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.

  • $\begingroup$ the derivative of $x^{s-1}$ is $x^{s-1} \log(x)$. The integral you wrote diverges $\endgroup$ – reuns Nov 19 '16 at 16:03
  • $\begingroup$ Thaks @user1952009 the improper integral $\int_0^\infty \frac{e^{-x}(x+\log (x))}{1-e^{-x}}dx$ diverges, but it is possible define PV, isn't ? Many thanks for your attention and patience (I say my last comment about what's a Euler product that I've deleted). $\endgroup$ – user243301 Nov 19 '16 at 16:19
  • $\begingroup$ No. ${}{}{}{}{}$ $\endgroup$ – reuns Nov 19 '16 at 16:21
  • $\begingroup$ I accept your words but Wolfram Alpha compute such here, or type integrate e^(-x)(x+log(x))/(1-e^(-x)) dx, from x=0 to infinite , in the online calculator @user1952009 When you refresh the code, it is calculated. $\endgroup$ – user243301 Nov 19 '16 at 16:25
  • $\begingroup$ It is the same as $\int_0^1 \frac{\log x}{x}dx$, it diverges, end of discussion. $\endgroup$ – reuns Nov 19 '16 at 16:29

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.

  • $\begingroup$ I accept your answer, I don't know what was about such calculation from the online calculator, then I was asking here. Many thanks. $\endgroup$ – user243301 Nov 19 '16 at 17:09
  • $\begingroup$ @user243301 my answer won't help if you don't take 2 months reading a real analysis course, explaining in detail how to see if an integral (or a series) diverges or converges (see your other question) $\endgroup$ – reuns Nov 19 '16 at 17:11

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