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This integral can be found in Cornel's book, (Almost) Impossible Integral, Sums and Series page $97$ where he showed that

$$\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dt=\frac14\left(\frac12\ln^2(1+x^2)-2\operatorname{Li}_2(x)+\frac12\operatorname{Li}_2(-x^2)+\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)\right)$$

by differentiating $\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)$ then integrating back. How magical is that?

My approach: \begin{align} I&=\int_0^x\frac{t\ln(1-t)}{1+t^2}\ dt\overset{IBP}{=}\frac12\ln(1+x^2)\ln(1-x)+\frac12\int_0^x\frac{\ln(1+t^2)}{1-t}\ dt\\ &\overset{1-x=u}{=}\frac12\ln(1+x^2)\ln(1-x)-\frac12\int_{1}^{1-x}\frac{\ln(2+2u+u^2)}{u}\ du\\ &=\frac12\ln(1+x^2)\ln(1-x)-\Re\int_{1}^{1-x}\frac{\ln((1+i)-u)}{u}\ du \end{align} and by using $\displaystyle\int \frac{\ln(a-x)}{x}\ dx=\ln(a)\ln x-\operatorname{Li}_2\left(\frac{x}{a}\right)\ $, we get \begin{align} I&=\frac12\ln(1+x^2)\ln(1-x)-\Re\left(\ln(1+i)\ln u-\operatorname{Li}_2\left(\frac{u}{1+i}\right)\right)_{u=1}^{u=1-x}\\ &=\frac12\ln(1+x^2)\ln(1-x)-\Re\left(\ln(1+i)\ln(1-x)-\operatorname{Li}_2\left(\frac{1-x}{1+i}\right)+\operatorname{Li}_2\left(\frac{1}{1+i}\right)\right)\\ &=\frac12\ln(1+x^2)\ln(1-x)+\frac12\ln2\ln(1-x)+\Re\operatorname{Li}_2\left(\frac{1-x}{1+i}\right)-\frac{\pi^2}{16}\\ &\boxed{=\frac12\ln\left(2(1+x^2)\right)\ln(1-x)+\Re\operatorname{Li}_2\left(\frac{1-x}{1+i}\right)-\frac{\pi^2}{16}} \end{align} where I used $\ \Re\ln(1+i)=\frac12\ln2\ $ and $\ \Re\operatorname{Li}_2\left(\frac{1}{1+i}\right)=\frac{\pi^2}{16}$.

I tested my solution numerically and all is good but as we can see it does not match Cornel's answer. any idea how to make it match?

Thanks,

Its worth to mention that by comparing the two results proved above, we find the new interesting identity:

$$\Re\operatorname{Li}_2\left(\frac{1-x}{1+i}\right) =\frac14\left(\frac12\ln^2(1+x^2)-2\ln\left(2(1+x^2)\right)\ln(1-x)-2\operatorname{Li}_2(x)+\frac12\operatorname{Li}_2(-x^2)\\+\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)+\frac{\pi^2}{4}\right)$$

Ali Shadhar
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    @ automaticallyGenerated yes because they both give the same results. the question is how to make them match. – Ali Shadhar Jul 05 '19 at 20:05
  • no problem.. I always do. – Ali Shadhar Jul 05 '19 at 20:19
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    Are you interested in other approaches? Because I don't have much knowldege about dilogarithms in order to match that. (Not saying that I have one now, but I'm thinking of an approach).// Btw, maybe you find this interesting: $$\int_0^x \frac{t\ln(1+t)}{1+t^2}dt=\frac12 \operatorname{Li}_{2}\left(\frac{1-x^2}{2}\right)+\frac12\ln(1-x^2)\ln\left(\frac{1+x^2}{2}\right)-\frac12 \operatorname{Li}_2\left(\frac12\right) -\frac14\left(\frac12\ln^2(1+x^2)-2\operatorname{Li}_2(x)+\frac12\operatorname{Li}_2(-x^2)+\operatorname{Li}_2\left(\frac{2x}{1+x^2}\right)\right)$$ – Zacky Jul 05 '19 at 21:04
  • @Zacky nice result. did you verify it? I just want to make my result match with Cornels'. – Ali Shadhar Jul 05 '19 at 21:13
  • @AliShather What is the range of $x$ for your formula?// At mine for $x=1$ doesn't seem to work, but it does for $\frac12$ and $\frac13$. Idk what's going on, I guess it works only for $x\in (0,1)$. – Zacky Jul 05 '19 at 21:31
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    Commenting since it doesn't work. My idea was to consider the following integrals:$$I=\int_0^x \frac{t\ln(1-t)}{1+t^2}dt, \quad J=\int_0^x \frac{t\ln(1+t)}{1+t^2}dt$$ $$I+J=\int_0^x\frac{t\ln(1-t^2)}{1+t^2}dt\overset{t^2=y}=\frac12\int_0^{x^2}\frac{\ln(1-y)}{1+y}dy$$ $$=\frac12 \left(\operatorname{Li}_{2}\left(\frac{1-x^2}{2}\right)+\ln(1-x^2)\ln\left(\frac{1+x^2}{2}\right)- \operatorname{Li}_2\left(\frac12\right)\right)$$ Similarly for $I-J$, just let $\frac{1-t}{1+t}$ and use partial fractions. Then $2I=(I+J)+(I-J)$. From the above the integral result that I mentioned easily follows. – Zacky Jul 05 '19 at 21:49
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    @Zacky $x$ can be any number. if we use my result or yours for $x=1$, it does not seem to work but its fine if we use Cornels'. now we can see the big difference between these results. – Ali Shadhar Jul 05 '19 at 21:50
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    nice idea Zacky. thats one of my fav techniques. – Ali Shadhar Jul 05 '19 at 21:52
  • @AliShather I just re-verified the formula from above, it works for any numbers too (including $x=1$). I don't know what went wrong the first time, but it works just fine. I don't know how to match it with that, but atleast we can get $3$ representations for it. – Zacky Jul 06 '19 at 10:39
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    @Zacky thats good . that means no problem with my result as well. maybe manipulating with dilogarithm identities would cancel out that $\ln(1-x)$.just a guess... have not tried yet. . – Ali Shadhar Jul 06 '19 at 23:22

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