How do I prove that $$\int_{0}^{1/2} \dfrac{\log (1-x)}{x} \mathrm{d}x=\dfrac{1}{2}\log^2{2}-\dfrac{\pi^2}{12}$$
I taylor expanded $\log(1-x)$ but ended up with a series I couldn't evaluate. Please help. Thank you.
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I don't think so that 1÷2(log2)^2 would be present. – Aditya Kumar Mar 08 '15 at 07:55
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@AdityaKumar yes it would be – RE60K Mar 08 '15 at 07:56
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$$\frac{\log(1-x)}x=-\frac{(x+x^2/2+x^3/3+x^4/4+...)}x=-(1+x/2+x^2/3+x^3/4+...)$$ Now: $$\int\frac{\log(1-x)}x{\rm d}x=-(x+x^2/2^2+x^3/3^3+...)=-{\rm Li}_2(x)$$ Now: $$\int_0^{1/2}\frac{\log(1-x)}x{\rm d}x={\rm Li}_2(0)-{\rm Li}_2(1/2)=\frac{\log^22}2-\frac{\pi^2}{12}$$ since ${\rm Li}_2(0)=0$ and by duplication formula: $$\mathrm{Li}_2(x)+\mathrm{Li}_2(1-x) =\frac{\pi^2}{6}-\log(x)\log(1-x)$$ Put $x=1/2$: $$2\mathrm{Li}_2(1/2)=\frac{\pi^2}{6}-\log(1/2)\log(1/2)\implies \mathrm{Li}_2(1/2)=\frac{\pi^2}{12}-\frac12\log^2(2)$$
