Evaluate $\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$

closed-formdefinite integralshypergeometric functionintegrationpolylogarithm

I encountered a hypergeometric integral while investigating harmonic sums
$$\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx$$
Based on my experience I suspect a nice closed-form exists but have found none. Any kind of help will be appreciated.

Update: To complete the solution following @Jack D'Aurizio's derivation,

$1$. Let $uz\to u$ in expression $f(z)=\int_{0}^{1}\frac{\arcsin\sqrt{uz}}{\sqrt{uz(1-uz)}}\log(1-u)\,du$

$2$. Apply Fubini to $\int_0^1 f(4x(1-x))dx$, then it become $\int_0^1 du \int_{\frac{1-\sqrt{1-u}}2}^{\frac{1+\sqrt{1-u}}2}dx\cdots$

$3$. Integrate w.r.t $x$ by brute force, then let $u\to \frac{4t^2}{1+2t^2+t^4}$

$4$. These integrals are evaluated using method of arXiv $2007.03957$. Whence

$$-\frac{1}4 \sum _{n=1}^{\infty } \left(\frac{4^n}{\binom{2 n}{n}}\right)^2\frac{ H_n}{n^3}=\int_0^1 \log (1-x)\ _3F_2\left(1,1,1;\frac{3}{2},\frac{3}{2};x\right) \, dx=-8 C^2+8 \pi C \log (2)-32 \pi \Im(\text{Li}_3(1+i))-16 \text{Li}_4\left(\frac{1}{2}\right)+\frac{413 \pi ^4}{360}-\frac{2}{3} \log ^4(2)+\frac{8}{3} \pi ^2 \log ^2(2)$$

Best Answer

We have $$\begin{eqnarray*} I=\int_{0}^{1}\log(1-x)\;\phantom{}_3 F_2\left(1,1,1;\tfrac{3}{2},\tfrac{3}{2};x\right)\,dx &=& \sum_{n\geq 0}\int_{0}^{1}\left(\frac{4^n}{\binom{2n}{n}}\right)^2\frac{x^n \log(1-x)}{(2n+1)^2}\,dx\\&=&-\sum_{n\geq 0}\left(\frac{4^n}{(2n+1)\binom{2n}{n}}\right)^2\frac{H_{n+1}}{n+1}\end{eqnarray*}$$ and $\frac{4^n}{(2n+1)\binom{2n}{n}}= 4^n B(n+1,n+1)$, so the integral is given by $$ I = -\int_{0}^{1} \sum_{n\geq 0}\frac{4^n H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}}(4x(1-x))^n\,dx $$ and the problem boils down to computing $$ f(z)=\sum_{n\geq 0}\frac{4^n H_{n+1}}{(n+1)(2n+1)\binom{2n}{n}}z^n=- \int_{0}^{1}\frac{\arcsin\sqrt{uz}}{\sqrt{uz(1-uz)}}\log(1-u)\,du$$ leading to $$\begin{eqnarray*} I &=& \iint_{(0,1)^2}\frac{\arcsin\sqrt{4xu(1-x)}}{\sqrt{4ux(1-x)(1-4ux(1-x))}}\log(1-u)\,du\,dx\\&=&2\iint_{(0,1)^2}\frac{\arcsin(ux)\log(1-u^2)}{\sqrt{1-(ux)^2}\sqrt{1-x^2}}\,du\,dx \end{eqnarray*}$$ This is a polylogarithmic integral which is expected to break down into Euler sums with weight four.
I guess that a reasonable approach for finishing the computation is to enforce the substitutions $x=\sin\vartheta, u=\sin\varphi $ and then exploit well-known Fourier series to convert the double integral into a combination of Euler sums.

Related Solutions

[Math] Proving that $\int_0^1 \frac{\log \left(\frac{1}{t}\right) \log (t+2)}{t+1} \, dt=\frac{13}{24} \zeta (3)$

The last series Special Addendum in the OP looks quite interesting! (just an observation, not an answer)

We have by symmetry of the series:

$$S = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^k }{k2^k n 2^n (k+n)} = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(-1)^n}{k2^k n 2^n (k+n)}$$

Hence, $$\begin{align}\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{1+(-1)^k+(-1)^n+(-1)^{k+n} }{k2^k n 2^n (k+n)} &= \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{(1+(-1)^k)(1+(-1)^n)}{k2^k n 2^n (k+n)}\\&= \frac{1}{2}\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{1}{k4^k n 4^n (k+n)}\end{align}$$

Since, the term $(1+(-1)^k)(1+(-1)^n)$ survives only when both $k,n$ are even.

