closed-formdefinite integralsintegrationpolylogarithmspecial functions
My friend offered to solve this integral.
$$I=\int_0^{\pi/2}x^2\frac{\ln(\sin x)}{\cos x}dx=\frac{\pi^4}{32}-{4G^2} $$
Where G is the Catalan's constant.
$$I=\int _0^{\infty }\frac{\arctan ^2\left(u\right)\ln \left(\frac{u}{\sqrt{1+u^2}}\right)}{\sqrt{1+u^2}}\:du\;.$$
Put $$\tan{x}=u$$
Inability to split into two integrals, they are divergent.
$\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{\mathrm{i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\Li}[1]{\,\mathrm{Li}_{#1}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\ol}[1]{\overline{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\int_{0}^{1}{\ln\pars{x}\ln\pars{1 - x} \over 1 + x^{2}}\,\dd x - \half\int_{0}^{1}{\ln\pars{x}\ln\pars{1 + x^{2}} \over 1 + x^{2}}\,\dd x = \half\,\ln\pars{2}\,\mrm{G}}$.
$\ds{\mrm{G}:\ \mbox{Catalan Constant.}}$
It turns out that the above integrals can be reduced to the form $$ \left.\int{\ln^{2}\pars{x} \over a - x}\,\dd x\,\right\vert_{\ a\ \not=\ 0} \,\,\,\,\,\stackrel{x\ =\ at}{=}\,\,\,\,\, \int{\ln^{2}\pars{at} \over 1 - t}\,\dd t = -\int\ln^{2}\pars{at}\,\dd\bracks{\ln\pars{1 - t}} $$ which can be easily evaluated by successive integration by parts: \begin{equation} \int{\ln^{2}\pars{x} \over a - x}\,\dd x = \left\lbrace\begin{array}{lcl} \ds{-\ln^{2}\pars{x}\ln\pars{1 - {x \over a}} - 2\ln\pars{x}\Li{2}\pars{x \over a} + 2\Li{3}\pars{x \over a}} & \mbox{if} & \ds{a \not= 0} \\ \ds{-\,{1 \over 3}\,\ln^{3}\pars{x}} & \mbox{if} & \ds{a = 0} \end{array}\right. \end{equation} $$ \begin{array}{|c|}\hline\mbox{}\\ \quad\mbox{Hereafter, we'll use this result to evaluate}\ \ds{\braces{\vphantom{\large A}\color{#f00}{\mc{J}_{k}}\,,\ k = 1,2,3,4}} \quad \\ \mbox{}\\ \hline \end{array} $$
With $\ds{r \equiv 1 + \ic}$:
Complete answer now!$$I=\int_0^\infty \frac{{\arctan(x^2)}}{x^4+x^2+1}dx {\overset{x=\frac1{t}}=} \int_0^\infty \frac{\arctan\left(\frac{1}{t^2}\right)}{\frac{1}{t^4}+\frac{1}{t^2}+1}\frac{dt}{t^2}\overset{t=x}=\int_0^\infty \frac{{x^2\left(\frac{\pi}{2}-\arctan(x^2)\right)}}{x^4+x^2+1}dx $$ Now if we add the result with the original integral $I$ we get: $$2I=\frac{\pi}{2}\int_0^\infty \frac{x^2}{x^4+x^2+1}dx+\int_0^\infty \frac{(1-x^2)\arctan(x^2)}{x^4+x^2+1}dx$$ $$\Rightarrow I = \frac12 \cdot \frac{\pi}{2}\cdot \frac{\pi}{2\sqrt 3}-\frac12 \int_0^\infty \frac{(x^2-1)\arctan(x^2)}{x^4+x^2+1}dx=\frac{\pi^2}{8\sqrt 3} -\frac12 J$$
