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Hi I have been trying to prove this $$ I:=\int \limits_{0}^{1} \left[ \frac{1}{x(x-1)} \bigg(2\mathrm{Li}_2\bigg(\frac{1-\sqrt{1-x}}{2}\bigg)-\log\bigg(\frac{1+\sqrt{1-x}}{2}\bigg)^2 \bigg) -\frac{\zeta(2)-2\log^2 2}{x-1} \right]{dx}=\sum_{k=2}^\infty \binom{2k}{k} \frac{1}{k^2 4^k} \sum_{j=1}^{k-1} \frac{1}{j}=\color{#00f}{\large% -{4 \over 3}\log^3 2-\frac{\pi^2}{3}\log 2+\frac{5}{2}\zeta(3) } $$ What a beautiful result!!!! I am trying to prove this. I am not sure of what to do, perhaps we could start with a change of variables $$ \xi=\frac{1-\sqrt{1-x}}{2}, $$ but I get stuck shortly after. This is strongly related to Mahler measures and integration. Thanks for your help.

I tried the following substitution but failed,

UPDATE: I tried a change of variables given above by $\xi$, we obtain $$ I=\int\limits_{0}^{1/2}\big(2\mathrm{Li}_2(\xi)-\log^2(1-\xi)\big)\left(\frac{4}{2\xi-1}-\frac{1}{\xi-1}-\frac{1}{\xi}\right)d\xi-4(\zeta(2)-2\log^2 2) \int\limits_0^{1/2}\frac{d\xi}{2\xi-1} $$ but the integral on the right diverges so I need to use another method now.

Thanks

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  • $\begingroup$ @O.L. thanks for fixing the title. $\endgroup$ – Jeff Faraci Apr 20 '14 at 16:48
  • $\begingroup$ My pleasure. Have you understood that you were actually asking about Poisson summation proof of the Jacobi imaginary transformation in your another question? $\endgroup$ – Start wearing purple Apr 20 '14 at 16:52
  • $\begingroup$ @O.L. I didn't know that. Thank you. I just am confused where the $\sqrt{\pi/\alpha}$ comes from. $\endgroup$ – Jeff Faraci Apr 20 '14 at 16:57
  • $\begingroup$ From the $(-i\tau)^{-1/2}$ factor in the formula (6) here. $\endgroup$ – Start wearing purple Apr 20 '14 at 17:00
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    $\begingroup$ @Integrals The main question involves two equalities. Are you unsure about how to prove both or only the last? $\endgroup$ – Meow Apr 20 '14 at 18:06
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Defining $I$ as the definite integral, $$I:= \int\limits_{0}^{1}\left[\frac{\zeta{(2)}-2\log^2{2}}{1-x}-\frac{1}{x(1-x)}\left(2\operatorname{Li}_2{\left(\frac{1-\sqrt{1-x}}{2}\right)}-\log^2{\left(\frac{1+\sqrt{1-x}}{2}\right)}\right)\right]\mathrm{d}x,$$ prove: $$I=\frac52\zeta{(3)}-2\zeta{(2)}\log{2}-\frac43\log^3{2}.$$


Substituting

$$\xi=\frac{1-\sqrt{1-x}}{2},$$

the integral becomes:

$$I= \int_{0}^{\frac12}\left[4\left(\zeta{(2)}-2\log^2{2}\right)-\frac{1}{\xi(1-\xi)}\left(2\operatorname{Li}_2{(\xi)}-\log^2{(1-\xi)}\right)\right]\frac{\mathrm{d}\xi}{1-2\xi}.$$

It turns out that the derivative of the expression $2\operatorname{Li}_2{(\xi)}-\log^2{(1-\xi)}$ is much simpler than the expression itself:

$$\frac{d}{d\xi}\left(2\operatorname{Li}_2{(\xi)}-\log^2{(1-\xi)}\right)=-\frac{2(1-2\xi)}{\xi (1-\xi)}\log{(1-\xi)}.$$

This suggests that we should integrate by parts.

$$\begin{align} I &=\int_{0}^{\frac12}\left[4\left(\zeta{(2)}-2\log^2{2}\right)-\frac{1}{\xi(1-\xi)}\left(2\operatorname{Li}_2{(\xi)}-\log^2{(1-\xi)}\right)\right]\frac{\mathrm{d}\xi}{1-2\xi}\\ &=\int_{0}^{\frac12}\left[4\left(\zeta{(2)}-2\log^2{2}\right)\xi(1-\xi)-\left(2\operatorname{Li}_2{(\xi)}-\log^2{(1-\xi)}\right)\right]\frac{\mathrm{d}\xi}{\xi(1-\xi)(1-2\xi)}\\ &=-\int_{0}^{\frac12}\left[4\left(\zeta{(2)}-2\log^2{2}\right)(1-2\xi)+\frac{2(1-2\xi)}{\xi (1-\xi)}\log{(1-\xi)}\right] \log{\frac{\xi(1-\xi)}{(1-2\xi)^2}} \mathrm{d}\xi\\ &=-2\int_{0}^{\frac12}\left[2\left(\zeta{(2)}-2\log^2{2}\right)+\frac{\log{(1-\xi)}}{\xi (1-\xi)}\right] (1-2\xi)\log{\frac{\xi(1-\xi)}{(1-2\xi)^2}} \mathrm{d}\xi\\ &=-4\left(\zeta{(2)}-2\log^2{2}\right)\int_{0}^{\frac12}(1-2\xi) \left[\log{\xi}+\log{(1-\xi)} - 2\log{(1-2\xi)}\right] \mathrm{d}\xi\\ &~~~~-2\int_{0}^{\frac12}\frac{(1-2\xi)}{\xi (1-\xi)} \log{(1-\xi)} \left[\log{\xi}+\log{(1-\xi)} - 2\log{(1-2\xi)}\right] \mathrm{d}\xi\\ &=4\left(\zeta{(2)}-2\log^2{2}\right)\int_{0}^{\frac12}(2\xi-1) \left[\log{\xi}+\log{(1-\xi)} - 2\log{(1-2\xi)}\right] \mathrm{d}\xi\\ &~~~~+2\int_{0}^{\frac12}\left(\frac{1}{1-x}-\frac{1}{x}\right) \left[\log{(1-\xi)}\log{\xi}+\log^2{(1-\xi)} - 2\log{(1-\xi)}\log{(1-2\xi)}\right] \mathrm{d}\xi. \end{align}$$

Now, distributing factors in the integrands and integrating term-by-term, we can write the integral $I$ as a sum of a dozen or so primitive integrals that each have anti-derivatives in terms of polylogarithms that may be easily evaluated and added up with the aid of a computer algebra program such as WolframAlpha, thus obtaining the desired result. However, such a solution leaves much to be desired in the way of elegance....

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