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Is it possible to show $$ \int_{0}^{1}\frac{K(k)\ln\left[\tfrac{\left ( 1+k \right)^3}{1-k} \right] }{k} \text{d}k=\frac{\pi^3}{4}\;\;? $$ where $K(k)$ is the complete elliptic integral of the first kind with modulus $k$. One proof is to compute the twins first: \begin{aligned} &\int_{0}^{1} \frac{K(k)\ln(1+k)}{k}\text{d}k =-2G\ln(2)-4\Im\operatorname{Li}_3\left ( \frac{1+i}{2} \right ) +\frac{5\pi^3}{32} +\frac\pi8\ln(2)^2,\\ &\int_{0}^{1} \frac{K(k)\ln(1-k)}{k}\text{d}k =-6G\ln(2)-12\Im\operatorname{Li}_3\left ( \frac{1+i}{2} \right ) +\frac{7\pi^3}{32} +\frac{3\pi}{8}\ln(2)^2, \end{aligned} where $G$ denotes Catalan's constant and $\text{Li}_3(.)$ trilogarithm.

The simplicity of the result make me believe that it could be obtained by some implicit approaches. I would appreciate if you could offer some insights or ideas.

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    $\begingroup$ The formulation of the results, very close to the link cited in the answer of the user user97357329, makes me belive that that article was a main source for the question. If this is the case, why don't you mention it? Please do this (here and in further posts). Also, it would be good for the community, for potential answers at any rate, to have with the question more related resources, where similar formulas are presented, where the framework and the methods are established, where proofs are given. (It's always a pain for me to find the related page i once knew in Hancock or Bailey or...) $\endgroup$
    – dan_fulea
    Commented Feb 24 at 3:41
  • $\begingroup$ Using Fourier Legendre Expansions you can convert the problem to, (but requires too much work) $$\sum_{n=0}^{\infty}\left[\frac{1}{4n+1}-\frac{1}{4n+3}-6\frac{(-1)^n}{2n+1}\right]H^{(2)}_n=\frac{\pi^3}{24}$$ $\endgroup$ Commented Feb 24 at 9:07

3 Answers 3

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Here's an elementary approach. Note that Landen's transformation gives us:

$$\int_0^1 \frac{K(x)\ln\left(\frac{1-x}{1+x}\right)}{x}dx=\int_0^1 \frac{K\left(\frac{2\sqrt x}{1+x}\right)\ln\left(\frac{1-x}{1+x}\right)}{x(1+x)}dx $$

$$\overset{\large \frac{2\sqrt x}{1+x}\to x}=\frac12\int_0^1 \frac{K(x)\ln(1-x^2)}{x}\left(1+\frac{1}{\sqrt{1-x^2}}\right)dx \tag{*}$$


Therefore, we can compute the original integral by rewriting $\frac{(1+x)^3}{1-x}$ as $(1-x^2)\frac{(1+x)^2}{(1-x)^2}$.

$$\int_0^1 \frac{K(x)\ln\left(\frac{(1+x)^3}{1-x}\right)}{x}dx\overset{(*)}=-\int_0^1 \frac{K(x)\ln(1-x^2)}{x\sqrt{1-x^2}}dx$$

$$\overset{1-x^2\to x^2}=-2\int_0^1 \frac{K'(x)\ln x}{1-x^2}dx \overset{(**)}=-\int_0^1 \frac{4}{1-x^2} \int_{x}^1 \frac{\ln y}{\sqrt{(1-y^2)(y^2-x^2)}}dydx $$

$$=-4\int_0^1 \frac{\ln y}{\sqrt{1-y^2}} \int_0^{y} \frac{1}{(1-x^2)\sqrt{y^2-x^2}}dxdy=-2\pi\int_0^1 \frac{\ln y}{1-y^2}dy=\frac{\pi^3}{4}$$


Above we used a more general variant of the result I've shown here, namely:

$$\int_{x}^1 \frac{\ln y}{\sqrt{(1-y^2)(y^2-x^2)}}dy\overset{y\to\frac{x}{y}}=\int_{x}^1 \frac{\ln\left(\frac{x}{y}\right)}{\sqrt{(1-y^2)(y^2-x^2)}} = \frac12 \ln (x) K'(x) \tag{**}$$