Now in order to calculate $J\,$ we start by performing IBP: $$J=\int_0^\infty \frac{(x^2-1)\arctan\left(x^2\right)}{x^4+x^2+1}dx =\int_0^\infty \arctan(x^2) \left(\frac12 \ln\left(\frac{x^2-x+1}{x^2+x+1}\right)\right)'dx=$$ $$=\underbrace{\frac{1}{2}\ln\left(\frac{x^2-x+1}{x^2+x+1}\right)\arctan(x^2)\bigg|_0^\infty}_{=0}+\int_0^\infty \frac{x}{1+x^4}\ln\left(\frac{x^2+x+1}{x^2-x+1}\right)dx$$ Substituting $x=\tan\left(t\right)$ and doing some simplifications yields: $$J=\int_0^\frac{\pi}{2} \frac{2\sin (2t)}{3+\cos(4t)}\ln\left(\frac{2+\sin (2t)}{2-\sin (2t)}\right)dt\overset{2t=x}=\int_0^\pi \frac{\sin x}{3+\cos(2x)}\ln\left(\frac{2+\sin x}{2-\sin x}\right)dx=$$ $$=2\int_0^\frac{\pi}{2}\frac{\sin x}{3+\cos(2x)}\ln\left(\frac{2+\sin x}{2-\sin x}\right)dx=\int_0^\frac{\pi}{2}\frac{\cos x}{1+\sin^2 x}\ln\left(\frac{2+\cos x}{2-\cos x}\right)dx =$$ $$=\frac12\int_0^\pi \frac{\cos x}{1+\sin^2 x}\ln\left(\frac{2+\cos x}{2-\cos x}\right)dx\overset{\large{\tan\left(\frac{x}{2}\right)=t}}=\int_0^\infty \frac{1-t^2}{t^4+6t^2+1}\ln\left(\frac{\color{blue}{t^2+3}}{\color{red}{3t^2+1}}\right)dt$$
Splitting the integral into two parts followed by the substitution $\,\displaystyle{t=\frac{1}{x}}\,$ in the second part gives: $$\int_0^\infty \frac{1-t^2}{t^4+6t^2+1}\ln(\color{red}{3t^2+1})dt =\int_0^\infty \frac{x^2-1}{x^4+6x^2+1}\ln\left(\color{red}{\frac{x^2+3}{x^2}}\right)dx$$ $$\Rightarrow J=\int_0^\infty \frac{1-x^2}{x^4+6x^2+1} \ln(\color{blue}{x^2+3})dx - \int_0^\infty \frac{1-x^2}{x^4+6x^2+1} {\left(\ln(\color{red}{x^2})-\ln(\color{red}{x^2+3})\right)}dx=$$ $$=2\int_0^\infty \frac{1-x^2}{x^4+6x^2+1}\ln\left(\color{purple}{\frac{x^2+3}{x}}\right)dx=2\int_0^\infty \left(\frac12\arctan\left(\frac{2x}{1+x^2}\right)\right)'\ln\left(\frac{x^2+3}{x}\right)dx=$$ $$=\underbrace{\arctan\left(\frac{2x}{1+x^2}\right)\ln\left(\frac{x^2+3}{x}\right)\bigg|_0^\infty}_{=0}-\int_0^\infty \arctan\left(\frac{2x}{1+x^2}\right)\left(\frac{2x}{x^2+3}-\frac{1}{x}\right)dx$$ $$\Rightarrow J=\int_0^\infty \arctan\left(\frac{2x}{1+x^2}\right)\frac{dx}{x}-\int_0^\infty \arctan\left(\frac{2x}{1+x^2}\right) \frac{2x}{x^2+3}dx=J_1-J_2$$
