Definite Integral – Evaluating $\int_0^1 \int_0^1 \frac{\frac{\pi}{4} dx dy}{\sin(\frac{\pi}{4}(1+x+y))}$

definite integralsintegration

I'm hoping to evaluate the definite integral $$I = \int_0^1 \int_0^1 \frac{\frac{\pi}{4} dx dy}{\sin(\frac{\pi}{4}(1+x+y))}$$

This integral arose in an attempt to evaluate $\int_{0}^\infty \left( \frac{x-1}{\log(x)} \right)^2 \frac{dx}{1+x^4}$; these integrals are equal and I will accept an evaluation of either.

It is straightforward to directly perform the integral on one of the coordinates via the antiderivative $\int \csc(x) dx = -\log\left( |\cot(x) + \csc(x)|\right)$. I suspect that there may not be a nice evaluation of the integral on the remaining coordinate, but I'm still hopeful.

Spelling out the integration over the coordinate $x$, we have

$$
\begin{split}
I &= \int_0^1 dy [ -\log(\cot(\frac{\pi}{4}(1+x+y)+\csc(\frac{\pi}{4}(1+x+y)) )] \bigg|_{x=0}^{x=1} \\
&= \int_{0}^1 dy \log \left(\frac{\cot(\frac{\pi}{4}(1+y)) +\csc(\frac{\pi}{4}(1+y)) }{\cot(\frac{\pi}{4}(2+y)) +\csc(\frac{\pi}{4}(2+y))} \right)
\end{split}$$

Best Answer

Make the substitution $u = \frac{x+y}2$ and $v=\frac{x-y}2$. $$I=\frac{\pi}{2}\int_{0}^{\frac{1}{2}}\int_{-u}^u \frac{1}{\sin(\frac\pi4 (1+2u))}dvdu+\frac{\pi}{2}\int_{\frac12}^{1}\int_{u-1}^{1-u} \frac{1}{\sin(\frac\pi4 (1+2u))}dvdu$$ Performing the innermost integral is now trivial. $$I=\pi\int_{0}^{\frac{1}{2}} \frac{u}{\sin(\frac\pi4 (1+2u))}du+\pi\int_{\frac12}^{1} \frac{1-u}{\sin(\frac\pi4 (1+2u))}du$$ Now, we substitute $u\to 1-t$ on the second integral. $$I=\pi\int_{0}^{\frac{1}{2}} \frac{u}{\sin(\frac\pi4 (1+2u))}du+\pi\int_{0}^{\frac12} \frac{t}{\sin(\frac\pi4 (3-2t))}dt$$ By using the trig identity $\sin(\pi-\theta)=\sin(\theta)$, we can see the integrands are the same.

$$I=2\pi\int_{0}^{\frac{1}{2}} \frac{u}{\sin(\frac\pi4 (1+2u))}du$$ Now substitute $\theta =\frac\pi4 (1+2u)$ and split the resulitng integrals. $$I=\frac{8}{\pi}\int_{\frac\pi4}^{\frac\pi2}\frac{\theta}{\sin(\theta)}d\theta-2\int_{\frac\pi4}^{\frac\pi2}\frac{1}{\sin(\theta)}d\theta$$ The second term is elementary, equaling $2\ln(\sqrt2 -1)$. However, I doubt the first term has an elementary form. But there is something nice we can do to it. Perform the half angle substitution: $z=\tan(\frac\theta2)$. $$I=\frac{16}{\pi}\int_{\sqrt{2}-1}^{1}\frac{\arctan(z)}{z}dz +2\ln(\sqrt2 -1)$$

The antiderivative of $\frac{\arctan(z)}{z}$ is the inverse tangent integral and is known to be transcendental as it can be written in terms of the dilogarithm. In particular, it has a nice power series. $$\mathrm{T}(z) =\sum_{k=0}^\infty \frac{(-1)^kz^{2k+1}}{(2k+1)^2} = z-\frac{z^3}{3^2}+\frac{z^5}{5^2}-\frac{z^7}{7^2}\dots$$ Notice that $\mathrm{T}(1)$ is Catalan's constant $G$. Therfore, $$I = \frac{16}{\pi}G + 2\ln(\sqrt2 -1) - \sum_{k=0}^\infty \frac{(-1)^k(\sqrt2 -1)^{2k+1}}{(2k+1)^2}$$

Related Solutions

Calculus – What is ?_0^1 ln(1-x) ln(ln(1/x)) dx?

We have the following closed form.

Proposition. $$ \int_0^1\log \left(1-x\right)\log \left(-\log x\right)\:dx=\gamma-\gamma_1+\gamma_1(1,0)\tag1 $$

where $\displaystyle \gamma$ is the Euler-Mascheroni constant, where $\gamma_1$ is the Stieltjes constant, $$\gamma_1 = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log n}n-\int_1^N\frac{\log t}t\:dt\right)$$ and where $\gamma_1(a,b)$ is the poly-Stieltjes constant (see here), $$\gamma_1(a,b) = \lim_{N\to+\infty}\left(\sum_{n=1}^N \frac{\log (n+a)}{n+b}-\int_1^N\frac{\log t}t\:dt\right)\!.$$

Proof. One may recall the classic integral representation of the Euler gamma function $$ \frac{\Gamma(s)}{(a+1)^s}=\int_0^\infty t^{s-1} e^{-(a+1)t}\:dt, \qquad s>0,\, a>-1. \tag2 $$ By differentiating $(2)$ with respect to $s$, putting $s=1$ and making the change of variable $x=e^{-t}$, we get $$ \int_0^1x^a\log\left(-\log x\right)\:dx=-\frac{\gamma+\log(a+1)}{a+1},\qquad a>-1, \tag3 $$

where $\displaystyle \gamma$ is the Euler-Mascheroni constant. We are allowed to insert the standard Taylor series expansion, $$ \log (1-x)= -\sum_{n=1}^{\infty} \frac{x^n}n, \qquad |x|<1,\tag4 $$ into the given integral, then using $(3)$ we obtain $$ \begin{align} \int_0^1\log \left(1-x\right)\log \left(-\log x\right)\:dx&=-\int_0^1\sum_{n=1}^{\infty}\frac{x^n}n \:\log (-\log x)\:dx\\ &=-\sum_{n=1}^{\infty} \frac1n\int_0^1 x^n\log (-\log x)\:dx\\ &=\sum_{n=1}^{\infty} \frac{\gamma+\log(n+1)}{n(n+1)}\\ &=\gamma \sum_{n=1}^{\infty}\frac1{n(n+1)}+\sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)}\\ &=\gamma +\sum_{n=1}^{\infty} \frac{\log(n+1)}{n(n+1)},\tag5 \end{align} $$ and we may conclude with Theorem $2$ here to get

$$ \begin{align} \sum_{n=1}^{\infty} \frac{\log (n+1)}{n(n+1)}=\gamma_1({1,0})-\gamma_1,\tag6 \end{align} $$

since $\gamma_1(1,1)=\gamma_1$.

Remark. I am inclined to believe that the poly-Stieltjes constants will turn out to be a tool for many of the considered integrals.

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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