Since, $$\sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \frac{a^{n+k}}{kn(k+n)} = \sum _{k=1}^{\infty } \sum _{n=1}^{\infty } \int_0^1 \frac{a^{n+k}x^{n+k}}{kn}\frac{dx}{x} = \int_0^1 \frac{\log^2 (1-ax)}{x}\,dx$$

We have: $$2S = -\int_0^1 \frac{\log^2(1-\frac{x}{2})}{x}\,dx-\int_0^1 \frac{\log^2(1+\frac{x}{2})}{x}\,dx+\frac{1}{2}\int_0^1 \frac{\log^2(1-\frac{x}{4})}{x}\,dx$$

The integrals for $0 < a < 1$:

$\displaystyle \begin{align} \int_0^1 \frac{\log^2 (1-ax)}{x}\,dx &= \int_0^a \frac{\log^2 (1-x)}{x}\,dx \\&= \log a \log^2(1-a)+2\int_0^a \frac{\log x \log(1-x)}{1-x}\,dx\\&= \log a \log^2(1-a)+2\log (1-a) \operatorname{Li}_2(1-a) - 2\operatorname{Li}_3(1-a)+2\operatorname{Li}_3(1)\end{align}$

and the integrals for $0 < a < 1$:

$\displaystyle \begin{align} \int_0^1 \frac{\log^2 (1+ax)}{x}\,dx &= \int_0^{a} \frac{\log^2 (1+x)}{x}\,dx \\&= \log a \log^2(1+a)-2\int_0^a \frac{\log x \log(1+x)}{1+x}\,dx\\&= \log a \log^2 (1+a) - \frac{2}{3}\log^3 (1+a)-2\log (1+a)\operatorname{Li}_2\left(\frac{1}{1+a}\right)-2\operatorname{Li}_3\left(\frac{1}{1+a}\right)+2\operatorname{Li}_3(1)\end{align}$

Wolfram alpha computes these three integrals as:

[1] $$\displaystyle \int_0^1 \log^2 \left(1+\frac{x}{2}\right)\,\frac{dx}{x}=-2 \text{Li}_3\left(\frac{2}{3}\right)-\text{Li}_2\left(\frac{2}{3}\right) \log \left(\frac{9}{4}\right)+2 \zeta (3)\\+\frac{4}{3} \log (18) \coth ^{-1}(5)^2$$

[2] $$\int_0^1 \log^2\left(1-\frac{x}{2}\right)\,\frac{dx}{x} = \frac{1}{4}\zeta(3) - \frac{1}{3}\log^3 (2)$$

[3] $$\int_0^1 \log^2\left(1-\frac{x}{4}\right)\,\frac{dx}{x} = -2 \text{Li}_3\left(\frac{3}{4}\right)-\text{Li}_2\left(\frac{3}{4}\right) \log \left(\frac{16}{9}\right)+2 \zeta (3)\\-\log^2(3)\log (4)+\log^2(4)\log\left(\frac{9}{4}\right)$$

Integration – Solving ??¹ (log(1-x²) sin?¹(x)² / x²) dx

(Please forgive the length of what follows. I wanted my response to be accessible to as broad of an audience as possible.)

Part 1:

Our main objective is to evaluate the following definite integral:

$$\mathcal{I}:=\int_{0}^{1}\frac{\ln{\left(1-x^{2}\right)}\arcsin^{2}{\left(x\right)}}{x^{2}}\,\mathrm{d}x.\tag{1}$$

It will be shown that the value of $\mathcal{I}$ is

$$\mathcal{I}=7\,\zeta{\left(3\right)}-\frac{\pi^{2}}{2}\ln{\left(2\right)}-\frac{\pi^{3}}{2}+8\ln{\left(2\right)}\,G+4\,{_4F_3}{\left(\frac12,\frac12,1,1;\frac32,\frac32,\frac32;1\right)}.$$

For our purposes here, it will be convenient to define elementary transcendental functions such as the logarithm and inverse trigonometric functions via their usual integral representations. In particular, for real argument $x\in\mathbb{R}$ we have

$$\ln{\left(x\right)}:=\int_{x}^{1}\frac{\left(-1\right)}{t}\,\mathrm{d}t;~~~\small{x>0},$$

and

$$\arcsin{\left(x\right)}:=\int_{0}^{x}\frac{\mathrm{d}t}{\sqrt{1-t^{2}}};~~~\small{\left|x\right|\le1}.$$

Consider the following antiderivative:

$$\begin{align} \int\frac{\ln{\left(1-x^{2}\right)}}{x^{2}}\,\mathrm{d}x &=\int\frac{\left(-1\right)\operatorname{Li}_{1}{\left(x^{2}\right)}}{x^{2}}\,\mathrm{d}x\\ &=\frac{\operatorname{Li}_{1}{\left(x^{2}\right)}}{x}-\int\frac{2\operatorname{Li}_{0}{\left(x^{2}\right)}}{x^{2}}\,\mathrm{d}x;~~~\small{I.B.P.s}\\ &=\frac{\operatorname{Li}_{1}{\left(x^{2}\right)}}{x}-\int\frac{2}{1-x^{2}}\,\mathrm{d}x\\ &=\frac{\operatorname{Li}_{1}{\left(x^{2}\right)}}{x}-2\operatorname{arctanh}{\left(x\right)}+\color{grey}{constant}.\tag{2}\\ \end{align}$$

