$$\int_{0}^{e} \frac{\operatorname{W(x)} - x}{\operatorname{W(x)} + x} dx = 2 \operatorname{Li_2(-e)} - e + \frac{\pi^2}{6} - \log(4) + 4 \log(1 + e)≈-0.819168$$

As usual I prefer to know if there is an antiderivative like here .So WA gives the result but I would like to understand better .I think that we can use the following substitution :


After I'm stuck because of the polylogarithm .

My question

How to solve this properly ?

Why we have $\zeta(2)$ in the formula ?

Thanks a lot for your comments or answers.


Performing the substitution $x=te^t$ we get :

$$\int_{0}^{1} \frac{t - te^t}{t + te^t} dte^t$$

Or $$\int_{0}^{1} \frac{1 - e^t}{1 + e^t}(e^t(1+t)) dt$$

Or : $$\int_{0}^{1} \frac{1 - e^t}{1 + e^t}(e^t)+te^t \frac{1 - e^t}{1 + e^t}dt$$

Or :

$$\int_{0}^{1} \frac{1 +e^t- 2e^t}{1 + e^t}(e^t)+te^t \frac{1+e^t - 2e^t}{1 + e^t}dt$$

Or: $$\int_{0}^{1} e^t +\frac{- 2e^{t}}{1 + e^t}(e^t)+te^t+ \frac{ - 2te^{2t}}{1 + e^t}dt$$

The problem is :

$$\int_{0}^{1} \frac{ - 2te^{2t}}{1 + e^t}dt$$

We integrate by parts to get :

$$\int_{0}^{1} \frac{ - 2te^{2t}}{1 + e^t}dt=[-2te^t\ln(1+e^t)]_0^1-\int_{0}^{1} - 2(t+1)e^{t}\ln(1 + e^t)dt$$

The problem is :

$$\int_{0}^{1} - 2(t+1)e^{t}\ln(1 + e^t)dt$$

After this I'm stuck again ... Oh If we perform the substitution $y=e^t$ in the last integral we get the integral of MHZ .

  • $\begingroup$ You are on the right track. What do you get after substituting $W(x)=t$? $\endgroup$
    – Zacky
    Mar 29, 2020 at 11:24
  • $\begingroup$ @Zacky I am on my phone but after the substitution and an integration by parts I can't evaluate this : $\int_{0}^{1}(x+1)e^x\log(e^x+1)dx$.I haven't the tools for (maybe power series?).Anyway thanks for your interest. $\endgroup$
    – Erik Satie
    Mar 29, 2020 at 12:15
  • $\begingroup$ Well, what do you get now if you substitute the obvious term $e^x=t$? Also you might need to mention what definition of dilogarithm are you using, I'm seeing a $\operatorname{Li}_2(-e)$ there. $\endgroup$
    – Zacky
    Mar 29, 2020 at 12:27
  • $\begingroup$ @Zacky I think we can use the substitution $y=-x$ and $t=e^{-x}$ and use an integration by parts with the integral representation of the dilogarithm and normally we are done.Is it ok ? $\endgroup$
    – Erik Satie
    Mar 29, 2020 at 14:24

1 Answer 1


The result holds since after some substitutions we have that:$$\int (\log (t)+1) \log (t+1) \, dt=\text{Li}_2(-t)+t+\log (t) ((t+1) \log (t+1)-t)$$ So $$\int_1^e (\log (t)+1) \log (t+1) \, dt=\text{Li}_2(-e)+\frac{\pi ^2}{12}-1+(1+e) \log (1+e)$$

  • $\begingroup$ How do you get these results ? $\endgroup$
    – Erik Satie
    Mar 30, 2020 at 11:14

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