# Evaluating integral $\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}$

To evaluate the integral $$$$\int_0^1 \frac{x-x^2}{\sin \pi x} dx = \frac{7 \zeta(3)}{\pi^3}$$$$ I tried to use the series for $$\sin \pi x$$ and maybe find something related to $$\zeta(3)$$. But it didn't work. I'm guessing this integral needs more than the little amount of calculus that I know.

• Laplace transforms, residue theory, and parameterizations come to mind, if you know any of those. May 24, 2018 at 3:55
• @Nitin While the residue theorem, no doubt, can integrate that, the OP added the real-analysis tag, indicating they want only real methods of integrating said integral. i.e no contour integration of that sort May 24, 2018 at 3:56
• You can use integration by parts and a simple variable substitution to get$$I=-\frac 8{\pi^3}\int\limits_0^{\pi/2}dx\, x\log\cot x$$ May 24, 2018 at 4:21

$\displaystyle f(a):=\int\limits_0^1 x e^{ax}dx = \frac{1+e^a(a-1)}{a^2}$

$\displaystyle g(a):=\int\limits_0^1 x^2 e^{ax}dx = \frac{-2+e^a(a^2-2a+2)}{a^3}$

$\displaystyle \int\limits_0^1 \frac{x^2-x}{\sin(\pi x)}dx = i2\int\limits_0^1\frac{x^2-x}{e^{i\pi x}-e^{-i\pi x}}dx = i2\sum\limits_{k=0}^\infty \int\limits_0^1 (x^2-x)e^{-i\pi x(2k+1)}dx$

$\displaystyle = i2\sum\limits_{k=0}^\infty (g(-i\pi(2k+1))-f(-i\pi(2k+1)))$

$\displaystyle = 2\sum\limits_{k=0}^\infty i\frac{-2+i\pi(2k+1) + e^{-i\pi(2k+1)}(2+i\pi(2k+1))}{(-i\pi(2k+1))^3} \enspace$ with $\enspace e^{-i\pi(2k+1)}=-1$

$\displaystyle = -8\sum\limits_{k=0}^\infty\frac{1}{(\pi(2k+1))^3}=-\frac{8}{\pi^3}(1-\frac{1}{2^3})\zeta(3)=-\frac{7\zeta(3)}{\pi^3}$