$$J_1=\int_0^\infty \arctan\left(\frac{2x}{1+x^2}\right)\frac{dx}{x}\overset{\large{x=\tan\left(\frac{t}{2}\right)}}=\int_0^\pi \frac{\arctan( \sin t)}{\sin t} dt\overset{t=x}=2\int_0^\frac{\pi}{2} \frac{\arctan( \sin x)}{\sin x} dx$$ In general, we have the following relation: $$\frac{\arctan x}{x}=\int_0^1 \frac{dy}{1+(xy)^2} \Rightarrow \color{red}{\frac{\arctan(\sin x)}{\sin x}=\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}$$ $$J_1 = 2\color{blue}{\int_{0}^{\frac{\pi}{2}}} \color{red}{\frac{\arctan\left(\sin x\right)}{\sin x}}\color{blue}{dx}=2\color{blue}{\int_0^\frac{\pi}{2}}\color{red}{\int_0^1 \frac{dy}{1+(\sin^2 x )y^2}}\color{blue}{dx}=2\color{red}{\int_0^1} \color{blue}{\int_0^\frac{\pi}{2}}\color{purple}{\frac{1}{1+(\sin^2 x )y^2}}\color{blue}{dx}\color{red}{dy}$$ $$=2\int_0^1 \frac{\arctan\left(\sqrt{1+y^2}\cdot\tan(x)\right) }{\sqrt{1+y^2}} \bigg|_0^\frac{\pi}{2}=\pi\int_0^1 \frac{dy}{\sqrt{1+y^2}}=\boxed{\pi\ln\left(1+\sqrt 2\right)}$$
In order to evaluate $J_2$ we return the integral before was integrated by parts. $$J_2=2\int_0^\infty \arctan\left(\frac{2 x} {x^2 +1}\right)\frac{x}{x^2 +3}dx=2\int_0^{\infty}\frac{(x^2-1)\ln(x^2+3)}{x^4+6x^2+1} dx=$$ $$=(\sqrt 2+1)\int_0^{\infty} \frac{\ln(x^2+3)}{x^2+\left(\sqrt 2+1\right)^2} \ dx - (\sqrt 2-1)\int_0^{\infty} \frac{\ln(x^2+3)}{x^2+\left(\sqrt 2-1\right)^2} dx$$ Using the following identity that is valid for $a\ge 0, b>0$:$$\int_0^{\infty} \frac{\ln(x^2+a^2)}{x^2+b^2} \ dx = \frac{\pi}{b}\ln(a+b)$$ $$\Rightarrow J_2=\pi\ln\left(\frac{\sqrt{3}+\sqrt{2}+1}{\sqrt{3}+\sqrt{2}-1}\right)=\boxed{\frac{\pi} {2}\ln(2+\sqrt 3)}$$ So we found that:$$J=\boxed{\pi \ln(1+\sqrt 2) - \frac{\pi} {2} \ln(2+\sqrt 3)}\Rightarrow I= \large\boxed{\frac{\pi^2} {8 \sqrt 3}+\frac{\pi}{4}\ln(2+\sqrt 3)-\frac{\pi}{2} \ln(1+\sqrt 2)}$$
Best Answer
A possible way to calculate it is starting with $x=\frac{\pi }{2}-t$, we then get $$\int _0^{\frac{\pi }{2}}x^2\sec \left(x\right)\ln \left(\sin \left(x\right)\right)\:dx=\frac{\pi ^2}{4}\int _0^{\frac{\pi }{2}}\csc \left(x\right)\ln \left(\cos \left(x\right)\right)\:dx$$ $$-\pi \int _0^{\frac{\pi }{2}}x\csc \left(x\right)\ln \left(\cos \left(x\right)\right)\:dx+\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\ln \left(\cos \left(x\right)\right)\:dx.\tag1$$ Notice that the first integral is very simple, as for the second one, it won't be necessary to calculate it if we focus on the third integral.
