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This question already has an answer here:

Calculate $$\int_0^1\frac{\ln(1+x)\ln(1-x)}{1+x}\,dx$$

My try :

Let : $$I(a,b)=\int_0^1\frac{\ln(1-ax)\ln(1+bx)}{1+x}\,dx$$

Then compute $\frac{d^2 I(a,b)}{dadb}$.

I'm happy to see ideas in order to kill this integral.

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marked as duplicate by Saad, Zacky, StubbornAtom, Joel Reyes Noche, Yanko Apr 24 at 12:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ The result should be $$\frac{1}{24} \left(3 \zeta (3)+8 \log ^3(2)-\pi ^2 \log (4)\right)$$ $\endgroup$ – Dr. Sonnhard Graubner Apr 24 at 10:58
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    $\begingroup$ By using the fact that: $$\int_a^bf(x)dx=\int_a^bf(a+b-x)dx$$ you can show that: $$\int_0^1\frac{\ln(1+x)\ln(1-x)}{1+x}dx=\int_0^1\frac{\ln(2-x)\ln(x)}{2-x}dx$$ Not sure if you can get anywhere with this or not $\endgroup$ – Henry Lee Apr 24 at 10:59
  • $\begingroup$ Thanks! But I don't know how I complete $\endgroup$ – user664780 Apr 24 at 11:07
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    $\begingroup$ The question has already been asked here $\endgroup$ – Mattos Apr 24 at 11:48
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We can take $\ 2ab= a^2+b^2-(a-b)^2$ to get: $$I= \frac12 \int_0^1 \frac{\ln^2 (1-x)}{1+x}dx+\frac12\int_0^1\frac{\ln^2(1+x)}{1+x}dx-\frac12 \int_0^1 \frac{\ln^2\left(\frac{1-x}{1+x}\right)}{1+x}dx$$ By letting $\frac{1-x}{1+x}=t$ and expanding into power series the last one we get: $$I=\frac12 J +\frac{\ln^3(1+x)}{6}\bigg|_0^1 -\frac12 \int_0^1 \frac{\ln^2 t}{1+t}dt=\frac12 J +\frac{\ln^3 2}{6}-\frac34\zeta(3) $$ $$J=\int_0^1 \frac{\ln^2(1-x)}{1+x}dx\overset{1-x\to x}=\frac12\int_0^1 \frac{\ln^2 x}{1-\frac{x}{2}}dx=\frac12 \sum_{n=0}^\infty \frac{1}{2^n} \int_0^1 x^{n}\ln^2 xdx$$ $$=\sum_{n=0}^\infty \frac{1}{2^n}\frac{1}{(n+1)^3}=2\operatorname{Li}_3 \left(\frac12\right)\Rightarrow \boxed{I=\operatorname{Li}_3 \left(\frac12\right)+\frac{\ln^3 2}{6}-\frac34\zeta(3)}$$ Of course one can rewrite the trilogarithm's value as seen from $(17)$ in this link, but this form is also valid.

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  • $\begingroup$ Wow!! Best way sir thanks! $\endgroup$ – user664780 Apr 24 at 13:17
  • $\begingroup$ Beautiful sir many thanks for you! $\endgroup$ – user664780 Apr 24 at 13:27
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My approach, here $t=x/2$

