))}{1-e^{-x}}dx.$$

Wolfram Alpha knows that is required to define the Cauchy principal value and find the result.

Question. Can you explain and compute this integral/value $$\text{PV}\int_0^\infty \frac{e^{-x}(x+\log (x))}{1-e^{-x}}dx?$$

Many thanks. I want learn these calculations since there are different integrands of thse Cauchy principal values that could be interesting. If you prefer only provide me hints, also you are welcome.

Answer 1

Read a course on the convergence of integrals and the principal value allowing us to say that $\int_{-1}^1 \frac{dx}{x} = 0$ but it doesn't change that $\int_0^1\frac{dx}{x} = \infty$

Also, WA should answer that it diverges, so maybe you have found a bug in WA, since in $\int_0^\infty e^{-x}\frac{x+\log(x)}{1-e^{-x}}dx$ everything converges except $\int_0^1 e^{-x}\frac{\log(x)}{1-e^{-x}}dx = \int_0^1 \frac{\log(x)}{x}dx+\int_0^1 \log(x)(\frac{1}{e^x-1}-\frac{1}{x})dx$ where the second integral converges, so it reduces to

Integrate[Log[x]/x, {x, 0, 1}, PrincipalValue -> True] 

where WA answers - as wanted - that it diverges.