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In a short presentation on RG, that is A Mesmerizing Integral Representation of $ζ^2(3)$ by C. I. Valean, we have, if I'm allowed, the following intriguing integral representation of $\zeta^2(3)$,

$$\zeta^2(3)$$ $$\small =\frac{32}{5} \int_0^1\left(2 \Re\left\{\text{Li}_3\left(\frac{1-x^2}{1+x^2}+i\frac{ 2 x}{1+x^2}\right)\right\}-\text{Li}_3\left(\frac{1}{1+x^2}\right)\right)\frac{\log (1-x) \log (1+x)}{x} \textrm{d}x.$$

The proof flows as follows: using that $\displaystyle \int_0^1 \frac{y\log(1-y)\log(1+y)}{x^2+y^2}\textrm{d}y=-\frac{3}{8}\zeta(3)+\frac{1}{4}\operatorname{Li}_3\left(\frac{1}{1+x^2}\right)-\frac{1}{2}\Re\biggr\{\operatorname{Li}_3\left(\frac{1-x^2}{1+x^2}+i\frac{2x}{1+x^2}\right)\biggr\}=-\frac{3}{8}\zeta(3)+\frac{1}{4}\operatorname{Li}_3(\cos^2(\theta))-\frac{1}{2}\sum_{n=1}^{\infty} \frac{\cos(2n\theta)}{n^3}, \ x=\tan(\theta), \theta \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)$, given in More (Almost) Impossible Integrals, Sums, and Series (2023), page $4$, together with the fact that $\displaystyle \int_0^1 \frac{\log (1-x) \log (1+x)}{x}\textrm{d}x=-\frac{5}{8}\zeta(3)$, also found (Almost) Impossible Integrals, Sums, and Series (2019), we have $$\frac{32}{5} \int_0^1\left(2 \Re\left\{\text{Li}_3\left(\frac{1-x^2}{1+x^2}+i\frac{ 2 x}{1+x^2}\right)\right\}-\text{Li}_3\left(\frac{1}{1+x^2}\right)\right)\frac{\log (1-x) \log (1+x)}{x} \textrm{d}x$$ $$=-\frac{32}{5} \int_0^1\left(\frac{3}{2}\zeta(3)+4\int_0^1 \frac{y\log(1-y)\log(1+y)}{x^2+y^2}\textrm{d}y\right)\frac{\log (1-x) \log (1+x)}{x} \textrm{d}x$$ $$=6\zeta^2(3)-\frac{128}{5} \int_0^1 \left( \int_0^1 \frac{y \overbrace{\log (1-x) \log (1+x)}^{\displaystyle f(x)} \overbrace{\log(1-y)\log(1+y)}^{\displaystyle f(y)}}{x(x^2+y^2)}\textrm{d}y\right)\textrm{d}x$$ $$\small=6\zeta^2(3)-\frac{128}{5} \int_0^1 \left( \int_0^1 \frac{y f(x) f(y)}{x(x^2+y^2)}\textrm{d}y\right)\textrm{d}x\overset{\substack{\text{use } \\ \text{the symmetry}}}{=}6\zeta^2(3)-\frac{128}{5} \int_0^1 \left( \int_0^1 \frac{x f(x) f(y)}{y(x^2+y^2)}\textrm{d}y\right)\textrm{d}x$$ $$\small=6\zeta^2(3)-\frac{64}{5} \int_0^1 \left( \int_0^1 \left(\frac{y}{x}+\frac{x}{y}\right) \frac{ f(x) f(y)}{x^2+y^2}\textrm{d}y\right)\textrm{d}x=6\zeta^2(3)-\frac{64}{5}\int_0^1 \frac{f(x)}{x}\left( \int_0^1 \frac{f(y) }{y}\textrm{d}y\right)\textrm{d}x$$ $$\small=6\zeta^2(3)-\frac{64}{5}\left(\int_0^1 \frac{\log(1-x)\log(1+x)}{x} \textrm{d}x\right)^2=\zeta^2(3),$$ which gives an end to the present proof.

Question $1$: Do we have in the mathematical literature challenging integrals with a similar structure of the integrand involving parts like $\displaystyle \text{Li}_n\left(\frac{1-x^2}{1+x^2}+i\frac{ 2 x}{1+x^2}\right), n\ge 2$? Any links or references would be highly appreciated.