Employing the well-known expansion for $\ln \left(\cos \left(x\right)\right)$ leads to $$\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\ln \left(\cos \left(x\right)\right)\:dx$$ $$=-\ln \left(2\right)\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\:dx-\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k}\underbrace{\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\cos \left(2kx\right)\:dx}_{I_k},\tag2$$ so in order to simplify $I_k$ we'll use the following strategy $$\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\cos \left(2kx\right)\:dx=\int _0^{\frac{\pi }{2}}x^2\frac{\cos \left(2kx\right)}{\sin \left(x\right)}\:dx$$ $$I_k=\int _0^{\frac{\pi }{2}}x^2\frac{\cos \left(2kx\right)}{\sin \left(x\right)}\:dx$$ $$I_{n+1}-I_n=\int _0^{\frac{\pi }{2}}x^2\frac{\cos \left(2\left(n+1\right)x\right)-\cos \left(2nx\right)}{\sin \left(x\right)}\:dx=-2\int _0^{\frac{\pi }{2}}x^2\sin \left(\left(2n+1\right)x\right)\:dx$$ $$=-\frac{2\pi \left(2n+1\right)\left(-1\right)^n-4}{\left(2n+1\right)^3}$$ $$I_k-I_0=-\sum _{n=0}^{k-1}\frac{2\pi \left(2n+1\right)\left(-1\right)^n-4}{\left(2n+1\right)^3}$$ $$\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\cos \left(2kx\right)\:dx=-2\pi \sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{\left(2n-1\right)^2}+4\sum _{n=1}^k\frac{1}{\left(2n-1\right)^3}+\int _0^{\frac{\pi }{2}}x^2\csc \left(x\right)\:dx.$$ Replacing this in $\left(2\right)$ yields $$=2\pi \sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k}\sum _{n=1}^k\frac{\left(-1\right)^{n+1}}{\left(2n-1\right)^2}-4\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k}\sum _{n=1}^k\frac{1}{\left(2n-1\right)^3}$$ $$=-2\pi \int _0^1\left(\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k}\left(1-\left(-x^2\right)^k\right)\right)\frac{\ln \left(x\right)}{1+x^2}\:dx$$ $$-2\int _0^1\left(\sum _{k=1}^{\infty }\frac{\left(-1\right)^k}{k}\left(1-x^{2k}\right)\right)\frac{\ln ^2\left(x\right)}{1-x^2}\:dx$$ $$=-2\pi \int _0^1\frac{\ln \left(x\right)\ln \left(\frac{1-x^2}{2}\right)}{1+x^2}\:dx-2\int _0^1\frac{\ln ^2\left(x\right)\ln \left(\frac{1+x^2}{2}\right)}{1-x^2}\:dx.$$ Lastly, notice that $$-2\int _0^1\frac{\ln \left(x\right)\ln \left(\frac{1-x^2}{2}\right)}{1+x^2}\:dx=\int _0^{\frac{\pi }{2}}x\csc \left(x\right)\ln \left(\cos \left(x\right)\right)\:dx$$ $$\int _0^1\frac{\ln ^2\left(x\right)\ln \left(\frac{1+x^2}{2}\right)}{1-x^2}\:dx=2G^2-\frac{45}{16}\zeta \left(4\right),$$ where in order to obtain such closed-form one can trivially reduce it to the derivatives of the Beta function or exploit symmetry with double integration.
Now that we know what $\left(2\right)$ transformed to, $\left(1\right)$ becomes $$\int _0^{\frac{\pi }{2}}x^2\sec \left(x\right)\ln \left(\sin \left(x\right)\right)\:dx$$ $$=\frac{\pi ^2}{4}\left(-\frac{\pi ^2}{8}\right)-\pi \int _0^{\frac{\pi }{2}}x\csc \left(x\right)\ln \left(\cos \left(x\right)\right)\:dx-2\pi \left(-\frac{1}{2}\int _0^{\frac{\pi }{2}}x\csc \left(x\right)\ln \left(\cos \left(x\right)\right)\:dx\right)$$ $$-2\left(2G^2-\frac{45}{16}\zeta \left(4\right)\right),$$ and finally $$\int _0^{\frac{\pi }{2}}x^2\sec \left(x\right)\ln \left(\sin \left(x\right)\right)\:dx=\frac{45}{16}\zeta \left(4\right)-4G^2.$$