Prove $\int_{0}^{1}\frac1k K(k)\ln\left[\frac{\left(1+k \right)^3}{1-k} \right]\text{d}k=\frac{\pi^3}{4}$

Question

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.

Answer 1

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).

Answer 2

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

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

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

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

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

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

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

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

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

Answer 3

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.