Using the above result, we begin to attack $\mathcal{I}$ with integration by parts. We obtain

$$\begin{align} \mathcal{I} &=\int_{0}^{1}\frac{\ln{\left(1-x^{2}\right)}\arcsin^{2}{\left(x\right)}}{x^{2}}\,\mathrm{d}x\\ &=\left[\left(\frac{\operatorname{Li}_{1}{\left(x^{2}\right)}}{x}-2\operatorname{arctanh}{\left(x\right)}\right)\arcsin^{2}{\left(x\right)}\right]_{x=0}^{x=1}\\ &~~~~~-\int_{0}^{1}\left[\frac{\operatorname{Li}_{1}{\left(x^{2}\right)}}{x}-2\operatorname{arctanh}{\left(x\right)}\right]\frac{2\arcsin{\left(x\right)}}{\sqrt{1-x^{2}}}\,\mathrm{d}x;~~~\small{I.B.P.s}\\ &=\arcsin^{2}{\left(1\right)}\lim_{x\to1}\left[-\frac{\left(1-x\right)\ln{\left(1-x\right)}+\left(1+x\right)\ln{\left(1+x\right)}}{x}\right]\\ &~~~~~+\int_{0}^{1}\frac{4\operatorname{arctanh}{\left(x\right)}\arcsin{\left(x\right)}}{\sqrt{1-x^{2}}}\,\mathrm{d}x-\int_{0}^{1}\frac{2\operatorname{Li}_{1}{\left(x^{2}\right)}\arcsin{\left(x\right)}}{x\sqrt{1-x^{2}}}\,\mathrm{d}x\\ &=-\frac{\pi^{2}}{2}\ln{\left(2\right)}\\ &~~~~~+\int_{0}^{1}\frac{2\left[\arcsin^{2}{\left(1\right)}-\arcsin^{2}{\left(x\right)}\right]}{1-x^{2}}\,\mathrm{d}x;~~~\small{I.B.P.s}\\ &~~~~~+\int_{0}^{1}\frac{2\ln{\left(1-x^{2}\right)}\arcsin{\left(x\right)}}{x\sqrt{1-x^{2}}}\,\mathrm{d}x\\ &=-\frac{\pi^{2}}{2}\ln{\left(2\right)}\\ &~~~~~+\int_{0}^{1}\frac{\left[\arcsin^{2}{\left(1\right)}-\arcsin^{2}{\left(\frac{1-y}{1+y}\right)}\right]}{y}\,\mathrm{d}y;~~~\small{\left[\frac{1-x}{1+x}=y\right]}\\ &~~~~~+\int_{0}^{\frac{\pi}{2}}\frac{2\theta\ln{\left(\cos^{2}{\left(\theta\right)}\right)}}{\sin{\left(\theta\right)}}\,\mathrm{d}\theta;~~~\small{\left[\arcsin{\left(x\right)}=\theta\right]}\\ &=-\frac{\pi^{2}}{2}\ln{\left(2\right)}+\int_{0}^{1}\frac{\frac{\pi^{2}}{4}-\arcsin^{2}{\left(\frac{1-y}{1+y}\right)}}{y}\,\mathrm{d}y\\ &~~~~~-4\int_{0}^{\frac{\pi}{2}}\theta\csc{\left(\theta\right)}\ln{\left(\sec{\left(\theta\right)}\right)}\,\mathrm{d}\theta.\tag{3}\\ \end{align}$$

Let $\mathcal{J}$ denote the final log-trig integral in the last line above:

$$\mathcal{J}:=\int_{0}^{\frac{\pi}{2}}\theta\csc{\left(\theta\right)}\ln{\left(\sec{\left(\theta\right)}\right)}\,\mathrm{d}\theta.\tag{4}$$

The integral $\mathcal{J}$ is by far the more difficult of the two remaining integrals we need to evaluate, so let's save that for last.

Now, the following trigonometric identity may be readily verified:

$$\arcsin{\left(\frac{1-y}{1+y}\right)}=\frac{\pi}{2}-2\arctan{\left(\sqrt{y}\right)};~~~\small{y\ge0}.\tag{5}$$