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    $\begingroup$ (+1) Nice. Essentially, you use the key result 4.242.1 (in a rearranged form) from Table of Integrals, Series and Products by I.S. Gradshteyn and I.M. Ryzhik, that is, $$\displaystyle \int_0^{\infty} \frac{\log(x)}{\sqrt{(a^2+x^2)(b^2+x^2)}}\textrm{d}x = \frac{1}{2a}\operatorname{K}\left(\frac{\sqrt{a^2-b^2}}{a}\right)\log(a b),$$ which immediately follows by exploiting $x\mapsto a b/x$ to get the initial integral with a different sign plus a constant multiplied by the integral in the next comment. (to be continued) $\endgroup$ Commented Feb 24 at 18:37
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    $\begingroup$ To continue, you need $$\displaystyle \int_0^{\infty} \frac{1}{\sqrt{(a^2+x^2)(b^2+x^2)}}\textrm{d}x=\frac{1}{b}\operatorname{K}\left(\frac{b^2-a^2}{b^2}\right),$$ which may be evaluated by letting the variable change $x\mapsto a \tan(x)$. In particular, you need the case $a=1$ of the main integral (in the previous comment). $\endgroup$ Commented Feb 24 at 18:38
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    $\begingroup$ @user97357329 thanks, you're right (that alongside the initial step with the Landen transformation are the key identities). I've looked up at the result from $4.242$, but I think $4.242.5$ is essentially the more simpler variant: $$\int_a^b \frac{\ln x}{\sqrt{a^2-x^2}{\sqrt{x^2-b^2}}}dx= \frac{1}{2a}K\left(\frac{\sqrt{a^2-b^2}}{a}\right)\ln(ab)$$ Of course, those are the same in disguise. $\endgroup$
    – Zacky
    Commented Feb 24 at 18:48
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    $\begingroup$ It is hard to beat the power of the identities with elliptic integrals (where elementary steps are discussable). Still, I am happy to see that very different ways are possible (e.g., using binoharmonic series combined with some trivial identities). $\endgroup$ Commented Feb 24 at 19:14
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To add onto @User Zacky's Brilliant Answer!

We have, $$I=-2\int_0^1K'\left[\frac{\ln k}{1-k^2}\right]dk$$

Here we can use a well known result,

$$\int_0^1K'f(k)dk=\int_0^{\pi/2}\int_0^{\pi/2}f(\sin x\sin y)\ dxdy$$ to get,

$$I=-2\int_0^{\pi/2}\int_0^{\pi/2}\frac{\ln(\sin x\sin y)}{1-(\sin x\sin y)^2}\ dxdy$$

Which reduces to,

$$I=-2\pi\int_0^{\pi/2}\frac{\ln(\sin x)}{\cos x}dx=\frac{\pi^3}{4}$$

EDIT:

The "well known" result mentioned here is in a Paper by ML Glasser, Equation $(16)$ to be exact.

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  • $\begingroup$ Very good, this can of course be further used to generate other interesting integrals. However about that "well known" result, is it really well known? Do you have a link/citation of something that mentions it? $\endgroup$
    – Zacky
    Commented Feb 24 at 17:15
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    $\begingroup$ @Zacky I am sorry I shouldn't have used "well known" here. I saw it first in a Paper by ML Glasser which has a interesting derivation but I guess a easier derivation would be considering the Moments of $K'$ (I noticed both of these evaluated to the same but I might be running around in circles here) $$\int_0^1k^nK'dk=\int_0^{\pi/2}\int_0^{\pi/2}(\sin x\sin y)^ndxdy$$ $\endgroup$ Commented Feb 24 at 17:28
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Binoharmonic Series with the Squared Central Binomial Coefficient And Their Integral Transformation Using Elliptic Integrals by Cornel Ioan Valean

EDIT: While it is true that Theorem 3 is consisted of three points which also contain the separate derivation of the integrals $\int_0^1 \frac{K(x)\log(1-x)}{x}\textrm{d}x$ and $\int_0^1 \frac{K(x)\log(1+x)}{x}\textrm{d}x$, and then it is pointed out that combining the two ones we get the main integral, at the very end of the proof of the mentioned theorem it is stated that: It is very important to note that the result from the point i) can be extracted directly by using the same strategy as before, involving transformations with binoharmonic series, and then employing Theorem 1. The whole point here is that, indeed, we can reduce the calculation of the integral $\int_0^1\frac{K(x)}{x}\log\left(\frac{( 1+x)^3}{1-x}\right)\textrm{d}x$ to the binoharmonic series at Theorem 1, that is, $\displaystyle \sum_{n=1}^{\infty} \frac{1}{2^{4n}}\binom{2n}{n}^2\frac{4H_{2n}-3H_n}{n}$, where happily the proof doesn't take into account a splitting of the series into the twins mentioned in the OP and calculating them separately, and thus avoiding completely the work with polylogarithms (e.g., nice to see we avoid completely the appearance of the polylogarithm with a complex argument).

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    $\begingroup$ Do you have a way to prove (If it can be done easily then we can use it to solve the problem) $$\int_0^1\int_0^1\frac{\ln(2t)-\ln(\sqrt{1-x^2t^2})}{x\sqrt{1-t^2}\sqrt{1-x^2t^2}}dxdt=\frac{\pi^3}{48}$$ $\endgroup$ Commented Feb 24 at 7:50

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