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I need to prove

$$I=\int_0^1\frac{\ln x\operatorname{Li}_2(x^2)}{\sqrt{1-x^2}}dx=\frac{5\pi}8\zeta(3)-\pi\ln2\zeta(2)+\pi\ln^32$$

to finish my solution for this probem.

What I did is the common sub $x=\sin\theta$ which gives us

$$I=\int_0^{\pi/2}\ln(\sin\theta)\operatorname{Li}_2(\sin^2\theta)d\theta$$ and I dont know how to continue. I am not sure if its helpful to use the dilogaritmic identity $\operatorname{Li}_2(z^2)=2\operatorname{Li}_2(z)+2\operatorname{Li}_2(-z)$ which results in

$$I=2\int_0^{\pi/2}\ln(\sin\theta)\operatorname{Li}_2(\sin\theta)d\theta+2\int_0^{\pi/2}\ln(\sin\theta)\operatorname{Li}_2(-\sin\theta)d\theta$$


Other try is to write $$\operatorname{Li}_2(x^2)=-\int_0^1\frac{x^2\ln u}{1-x^2u}du$$

but this technique seems to make the problem even harder. So any idea how to crack this integral?

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A solution in large steps by Cornel

By combining simple trigonometric formulae and Landen's Identity, we get

$$\mathcal{I}=\int_0^{\pi/2}\log(\sin(x))\operatorname{Li}_2(\sin^2(x))\textrm{d}x=\int_0^{\pi/2}\log(\sin(x))\operatorname{Li}_2\left(\frac{\tan^2(x)}{1+\tan^2(x)}\right)\textrm{d}x$$ $$=-\underbrace{\int_0^{\pi/2}\log(\sin (x))\operatorname{Li}_2(-\tan^2(x))\textrm{d}x}_{\mathcal{J}}-2\underbrace{\int_0^{\pi/2}\log(\sin(x))\log^2(\cos(x))\textrm{d}x}_{\text{Beta function: $\pi \zeta(3)/8-\pi\log ^3(2)/2$}} \tag1.$$

Now, based on the result in $(1.12)$ from the book, (Almost) Impossible Integrals, Sums, and Series, we have that $\displaystyle \operatorname{Li}_2(-\tan^2(x))=\int_0^1 \frac{\tan ^2(x) \log (y)}{1+\tan ^2(x)y} \textrm{d}y$, and then we write

$$\mathcal{J}=\int_0^{\pi/2}\left(\int_0^1 \frac{\log(\sin(x))\tan ^2(x) \log (y)}{1+\tan ^2(x)y} \textrm{d}y \right)\textrm{d}x=\int_0^1\left(\int_0^{\pi/2} \frac{\log(\sin(x))\tan ^2(x) \log (y)}{1+\tan ^2(x)y} \textrm{d}x \right)\textrm{d}y,$$ and if we make the change of variable $\tan(x)\mapsto x$, we arrive at $$\mathcal{J}=\int_0^1\left(\int_0^{\infty}\frac{x^2 \log (x)\log (y)}{(1+x^2)(1+y x^2)}\textrm{d}x\right)\textrm{d}y-2\int_0^1\left(\underbrace{\int_0^{\infty}\frac{y\log(y) x^2 \log(1+x^2) }{(1+x^2)(1+y^2 x^2)}\textrm{d}x}_{\displaystyle f(y)}\right)\textrm{d}y$$ or $$\mathcal{J}=\frac{\pi}{4}\int_0^1\frac{\log ^2(y)}{(y-1)\sqrt{y}}\textrm{d}y-2\log(2)\pi\int_0^1\frac{y\log (y)}{y^2-1}\textrm{d}y-2\pi\int_0^1\frac{\log^2 (y)}{y^2-1}\textrm{d}y$$ $$+\log(2)\pi\int_0^1\frac{\log (y)}{y-1}\textrm{d}y-\pi\int_0^1\frac{\log(1+y) \log (y)}{1+y}\textrm{d}y$$ $$+\pi\int_0^1\frac{\log \left(\frac{1+y}{2}\right) \log (y)}{y-1}\textrm{d}y$$ $$=\frac{\pi ^3}{6} \log (2)-\frac{7 }{8}\pi \zeta (3),\tag2$$ where all the resulting integrals are trivial, excepting the last one which is slightly more difficult, but it can be reduced to the calculations with Beta function, or one can use the generalization from A simple strategy of calculating two alternating harmonic series generalizations by Cornel Ioan Valean.

