$$\int_{0}^{e} \frac{\operatorname{W(x)} – x}{\operatorname{W(x)} + x} dx = 2 \operatorname{Li_2(-e)} – e + \frac{\pi^2}{6} – \log(4) + 4 \log(1 + e)≈-0.819168$$
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 :
$$t=xe^x$$
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.
Update
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 .
Best Answer
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 $$\int_1^e (\log (t)+1) \log (t+1) \, dt=\text{Li}_2(-e)+\frac{\pi ^2}{12}-1+(1+e) \log (1+e)$$