¿Qué es? $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$ ?

Question

Me encontré con la integral $$\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx = -0.9393323982...$$ mientras investigaba la evaluación de sumas armónicas. Mathematica 11 no es capaz de evaluar esta integral, y no es capaz de evaluar la integral indefinida subyacente.

He considerado utilizar la serie Maclaurin para la expresión $\sin ^{-1}(x)^2$ para evaluar esta integral. Usando esta serie de Maclaurin, se ve fácilmente que el problema de evaluar la integral anterior es equivalente al problema de evaluar la serie $$\sum _{n=1}^{\infty } \frac{2^{2 n-1} H_{n-\frac{1}{2}}}{(1-2 n) n^2 \binom{2 n}{n}}=-0.9393323982...$$ que Mathematica 11 no es capaz de evaluar directamente.

También consideré utilizar la serie Maclaurin para la expresión $\log \left(1-x^2\right)$ para evaluar la integral anterior. Utilizando esta serie de Macluarin, se ve fácilmente que esta integral es igual a $$\frac{1}{4} \left(16 \pi C-\pi ^3-\pi ^2 \log (4)\right)+\sum _{n=1}^{\infty } \frac{\, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},n+1;1\right)}{n^2-2 n^3}=-0.939332...,$$ pero Mathematica no puede evaluar la serie infinita anterior.

¿Qué es una evaluación de forma cerrada de $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$ ? ¿Qué técnicas se pueden aplicar para evaluar esta integral definida?

Answer 1

(Por favor, perdonen la longitud de lo que sigue. Quería que mi respuesta fuera accesible para un público lo más amplio posible).

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Parte 1:

Nuestro objetivo principal es evaluar la siguiente integral definida:

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

Se demostrará que el valor de $\mathcal{I}$ es

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

Para nuestros propósitos aquí, será conveniente definir funciones trascendentales elementales como el logaritmo y las funciones trigonométricas inversas a través de sus representaciones integrales habituales. En particular, para un argumento real $x\in\mathbb{R}$ tenemos

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

y

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

Consideremos la siguiente antiderivada:

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

Utilizando el resultado anterior, comenzamos a atacar $\mathcal{I}$ con integración por partes. Obtenemos

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

Dejemos que $\mathcal{J}$ denotan la integral log-trig final en la última línea anterior:

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

La integral $\mathcal{J}$ es con mucho la más difícil de las dos integrales restantes que tenemos que evaluar, así que vamos a dejarla para el final.

Ahora, la siguiente identidad trigonométrica puede ser fácilmente verificada:

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

Continuando con nuestro cálculo principal,

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

donde $\operatorname{Ti}_{2}{\left(z\right)}$ es una función polilogarítmica auxiliar conocida como integral tangente inversa . Su valor en la unidad es simplemente la constante de Catalan, que denotaré por $G$ .

En cuanto a la integral $\int_{0}^{1}\frac{8\arctan^{2}{\left(x\right)}}{x}\,\mathrm{d}x$ En el primer apéndice se muestra que

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

Entonces,

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

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Segunda parte:

Suponiendo que $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$ la siguiente integral definida puede ser evaluada en términos de la Función chi de Legendre :

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

La función chi de Legendre evaluada en la unidad toma el valor especial,

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

Utilizando la integral anterior para integrar $\mathcal{J}$ por partes, encontramos

$$\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}$$

donde en la última línea anterior hemos definido implícitamente $\mathcal{K}$ para denotar la integral

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

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Parte 3:

Ahora encontramos una evaluación de forma cerrada para la integral $\mathcal{K}$ en términos de hipergeometría generalizada. Para empezar, podemos reducir $\mathcal{K}$ a una integral triple de un integrando algebraico simple (relativamente hablando) sobre el cubo unitario:

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

Utilizando la maquinaria de las funciones hipergeométricas para resolver las integrales, llegamos a

$$\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}$$

El ${_3F_2}$ resulta ser simplemente un múltiplo entero de la constante de Catalan. Su valor se deduce en el segundo apéndice.

Así,

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

Finalmente,

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

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Anexo 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}$$

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Anexo 2:

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

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Answer 2

Como adición a la respuesta anterior, se puede demostrar (a través de la expansión FL de $\frac{\arcsin\sqrt{x}}{\sqrt{x}}$ ) que

$$\phantom{}_4 F_3\left(\tfrac{1}{2},\tfrac{1}{2},1,1;\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2};1\right)=4\text{ Im } \text{Li}_3\left(\frac{1+i}{2}\right)-\frac{\pi^3}{32}-\frac{\pi}{8}\log^2(2).$$