Combining $(1)$ and $(2)$ we arrive at the desired result.

A short note: For a fast calculation of $f(y)$ we may use the differentiation under the integral sign (with respect to $a$) e.g. $\displaystyle f(y,a)=\int_0^{\infty}\frac{y\log(y) x^2 \log(1+a^2 x^2) }{(1+x^2)(1+y^2 x^2)}\textrm{d}x$.

A BONUS: Using the integral $\mathcal{I}$, we observe that $$\mathcal{I}=\int_0^{\pi/2}\log(\sin(x))\operatorname{Li}_2(\sin^2(x))\textrm{d}x=\int_0^{\pi/2}\log(\sin(x))\operatorname{Li}_2(1-\cos^2(x))\textrm{d}x,$$ and if we combine the last integral with Dilogarithm reflection formula here, we get $$\mathcal{K}=\int_0^{\pi/2}\log(\sin(x))\operatorname{Li}_2(\cos^2(x))\textrm{d}x$$ $$=\zeta(2)\underbrace{\int_0^{\pi/2} \log (\sin(x))\textrm{d}x}_{-\log(2)\pi/2}-4\underbrace{\int_0^{\pi/2} \log ^2(\sin (x)) \log (\cos (x)) \textrm{d}x}_{\text{Beta function: $\pi \zeta(3)/8-\pi\log ^3(2)/2$}}$$ $$-\underbrace{\int_0^{\pi/2} \log (\sin (x))\operatorname{Li}_2(\sin ^2(x))\textrm{d}x}_{\displaystyle \mathcal{I}}$$ $$=\log ^3(2)\pi+\frac{1}{12}\log(2)\pi ^3-\frac{9 }{8}\pi \zeta (3). $$

The calculation of the integral $\mathcal{K}$ can also be achieved by similar steps to the ones used for the evaluation of the integral $\mathcal{I}$.

To conclude, we obtain that

$$\int_0^{\pi/2}\log(\sin(x))\operatorname{Li}_2(\sin^2(x))\textrm{d}x=\int_0^{\pi/2}\log(\cos(x))\operatorname{Li}_2(\cos^2(x))\textrm{d}x$$ $$=\log^3(2)\pi-\frac{1}{6}\log(2)\pi^3+\frac{5}{8}\pi\zeta(3);$$ $$\int_0^{\pi/2}\log(\sin(x))\operatorname{Li}_2(\cos^2(x))\textrm{d}x=\int_0^{\pi/2}\log(\cos(x))\operatorname{Li}_2(\sin^2(x))\textrm{d}x$$ $$=\log ^3(2)\pi+\frac{1}{12}\log(2)\pi ^3-\frac{9 }{8}\pi \zeta (3). $$

Another short note: Alternatively, one can also build a system of relations with $\displaystyle \mathcal{I}=\int_0^{\pi/2}\log(\sin(x))\operatorname{Li}_2(\sin^2(x))\textrm{d}x$ and $\displaystyle \mathcal{K}=\int_0^{\pi/2}\log(\cos(x))\operatorname{Li}_2(\sin^2(x))\textrm{d}x$ and then calculate $\mathcal{I}+\mathcal{K}$ and $\mathcal{I}-\mathcal{K}$.

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  • 2
    $\begingroup$ That's really amazing solution (+1) $\endgroup$ – Ali Shather Nov 7 '19 at 23:23
  • $\begingroup$ I like the bonus one... very creative. $\endgroup$ – Ali Shather Nov 8 '19 at 16:44

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