$$\begin{align} & \int_0^1 \frac{\ln(1+x)\ln(1-x)}{1+x} \mathrm{d}x = \int_0^1 \frac{\ln(2-x)\ln x}{2-x} \mathrm{d}x\\ = & \int_0^1 \frac{(\ln(1-\tfrac{x}{2})+\ln2)(\ln\tfrac{x}{2}+\ln2)}{2-x} \mathrm{d}x\\ = &\> \ln^22\int_0^1 \frac{\mathrm{d}x}{2-x} + \ln2\int_0^{1/2} \frac{\ln t}{1-t}\mathrm{d}t + \ln2\int_0^{1/2} \frac{\ln(1-t)}{1-t}\mathrm{d}t + \int_0^{1/2} \frac{\ln(1-t)\ln t}{1-t}\mathrm{d}t\\ = &\> \ln^22(-\ln(2-x))|_{x=0}^{1}+\ln2(\operatorname{Li}_{2}(\tfrac1{2})-\operatorname{Li}_{2}(1))-\frac{\ln2\ln^2(1-t)}{2}\bigg|_{0}^{1/2}\\ & -\frac{\ln^2(1-t)\ln t}{2}\bigg|_{0}^{1/2} + \frac1{2}\int_0^{1/2}\frac{\ln^2(1-t)}{t}\mathrm{d}t \end{align}$$

Last integral is not trivial, but you can finish it with series, which is

$$\int_0^{1/2}\frac{\ln^2(1-t)}{t}\mathrm{d}t = -\frac{\ln^32}{3}+\frac{\zeta(3)}{4}$$

and $\operatorname{Li}_{2}(1) = \frac{\pi^2}{6}$, $\operatorname{Li}_{2}(\tfrac1{2}) = \frac{\pi^2}{12}-\frac{\ln^22}{2}$

thus

$$\int_0^1 \frac{\ln(1+x)\ln(1-x)}{1+x} \mathrm{d}x = \frac{\ln^32}{3} - \frac{\pi^2}{12}\ln2 + \frac{\zeta(3)}{8}$$

Some supplementary for Last integral with only elementary resources

$$\int_0^{1/2}\frac{\ln^2(1-t)}{t}\mathrm{d}t = \int_{1/2}^{1}\frac{\ln^2t}{1-t}\mathrm{d}t = \int_{0}^{1}\frac{\ln^2t}{1-t}\mathrm{d}t - \int_{0}^{1/2}\frac{\ln^2t}{1-t}\mathrm{d}t$$

Let $t=e^{-u}$, using $\frac1{1-t} = \sum_{n=0}^{\infty}t^n$

$$\begin{align} \int_{0}^{1/2}\frac{\ln^2t}{1-t}\mathrm{d}t & = \sum_{n=0}^{\infty}\int_{\ln2}^{\infty}u^2e^{-(n+1)u}\>\mathrm{d}u\\ & = 2\sum_{n=0}^{\infty}\frac{e^{-(n+1)\ln2}}{(n+1)^3} + 2\ln2\sum_{n=0}^{\infty}\frac{e^{-(n+1)\ln2}}{(n+1)^2} + \ln^22\sum_{n=0}^{\infty}\frac{e^{-(n+1)\ln2}}{n+1}\\ & = 2\operatorname{Li}_{3}(\tfrac1{2}) + 2\ln2\operatorname{Li}_{2}(\tfrac1{2}) + \ln^22\operatorname{Li}_{1}(\tfrac1{2}) \end{align}$$

you can calculate $\int_{0}^{1/2}\frac{\ln^2t}{1-t}\mathrm{d}t$ with $\operatorname{Li}_{3}(\tfrac1{2}) = \frac{\ln^32}{6}-\frac{\pi^2}{12}\ln2+\frac7{8}\zeta(3)$, $\operatorname{Li}_{1}(\tfrac1{2}) = \ln2$. And by the same fashion, easily have $\int_{0}^{1}\frac{\ln^2t}{1-t}\mathrm{d}t = 2\zeta(3)$ to finish this solution.

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  • $\begingroup$ Very nice sir 😇😇 $\endgroup$ – user664780 Apr 24 at 13:16
  • $\begingroup$ Many thanks for you sir! $\endgroup$ – user664780 Apr 24 at 13:27
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I'd differentiate the $\ln(1-x)$ and integrate the rest giving $\ln(\ln(1+x))$. After that you could do a series expansion for $(1-x)^{-1}$

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  • $\begingroup$ Can you complete solution plz $\endgroup$ – user664780 Apr 24 at 11:08