Continuing we our main calculation,

$$\begin{align} \mathcal{I} &=\small{-\frac{\pi^{2}}{2}\ln{\left(2\right)}+\int_{0}^{1}\frac{\frac{\pi^{2}}{4}-\arcsin^{2}{\left(\frac{1-y}{1+y}\right)}}{y}\,\mathrm{d}y-4\int_{0}^{\frac{\pi}{2}}\theta\csc{\left(\theta\right)}\ln{\left(\sec{\left(\theta\right)}\right)}\,\mathrm{d}\theta}\\ &=-4\,\mathcal{J}-\frac{\pi^{2}}{2}\ln{\left(2\right)}+\int_{0}^{1}\frac{\frac{\pi^{2}}{4}-\left[\frac{\pi}{2}-2\arctan{\left(\sqrt{y}\right)}\right]^{2}}{y}\,\mathrm{d}y\\ &=-4\,\mathcal{J}-\frac{\pi^{2}}{2}\ln{\left(2\right)}+\int_{0}^{1}\frac{4\arctan{\left(\sqrt{y}\right)}\left[\frac{\pi}{2}-\arctan{\left(\sqrt{y}\right)}\right]}{y}\,\mathrm{d}y\\ &=-4\,\mathcal{J}-\frac{\pi^{2}}{2}\ln{\left(2\right)}+\int_{0}^{1}\frac{2\pi\arctan{\left(\sqrt{y}\right)}-4\arctan^{2}{\left(\sqrt{y}\right)}}{y}\,\mathrm{d}y\\ &=-4\,\mathcal{J}-\frac{\pi^{2}}{2}\ln{\left(2\right)}+\int_{0}^{1}\frac{4\pi\arctan{\left(x\right)}-8\arctan^{2}{\left(x\right)}}{x}\,\mathrm{d}x;~~~\small{\left[\sqrt{y}=x\right]}\\ &=-\frac{\pi^{2}}{2}\ln{\left(2\right)}+4\pi\int_{0}^{1}\frac{\arctan{\left(x\right)}}{x}\,\mathrm{d}x-8\int_{0}^{1}\frac{\arctan^{2}{\left(x\right)}}{x}\,\mathrm{d}x-4\,\mathcal{J}\\ &=-\frac{\pi^{2}}{2}\ln{\left(2\right)}+4\pi\operatorname{Ti}_{2}{\left(1\right)}-\int_{0}^{1}\frac{8\arctan^{2}{\left(x\right)}}{x}\,\mathrm{d}x-4\,\mathcal{J},\\ \end{align}$$

where $\operatorname{Ti}_{2}{\left(z\right)}$ is an auxiliary polylogarithmic function known as the inverse tangent integral. It's value at unity is simply Catalan's constant, which I'll be denoting by $G$.

As for the integral $\int_{0}^{1}\frac{8\arctan^{2}{\left(x\right)}}{x}\,\mathrm{d}x$, it is shown below in the first appendix that

$$\int_{0}^{1}\frac{8\arctan^{2}{\left(x\right)}}{x}\,\mathrm{d}x=4\pi\,G-7\,\zeta{\left(3\right)}.$$

Then,

$$\begin{align} \mathcal{I} &=-\frac{\pi^{2}}{2}\ln{\left(2\right)}+4\pi\operatorname{Ti}_{2}{\left(1\right)}-\int_{0}^{1}\frac{8\arctan^{2}{\left(x\right)}}{x}\,\mathrm{d}x-4\,\mathcal{J}\\ &=-\frac{\pi^{2}}{2}\ln{\left(2\right)}+4\pi\,G-\left[4\pi\,G-7\,\zeta{\left(3\right)}\right]-4\,\mathcal{J}\\ &=7\,\zeta{\left(3\right)}-3\ln{\left(2\right)}\,\zeta{\left(2\right)}-4\,\mathcal{J}.\tag{6}\\ \end{align}$$

Part 2:

Assuming $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$, the following definite integral may be evaluated in terms of the Legendre chi function:

$$\begin{align} \int_{0}^{\theta}\csc{\left(\varphi\right)}\ln{\left(\sec{\left(\varphi\right)}\right)}\,\mathrm{d}\varphi &=\int_{0}^{\tan{\left(\frac{\theta}{2}\right)}}t^{-1}\ln{\left(\frac{1+t^{2}}{1-t^{2}}\right)}\,\mathrm{d}t;~~~\small{\left[\tan{\left(\frac{\varphi}{2}\right)}=t\right]}\\ &=\int_{0}^{\tan^{2}{\left(\frac{\theta}{2}\right)}}\frac{\ln{\left(\frac{1+x}{1-x}\right)}}{2x}\,\mathrm{d}x;~~~\small{\left[t^{2}=x\right]}\\ &=\int_{0}^{\tan^{2}{\left(\frac{\theta}{2}\right)}}\frac{\operatorname{arctanh}{\left(x\right)}}{x}\,\mathrm{d}x\\ &=\chi_{2}{\left(\tan^{2}{\left(\frac{\theta}{2}\right)}\right)}.\\ \end{align}$$

