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I encountered the integral $$\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx = -0.9393323982...$$ while researching the evaluation of harmonic sums. Mathematica 11 is not able to evaluate this integral, and is not able to evaluate the underlying indefinite integral.

I considered using the Maclaurin series for the expression $\sin ^{-1}(x)^2$ to evaluate this integral. Using this Maclaurin series, it is easily seen that the problem of evaluating the above integral is equivalent to the problem of evaluating the series $$\sum _{n=1}^{\infty } \frac{2^{2 n-1} H_{n-\frac{1}{2}}}{(1-2 n) n^2 \binom{2 n}{n}}=-0.9393323982...$$ which Mathematica 11 is not able to evaluate directly.

I also considered using the Maclaurin series for the expression $\log \left(1-x^2\right)$ to evaluate the above integral. Using this Macluarin series, it is easily seen that this integral is equal to $$\frac{1}{4} \left(16 \pi C-\pi ^3-\pi ^2 \log (4)\right)+\sum _{n=1}^{\infty } \frac{\, _3F_2\left(\frac{1}{2},\frac{1}{2},1;\frac{3}{2},n+1;1\right)}{n^2-2 n^3}=-0.939332...,$$ but Mathematica is unable to evaluate the above infinite series.

What is a closed-form evaluation of $\int_0^1 \frac{\log \left(1-x^2\right) \sin ^{-1}(x)^2}{x^2} \, dx$? What techniques can be applied to evaluate this definite integral?

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    $\begingroup$ Inverse symbolic calculators do not find anything. $\endgroup$ – Claude Leibovici Aug 23 '16 at 5:25
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    $\begingroup$ $$I=-\int_0^1 \int_0^1 \int_0^1 \int_0^1 \frac{x^2 dx~ dy~ dz~ dt}{(1-tx^2 )\sqrt{(1-y^2 x^2)(1-z^2 x^2)}}$$ looks ugly... $\endgroup$ – Yuriy S Aug 23 '16 at 10:32
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    $\begingroup$ $$\frac{\pi^3}{4} + \pi \ln^2 2 - \frac{\pi^2}{2}\, \ln2 + 8 \, G \, \ln2 + 7\, \zeta(3) - 16 \operatorname{Im} \text{Li}_3(1 + i ). $$ $G$ is Catalan's constant. $\endgroup$ – nospoon Aug 25 '16 at 12:22
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    $\begingroup$ @nospoon Thank you for your comment. Would you add this as an answer and briefly explain how you arrived at the expression given in your comment? Also, can the expression $\text{Im}\text{Li}_{3}(1+i)$ be simplified somehow? $\endgroup$ – John M. Campbell Aug 25 '16 at 17:25
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(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:

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


$\endgroup$
  • 1
    $\begingroup$ Could we by any chance rewrite that hypergeometric series in terms of polylogarithms? $\endgroup$ – Lucian Sep 10 '16 at 0:09
  • 1
    $\begingroup$ @Lucian Only if we resort to polylogarithms at imaginary arguments. See this problem. $\endgroup$ – David H Sep 10 '16 at 2:36
  • 1
    $\begingroup$ @Lucian Me neither, truthfully. It's just that at this point it's kinda subjective as to which representation is simpler. One reason to prefer the hypergeometric representation is that it is more flexible for finding equivalent representations. $\endgroup$ – David H Sep 10 '16 at 2:53
  • 1
    $\begingroup$ This answer deserves more upvotes. Excellent(!) work. $\endgroup$ – edmz Sep 7 '17 at 11:18
  • 1
    $\begingroup$ @black Thank you for your kind words. I put considerable effort into this answer in the hopes that it make for a good teaching tool, and it's always nice to be noticed :) $\endgroup$ – David H Sep 14 '17 at 2:22

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