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While playing around with Mathematica, I found that

$$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx = \frac{1}{3}\log^3(2)-\frac{\pi^2}{12}\log(2)+\frac{\zeta(3)}{8}$$

Please help me prove this result.

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Let $u=\log(1+x)$, $du=1/(1+x)\,dx$ and $1-x=2-e^u$ transforms our integral into $$\int_0^{\log2}u\log(2-e^u)\,\mathrm{d}u$$... not sure what to do from there, however. –  oldrinb Jun 1 '13 at 7:36
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you can see:math.stackexchange.com/questions/405356/… –  math110 Jun 1 '13 at 7:47
    
@math110: I think I just understood what you are trying to say. We can relate it to the Euler Sum in your link. $$\int_0^1 \frac{\log(1+x)\log(1-x)}{1+x}dx=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}(H_n)^2$$ –  shobhit.iands Jun 1 '13 at 8:16
    
For a more generalized form of this integral, where the integral is over the range 0 to z rather than 0 to 1, see here -->> [mathhelpboards.com/questions-other-sites-52/… –  Dream Weaver Oct 1 '13 at 19:32
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1 Answer

up vote 4 down vote accepted

Use your favorite program to compute the indefinite integral in terms of polylogarithms $$\int\frac{\ln(1+x)\ln(1-x)\,dx}{1+x}=\frac{\ln2}{2}\ln^2(1+x)-\ln(1+x)\,\mathrm{Li}_2\left(\frac{1+x}{2}\right)+\mathrm{Li}_3\left(\frac{1+x}{2}\right).$$ [This can be verified by straightforward differentiation].

To compute the definite integral, it suffices to know $\mathrm{Li}_{2,3}\left(\frac12\right)$ and $\mathrm{Li}_{2,3}(1)$. However, the definition of polylogarithm immediately implies $\mathrm{Li}_s(1)=\zeta(s)$. Also, the values $\mathrm{Li}_{2,3}\left(\frac12\right)$ can be found here (formulas (16), (17)).

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I can verify that the anti-derivative can be computed through some tedious integration by parts. –  Potato Jun 1 '13 at 8:46
    
@Potato It is maybe not easy to guess the form of the antiderivative. But once you have a correct guess, to check it is a one-line calculation using that $\mathrm{Li}_2'(z)=-\ln(1-z)/z$ and $\mathrm{Li}_3'(z)=\mathrm{Li}_2(z)/z$. –  O.L. Jun 1 '13 at 8:50
    
@Potato: I think I understood what you mean. One integration by parts gives roughly $\mathrm{Li}_2(smth)$ instead of $\frac{\ln(1-z)}{1+z}$ and $1/(1+z)$ instead of $\ln(1+z)$. Subsequent integration gives $\mathrm{Li}_3$. –  O.L. Jun 1 '13 at 8:56
    
Thank O.L.! I did not realize that the indefinite integral would be so easy. –  shobhit.iands Jun 1 '13 at 9:40
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