The Legendre chi function evaluated at unity takes the special value,

$$\chi_{2}{\left(1\right)}=\frac34\,\zeta{\left(2\right)}=\frac{\pi^{2}}{8}.$$

Using the above integral to integrate $\mathcal{J}$ by parts, we find

$$\begin{align} \mathcal{J} &=\int_{0}^{\frac{\pi}{2}}\theta\csc{\left(\theta\right)}\ln{\left(\sec{\left(\theta\right)}\right)}\,\mathrm{d}\theta\\ &=\lim_{\theta\to\frac{\pi}{2}}\theta\int_{0}^{\theta}\csc{\left(\varphi\right)}\ln{\left(\sec{\left(\varphi\right)}\right)}\,\mathrm{d}\varphi-\int_{0}^{\frac{\pi}{2}}\mathrm{d}\theta\int_{0}^{\theta}\mathrm{d}\varphi\,\csc{\left(\varphi\right)}\ln{\left(\sec{\left(\varphi\right)}\right)}\\ &=\lim_{\theta\to\frac{\pi}{2}}\theta\,\chi_{2}{\left(\tan^{2}{\left(\frac{\theta}{2}\right)}\right)}-\int_{0}^{\frac{\pi}{2}}\chi_{2}{\left(\tan^{2}{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta\\ &=\frac{\pi}{2}\,\chi_{2}{\left(\tan^{2}{\left(\frac{\pi}{4}\right)}\right)}-\int_{0}^{1}\frac{2\,\chi_{2}{\left(t^{2}\right)}}{1+t^{2}}\,\mathrm{d}t;~~~\small{\left[\tan{\left(\frac{\theta}{2}\right)}=t\right]}\\ &=\frac{\pi}{2}\,\chi_{2}{\left(1\right)}-\int_{0}^{1}\frac{\chi_{2}{\left(u\right)}}{\left(1+u\right)\sqrt{u}}\,\mathrm{d}u;~~~\small{\left[t^{2}=u\right]}\\ &=\frac{\pi}{2}\,\chi_{2}{\left(1\right)}-\int_{0}^{1}\frac{\chi_{2}{\left(\frac{1-x}{1+x}\right)}}{\sqrt{1-x^{2}}}\,\mathrm{d}x;~~~\small{\left[u=\frac{1-x}{1+x}\right]}\\ &=\frac{\pi}{2}\,\chi_{2}{\left(1\right)}-\int_{0}^{1}\frac{\chi_{2}{\left(1\right)}+\ln{\left(x\right)}\operatorname{arctanh}{\left(x\right)}-\chi_{2}{\left(x\right)}}{\sqrt{1-x^{2}}}\,\mathrm{d}x\\ &=\frac{\pi}{2}\,\chi_{2}{\left(1\right)}-\chi_{2}{\left(1\right)}\int_{0}^{1}\frac{\mathrm{d}x}{\sqrt{1-x^{2}}}-\int_{0}^{1}\frac{\ln{\left(x\right)}\operatorname{arctanh}{\left(x\right)}}{\sqrt{1-x^{2}}}\,\mathrm{d}x+\int_{0}^{1}\frac{\chi_{2}{\left(x\right)}}{\sqrt{1-x^{2}}}\,\mathrm{d}x\\ &=\frac{\pi}{2}\,\chi_{2}{\left(1\right)}-\frac{\pi}{2}\,\chi_{2}{\left(1\right)}\\ &~~~~~+\int_{0}^{1}\frac{\operatorname{arctanh}{\left(x\right)}\arcsin{\left(x\right)}}{x}\,\mathrm{d}x+\int_{0}^{1}\frac{\ln{\left(x\right)}\arcsin{\left(x\right)}}{1-x^{2}}\,\mathrm{d}x;~~~\small{I.B.P.s}\\ &~~~~~+\int_{0}^{1}\frac{\chi_{2}{\left(x\right)}}{\sqrt{1-x^{2}}}\,\mathrm{d}x\\ &=\int_{0}^{1}\left[\frac{\operatorname{arctanh}{\left(x\right)}\arcsin{\left(x\right)}}{x}+\frac{\chi_{2}{\left(x\right)}}{\sqrt{1-x^{2}}}\right]\,\mathrm{d}x+\int_{0}^{1}\frac{\ln{\left(x\right)}\arcsin{\left(x\right)}}{1-x^{2}}\,\mathrm{d}x\\ &=\frac{\pi}{2}\,\chi_{2}{\left(1\right)}-\int_{0}^{1}\frac{\ln{\left(\frac{1}{x}\right)}\arcsin{\left(x\right)}}{1-x^{2}}\,\mathrm{d}x\\ &=:\frac{\pi}{2}\,\chi_{2}{\left(1\right)}-\mathcal{K},\\ \end{align}$$

where in the last line above we've implicitly defined $\mathcal{K}$ to denote the integral

$$\mathcal{K}:=\int_{0}^{1}\frac{\ln{\left(\frac{1}{x}\right)}\arcsin{\left(x\right)}}{1-x^{2}}\,\mathrm{d}x.$$

Part 3:

We now find a closed form evaluation for the integral $\mathcal{K}$ in terms of generalized hypergeometrics. To begin, we can reduce $\mathcal{K}$ to a triple integral of a simple (relatively speaking) algebraic integrand over the unit cube:

