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.


  • $\begingroup$ @O.L. thanks for fixing the title. $\endgroup$ 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? $\endgroup$ 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. $\endgroup$ Apr 20 '14 at 16:57
  • $\begingroup$ From the $(-i\tau)^{-1/2}$ factor in the formula (6) here. $\endgroup$ Apr 20 '14 at 17:00
  • 1
    $\begingroup$ @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

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: $$I=\frac52\zeta{(3)}-2\zeta{(2)}\log{2}-\frac43\log^3{2}.$$



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....


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.