Question $2$: I would love to see different approaches to this integral, this time not for the sake of seeing more proofs, but out of the curiosity of possible interesting and subtle connections with integrals and series that are out of my sight at the moment. Therefore, even some unfinished solutions could be very interesting.

Update $1$: Another very nice version presented by the same author is

$$\zeta^2(3)$$ $$=\frac{32}{19} \int_0^1 \biggr(\Re\left\{\operatorname{Li}_3\left(\frac{1-x^2}{1+x^2}+i\frac{ 2 x}{1+x^2}\right)\right\}\left(\log ^2(1-x)+\log ^2(1+x)\right)$$ $$-\operatorname{Li}_3\left(\frac{1}{1+x^2}\right) \log (1-x) \log (1+x)\biggr)\frac{1}{x} \textrm{d}x,$$

which combines the previous result and the particular case of a result given in More (Almost) Impossible Integrals, Sums, and Series (2023).

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    $\begingroup$ Way too complicated to be intriguing. Obviously $Li_3$ is related to $\zeta(3)$. It would be much more interesting if the integral contained elementary functions only $\endgroup$
    – Yuriy S
    Oct 16, 2023 at 20:49
  • $\begingroup$ @YuriyS I see your point. Perhaps you might think of something simpler like the integral $K$ here: math.stackexchange.com/q/3783303. $\endgroup$ Oct 16, 2023 at 21:06
  • $\begingroup$ Well, this is an almost impossible to digest integral, i do understand that integrals are interesting and getting new ideas to compute "periods" (best related to other algebraic objects like algebraic curves) are always welcome. However, in this case i see an integral that has most chances to appear by writing its proof in reverse order, and thus starting with its end. There is nothing intriguing or mesmerizing in the integral. Rather i can find something irritating and something messy in it. Now to the questions. Q1: No. Q2: Not a question. $\endgroup$
    – dan_fulea
    Nov 13, 2023 at 18:12
  • $\begingroup$ @dan_fulea There are two different things to ponder over well in the context of this precise example above 1) constructing such an integral in reverse order and the truly important one 2) receiving such an integral and naturally being able to recognize the possibility of exploiting the auxiliary integral stated at the beginning of the proof supposing that you are aware of it. Well, as regards 2), you have already witnessed that finding such a solution is perfectly possible, right to your integral here math.stackexchange.com/questions/3905908. $\endgroup$ Nov 13, 2023 at 19:49
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    $\begingroup$ Q1: I think that searching in literature for integrals involving the Clausen function might give better results. This is since the generalized Clausen function can be written as: $$\Im \text{Li}_{2n}\left(\frac{1-x^2}{1+x^2}+i\frac{ 2 x}{1+x^2}\right) = \text{Cl}_{2n}\left(2\arctan x\right)$$ $$\Re \text{Li}_{2n+1}\left(\frac{1-x^2}{1+x^2}+i\frac{ 2 x}{1+x^2}\right) = \text{Cl}_{2n+1}\left(2\arctan x\right)$$ Here is one example of such integral (squared). $\endgroup$
    – Zacky
    Feb 8 at 11:57

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These ideas are well explained here :

( the question ) Is there an integral for $\frac{1}{\zeta(3)} $?

(the answer ) Understanding when the Multiplication of two Definite Integrals gives Unity

Your question is an analogue.

In general we can put any given function $f(x)$ of a constant $c$, where $c$ i given by an integral , in an integral form.

So for $f(x) = 1/x $ we get things like

$$\frac{1}{\gamma}=\int_1^{\infty } -\frac{\log \left(\log \left(\frac{1}{x}\right)\right)}{\left(-\text{Ei}\left(-\log \left(\frac{1}{x}\right)\right)+x \log \left(\log \left(\frac{1}{x}\right)\right)-i \pi \right)^2} \, dx$$

$$\frac{1}{\zeta (3)}=\frac{3}{2} \int_1^{\infty } \frac{\log ^2(x)}{(x+1) \left(-2 \text{Li}_3(-x)+2 \text{Li}_2(-x) \log (x)+\log (x+1) \log ^2(x)\right){}^2} \, dx$$

by the techniques in the links above.

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  • $\begingroup$ One could notice that these are a kind of 'double integrals ' since we use special functions defined by integrals in the integrand. I think this is often unavoidable. But one might wonder if these special functions can be avoided in the integrand. I think in general it is not avoidable. $\endgroup$
    – mick
    Nov 16, 2023 at 23:51

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