$$\begin{align} \mathcal{K} &=\int_{0}^{1}\frac{\ln{\left(\frac{1}{x}\right)}\arcsin{\left(x\right)}}{1-x^{2}}\,\mathrm{d}x\\ &=-\int_{0}^{1}\frac{\ln{\left(x\right)}\arcsin{\left(x\right)}}{1-x^{2}}\,\mathrm{d}x\\ &=-\int_{0}^{1}\mathrm{d}x\,\frac{\arcsin{\left(x\right)}}{\left(1-x^{2}\right)}\int_{x}^{1}\mathrm{d}z\,\frac{\left(-1\right)}{z}\\ &=\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\mathrm{d}z\,\frac{\arcsin{\left(x\right)}}{\left(1-x^{2}\right)z}\\ &=\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\mathrm{d}z\,\frac{1}{\left(1-x^{2}\right)z}\int_{0}^{x}\mathrm{d}y\,\frac{1}{\sqrt{1-y^{2}}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{x}^{1}\mathrm{d}z\int_{0}^{x}\mathrm{d}y\,\frac{1}{\left(1-x^{2}\right)z\sqrt{1-y^{2}}}\\ &=\int_{0}^{1}\mathrm{d}z\int_{0}^{z}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{1}{z\left(1-x^{2}\right)\sqrt{1-y^{2}}}\\ &=\int_{0}^{1}\mathrm{d}z\int_{0}^{1}\mathrm{d}t\int_{0}^{zt}\mathrm{d}y\,\frac{1}{\left(1-z^{2}t^{2}\right)\sqrt{1-y^{2}}};~~~\small{\left[x=zt\right]}\\ &=\int_{0}^{1}\mathrm{d}z\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\,\frac{zt}{\left(1-z^{2}t^{2}\right)\sqrt{1-z^{2}t^{2}u^{2}}};~~~\small{\left[y=ztu\right]}\\ &=\frac12\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\,\frac{t}{\left(1-xt^{2}\right)\sqrt{1-xt^{2}u^{2}}};~~~\small{\left[z^{2}=x\right]}\\ &=\frac14\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}u\,\frac{1}{\left(1-xy\right)\sqrt{1-xyu^{2}}};~~~\small{\left[t^{2}=y\right]}\\ &=\frac18\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}y\int_{0}^{1}\mathrm{d}z\,\frac{1}{\left(1-xy\right)\sqrt{z\left(1-xyz\right)}};~~~\small{\left[u^{2}=z\right]}\\ &=\frac18\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}z\int_{0}^{1}\mathrm{d}y\,\frac{1}{\left(1-xy\right)\sqrt{z\left(1-zxy\right)}}\\ &=\frac18\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}z\,\frac{1}{\sqrt{z}}\int_{0}^{1}\frac{\mathrm{d}y}{\left(1-xy\right)\sqrt{1-zxy}}.\\ \end{align}$$

Using the machinery of hypergeometric functions to solve the integrals, we arrive at

$$\begin{align} \mathcal{K} &=\frac18\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}z\,\frac{1}{\sqrt{z}}\int_{0}^{1}\frac{\mathrm{d}y}{\left(1-xy\right)\sqrt{1-zxy}}\\ &=\frac18\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}z\,\frac{1}{\sqrt{z}}\,F_{1}{\left(1;1,\frac12;2;x,zx\right)}\\ &=\small{\int_{0}^{1}\frac{\mathrm{d}x}{4x}\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{z\left(1-xz\right)}}\left[{_2F_1}{\left(1,\frac12;\frac32;\frac{1-z}{1-xz}\right)}-\sqrt{1-xz}\,{_2F_1}{\left(1,\frac12;\frac32;1-z\right)}\right]}\\ &=\small{\frac14\int_{0}^{1}\frac{\mathrm{d}x}{x}\,\left[\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{z\left(1-xz\right)}}\,{_2F_1}{\left(1,\frac12;\frac32;\frac{1-z}{1-xz}\right)}-\int_{0}^{1}\frac{\mathrm{d}z}{\sqrt{z}}\,{_2F_1}{\left(1,\frac12;\frac32;1-z\right)}\right]}\\ &=\frac14\int_{0}^{1}\frac{\mathrm{d}x}{x}\,\bigg{[}\int_{0}^{1}\mathrm{d}w\,\frac{\sqrt{1-x}}{\left(1-xw\right)\sqrt{1-w}}\,{_2F_1}{\left(1,\frac12;\frac32;w\right)};~~~\small{\left[\frac{1-z}{1-xz}=w\right]}\\ &~~~~~-\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t}}\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\bigg{]};~~~\small{\left[z=1-t\right]}\\ &=\frac14\int_{0}^{1}\frac{\mathrm{d}x}{x}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t}}\,\left[\frac{\sqrt{1-x}}{\left(1-xt\right)}-1\right]\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\\ &=\frac14\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t}}\int_{0}^{1}\mathrm{d}x\,\left[\frac{1-x}{x\left(1-tx\right)\sqrt{1-x}}-\frac{1}{x}\right]\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\\ &=\frac14\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t}}\int_{0}^{1}\mathrm{d}x\,\left[\frac{1}{x\sqrt{1-x}}-\frac{1-t}{\left(1-tx\right)\sqrt{1-x}}-\frac{1}{x}\right]\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\\ &=\frac14\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t}}\,\left[\int_{0}^{1}\mathrm{d}x\,\frac{1-\sqrt{1-x}}{x\sqrt{1-x}}-\int_{0}^{1}\mathrm{d}x\,\frac{1-t}{\left(1-tx\right)\sqrt{1-x}}\right]\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\\ &=\frac14\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t}}\,\left[\int_{0}^{1}\mathrm{d}y\,\frac{2}{2-y}-2\left(1-t\right)\int_{0}^{1}\frac{\mathrm{d}x}{2\left(1-tx\right)\sqrt{1-x}}\right]\,\\ &~~~~~\times{_2F_1}{\left(1,\frac12;\frac32;t\right)};~~~\small{\left[1-\sqrt{1-x}=y\right]}\\ &=\frac12\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t}}\,\left[\int_{0}^{1}\frac{\mathrm{d}y}{2-y}-\left(1-t\right)\,{_2F_1}{\left(1,1;\frac32;t\right)}\right]\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\\ &=\frac12\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t}}\,\left[\ln{\left(2\right)}-\left(1-t\right)\,{_2F_1}{\left(1,1;\frac32;t\right)}\right]\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\\ &=\frac12\ln{\left(2\right)}\int_{0}^{1}\frac{\mathrm{d}t}{\sqrt{1-t}}\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\\ &~~~~~-\frac12\int_{0}^{1}\mathrm{d}t\,\sqrt{1-t}\,{_2F_1}{\left(1,\frac12;\frac32;t\right)}\,{_2F_1}{\left(1,1;\frac32;t\right)}\\ &=\ln{\left(2\right)}\,{_3F_2}{\left(1,\frac12,1;\frac32,\frac32;1\right)}-\frac12\int_{0}^{1}\mathrm{d}t\,{_2F_1}{\left(\frac12,1;\frac32;t\right)}\,{_2F_1}{\left(\frac12,\frac12;\frac32;t\right)}\\ &=\ln{\left(2\right)}\,{_3F_2}{\left(\frac12,1,1;\frac32,\frac32;1\right)}-\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\operatorname{arctanh}{\left(\sqrt{t}\right)}\arcsin{\left(\sqrt{t}\right)}}{t}\\ &=\ln{\left(2\right)}\,{_3F_2}{\left(\frac12,1,1;\frac32,\frac32;1\right)}-\int_{0}^{1}\mathrm{d}x\,\frac{\operatorname{arctanh}{\left(x\right)}\arcsin{\left(x\right)}}{x};~~~\small{\left[\sqrt{t}=x\right]}\\ &=\ln{\left(2\right)}\,{_3F_2}{\left(\frac12,1,1;\frac32,\frac32;1\right)}-\frac{\pi}{2}\,\chi_{2}{\left(1\right)}+\int_{0}^{1}\mathrm{d}x\,\frac{\chi_{2}{\left(x\right)}}{\sqrt{1-x^{2}}};~~~\small{I.B.P.s}\\ &=\ln{\left(2\right)}\,{_3F_2}{\left(\frac12,1,1;\frac32,\frac32;1\right)}-\frac{\pi}{2}\,\chi_{2}{\left(1\right)}+\int_{0}^{1}\mathrm{d}t\,\frac{\chi_{2}{\left(\sqrt{t}\right)}}{2\sqrt{t\left(1-t\right)}};~~~\small{\left[x=\sqrt{t}\right]}\\ &=\ln{\left(2\right)}\,{_3F_2}{\left(\frac12,1,1;\frac32,\frac32;1\right)}-\frac{\pi}{2}\,\chi_{2}{\left(1\right)}+\int_{0}^{1}\mathrm{d}t\,\frac{1}{2\sqrt{1-t}}\,{_3F_2}{\left(\frac12,\frac12,1;\frac32,\frac32;t\right)}\\ &=-\frac{\pi}{2}\,\chi_{2}{\left(1\right)}+\ln{\left(2\right)}\,{_3F_2}{\left(\frac12,1,1;\frac32,\frac32;1\right)}+{_4F_3}{\left(\frac12,\frac12,1,1;\frac32,\frac32,\frac32;1\right)}.\tag{7}\\ \end{align}$$

The ${_3F_2}$ turns out to be simply an integer multiple of Catalan's constant. It's value is derived in the second appendix below.

Thus,

$$\begin{align} \mathcal{J} &=\frac{\pi}{2}\,\chi_{2}{\left(1\right)}-\mathcal{K}\\ &=\pi\,\chi_{2}{\left(1\right)}-\ln{\left(2\right)}\,{_3F_2}{\left(\frac12,1,1;\frac32,\frac32;1\right)}-{_4F_3}{\left(\frac12,\frac12,1,1;\frac32,\frac32,\frac32;1\right)}\\ &=\pi\,\chi_{2}{\left(1\right)}-2\ln{\left(2\right)}\,G-{_4F_3}{\left(\frac12,\frac12,1,1;\frac32,\frac32,\frac32;1\right)}.\\ \end{align}$$

Finally,

$$\begin{align} \mathcal{I} &=7\,\zeta{\left(3\right)}-3\ln{\left(2\right)}\,\zeta{\left(2\right)}-4\,\mathcal{J}\\ &=7\,\zeta{\left(3\right)}-\frac{\pi^{2}}{2}\ln{\left(2\right)}-4\left[\pi\,\chi_{2}{\left(1\right)}-2\ln{\left(2\right)}\,G-{_4F_3}{\left(\frac12,\frac12,1,1;\frac32,\frac32,\frac32;1\right)}\right]\\ &=7\,\zeta{\left(3\right)}-\frac{\pi^{2}}{2}\ln{\left(2\right)}-\frac{\pi^{3}}{2}+8\ln{\left(2\right)}\,G+4\,{_4F_3}{\left(\frac12,\frac12,1,1;\frac32,\frac32,\frac32;1\right)}.\blacksquare\\ \end{align}$$

Appendix 1:

$$\begin{align} \int_{0}^{1}\frac{8\arctan^{2}{\left(x\right)}}{x}\,\mathrm{d}x &=\int_{0}^{\frac{\pi}{2}}\theta^{2}\cot{\left(\frac{\theta}{2}\right)}\sec^{2}{\left(\frac{\theta}{2}\right)}\,\mathrm{d}\theta;~~~\small{\left[2\arctan{\left(x\right)}=\theta\right]}\\ &=\int_{0}^{\frac{\pi}{2}}\frac{2\theta^{2}}{2\sin{\left(\frac{\theta}{2}\right)}\cos{\left(\frac{\theta}{2}\right)}}\,\mathrm{d}\theta\\ &=\int_{0}^{\frac{\pi}{2}}\frac{2\theta^{2}}{\sin{\left(\theta\right)}}\,\mathrm{d}\theta\\ &=-\int_{0}^{\frac{\pi}{2}}4\theta\ln{\left(\tan{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta\\ &=-\int_{0}^{\frac{\pi}{2}}4\theta\ln{\left(\sin{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta+\int_{0}^{\frac{\pi}{2}}4\theta\ln{\left(\cos{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta\\ &=-\int_{0}^{\frac{\pi}{2}}4\theta\ln{\left(\sin{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta+\int_{\frac{\pi}{2}}^{\pi}4\left(\pi-\theta\right)\ln{\left(\sin{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta\\ &=-\int_{0}^{\frac{\pi}{2}}4\theta\ln{\left(\sin{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta-\int_{\frac{\pi}{2}}^{\pi}4\theta\ln{\left(\sin{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta\\ &~~~~~+\int_{\frac{\pi}{2}}^{\pi}4\pi\ln{\left(\sin{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta\\ &=-4\int_{0}^{\pi}\theta\ln{\left(\sin{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta+4\pi\int_{\frac{\pi}{2}}^{\pi}\ln{\left(\sin{\left(\frac{\theta}{2}\right)}\right)}\,\mathrm{d}\theta\\ &=4\ln{\left(2\right)}\int_{0}^{\pi}\theta\,\mathrm{d}\theta+4\int_{0}^{\pi}\theta\left[-\ln{\left(2\sin{\left(\frac{\theta}{2}\right)}\right)}\right]\,\mathrm{d}\theta\\ &~~~~~-4\pi\ln{\left(2\right)}\int_{\frac{\pi}{2}}^{\pi}\mathrm{d}\theta-4\pi\int_{\frac{\pi}{2}}^{\pi}\left[-\ln{\left(2\sin{\left(\frac{\theta}{2}\right)}\right)}\right]\,\mathrm{d}\theta\\ &=2\pi^{2}\ln{\left(2\right)}-2\pi^{2}\ln{\left(2\right)}-4\pi\left[\operatorname{Cl}_{2}{\left(\pi\right)}-\operatorname{Cl}_{2}{\left(\frac{\pi}{2}\right)}\right]\\ &~~~~~-4\int_{0}^{\pi}\operatorname{Cl}_{2}{\left(\theta\right)}\,\mathrm{d}\theta;~~~\small{I.B.P.s}\\ &=4\operatorname{Cl}_{3}{\left(\pi\right)}+4\pi\operatorname{Cl}_{2}{\left(\frac{\pi}{2}\right)}-4\,\zeta{\left(3\right)}\\ &=4\pi\,G-7\,\zeta{\left(3\right)}.\blacksquare\\ \end{align}$$

Appendix 2:

$$\begin{align} {_3F_2}{\left(\frac12,1,1;\frac32,\frac32;1\right)} &={_3F_2}{\left(1,1,\frac12;\frac32,\frac32;1\right)}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,t^{-\frac12}\,{_2F_1}{\left(1,1;\frac32;t\right)}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{1}{\sqrt{t\left(1-t\right)}}\,{_2F_1}{\left(\frac12,\frac12;\frac32;t\right)}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,\frac{\arcsin{\left(\sqrt{t}\right)}}{t\sqrt{1-t}}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{\arcsin{\left(u\right)}}{u\sqrt{1-u^{2}}};~~~\small{\left[t=u^{2}\right]}\\ &=\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\frac{\varphi}{\sin{\left(\varphi\right)}};~~~\small{\left[\arcsin{\left(u\right)}=\varphi\right]}\\ &=-\int_{0}^{\frac{\pi}{2}}\mathrm{d}\varphi\,\ln{\left(\tan{\left(\frac{\varphi}{2}\right)}\right)};~~~\small{I.B.P.s}\\ &=-\int_{0}^{1}\mathrm{d}t\,\frac{2\ln{\left(t\right)}}{1+t^{2}};~~~\small{\left[\tan{\left(\frac{\varphi}{2}\right)}=t\right]}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{2\arctan{\left(t\right)}}{t};~~~\small{I.B.P.s}\\ &=2\operatorname{Ti}_{2}{\left(1\right)}\\ &=2\,G.\blacksquare\\ \end{align}$$

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