First, denote the integral below as $I$$I=\int\limits_0^1dx\space\frac {x(1-x)}{\sin\pi x}$$and through integration by parts on$u=x-x^2, then we have \begin{align*}I & =-\frac 1{\pi}(x-x^2)\log\cot\left(\frac {\pi x}2\right)\,\Biggr\rvert_0^1+\frac 1{\pi}\int\limits_0^1dx\, (1-2x)\log\cot\left(\frac {\pi x}2\right)\\ & =\frac 1{\pi}\int\limits_0^1dx\,\log\cot\left(\frac {\pi x}2\right)-\frac 2{\pi}\int\limits_0^1dx\, x\log\cot\left(\frac {\pi x}2\right)\\ & =-\frac 8{\pi^3}\int\limits_0^{\pi/2}dx\, x\log\cot x\tag1\end{align*} where equation (1$) comes from making the substitution$x\mapsto\frac {\pi x}2$. The latter integral can be evaluated by splitting up the natural log into two separate integrals and using the Fourier series for$\log\sin x$and$\log\cos x, which I have included below \begin{align*}\log\cos x & =\sum\limits_{k\geq1}(-1)^{k-1}\frac {\cos2kx}{k}-\log 2\tag2\\\log\sin x & =-\sum\limits_{k\geq1}\frac {\cos 2kx}k-\log 2\tag3\end{align*} Expanding (1$) gives $$I=-\frac 8{\pi^3}\underbrace{\int\limits_0^{\pi/2}dx\, x\log\cos x}_{I_1}+\frac 8{\pi^3}\underbrace{\int\limits_0^{\pi/2}dx\, x\log\sin x}_{I_2}\tag4$$ Call the first and second integrals$I_1$and$I_2$respectively. Using ($2$) and ($3) gives the following two identities \begin{align*}I_1 & =\int\limits_0^{\pi/2}dx\,\left(\sum\limits_{k\geq1}\frac {(-1)^{k-1}\cos 2kx}k-x\log 2\right)\\ & =\sum\limits_{k\geq1}\frac {(-1)^{k-1}}k\left[\frac {\pi}{4k^2}\sin\pi k+\frac 1{4k^3}\cos\pi k-\frac 1{4k^2}\right]-\frac {\pi^2}8\log2\\ & =\frac 14\sum\limits_{k\geq1}\frac {(-1)^{k-1}}{k^3}\cos\pi k-\frac 14\sum\limits_{k\geq1}\frac {(-1)^{k-1}}{k^3}-\frac {\pi^2}8\log 2\\ & \color{blue}{=-\frac 14\zeta(3)-\frac 3{16}\zeta(3)-\frac {\pi^2}8\log 2}\tag5\end{align*} As a side note, the infinite sum with\sin\pi k$vanishes because$\sin\pi k=0$for$k\in\mathbb{Z}$. In a similar manner,$I_2can be integrated as follows \begin{align*}I_2 & =-\int\limits_0^{\pi/2}dx\,\left(\sum\limits_{k\geq1}\frac {\cos 2kx}k+x\log 2\right)\\ & =-\sum\limits_{k\geq1}\frac 1k\left[\frac {\pi}{4k}\sin\pi k+\frac 1{4k^2}\cos\pi k-\frac 1{4k^2}\right]-\frac {\pi^2}8\log 2\\ & =-\frac 14\sum\limits_{k\geq1}\frac {\cos\pi k}{k^3}+\frac 14\sum\limits_{k\geq1}\frac 1{k^3}-\frac {\pi^2}8\log 2\\ & \color{red}{=\frac 14\zeta(3)+\frac 3{16}\zeta(3)-\frac {\pi^2}8\log 2}\tag6\end{align*} Substituting the results for (5$) and ($6$) into ($4) leaves us with \begin{align*}I & =-\frac 8{\pi^3}\left[\color{blue}{-\frac 14\zeta(3)-\frac 3{16}\zeta(3)}\color{red}{-\frac 14\zeta(3)-\frac 3{16}\zeta(3)}\right]\\ & =\frac 7{\pi^3}\zeta(3)\end{align*} Multiply by-1$to get the integral under question $$\int\limits_0^1dx\space\frac {x^2-x}{\sin\pi x}\color{brown}{=-\frac 7{\pi^3}\zeta(3)}$$ • This is quite the answer, (+1+eps) May 24, 2018 at 6:37 • @Nitin Thanks! This means a lot! May 24, 2018 at 13:40 Probably not an answer. For the antiderivative $$I=2 \pi^3\int \frac{x^2-x}{\sin (\pi x)}\, dx$$ a CAS give the ugly $$I=-i \pi (2 x-1) \left(4 \text{Li}_2\left(e^{i \pi x}\right)-\text{Li}_2\left(e^{2 i \pi x}\right)\right)+8 \text{Li}_3\left(e^{i \pi x}\right)-\text{Li}_3\left(e^{2 i \pi x}\right)-$$ $$4 \pi ^2 (x-1) x \tanh ^{-1}\left(e^{i \pi x}\right)$$ $$\lim_{x\to 1} \, I=-7 \zeta (3)+i\frac{ \pi ^3}{2} \qquad \text{and} \qquad \lim_{x\to 0} \, I=7 \zeta (3)+i\frac{ \pi ^3}{2}$$ What is interesting is that a rather good approximation could be obtained using a$[2,2]$Padé approximant built at$x=\frac 12$making $$\frac{x^2-x}{\sin (\pi x)}=\frac{-\frac 14+ a(x-\frac 12)^2 }{1+ b(x-\frac 12)^2 }$$ where $$a=-\frac{384-48 \pi ^2+\pi ^4}{48 \left(\pi ^2-8\right)} \qquad \text{and} \qquad b=-\frac{5 \pi ^4-48 \pi ^2}{12 \left(\pi ^2-8\right)}$$ making the definite integral easy to solve (leading to a value of$\approx -0.271415$while the exact value is$\approx -0.271377$). Still more amazing (at least to me), the approximation $$\sin(y) \simeq \frac{16 (\pi -y) y}{5 \pi ^2-4 (\pi -y) y}\qquad (0\leq y\leq\pi)$$ proposed by Mahabhaskariya of Bhaskara I, a seventh-century Indian mathematician would lead to$-\frac{13}{48} \approx -0.270833$. • +1 for mention of Indian mathematician Mahabhaskariya. As always your insight into numerical analysis is very amazing. May 25, 2018 at 15:55 • @ParamanandSingh. May I confess that it has been one of my big emotions, as a scientist, to "discover" this (given as a comment to one of my questions on MSE). From that day, I started looking at the works of Indian mathematicians and I am just fascinated. Cheers. May 26, 2018 at 2:53 An elementary evaluation: $$\int_0^1 \frac{x-x^2}{\sin \pi x} dx \overset{ibp} =\int_0^1 \frac{2x-1}\pi\ln \tan \frac{\pi x}2 dx \overset{t=\tan\frac{\pi x}2}= \frac8{\pi^3}\int_0^\infty \frac{\ln t\tan^{-1}t}{1+t^2}dt$$ Let $$J(a)=\int_0^\infty \frac{\ln t\tan^{-1}a t}{1+t^2}dt$$ $$J’(a)= \int_0^\infty \frac{t\ln t \>dt}{(1+t^2)(1+a^2t^2)} \overset{y=\frac1{at}}=\int_0^\infty \frac{ - \ln a \ y\ dy}{(1+y^2)(1+a^2y^2)}= \frac{\ln^2a}{2(1-a^2)}$$ Then, with $$\int_0^1 \frac{\ln^2a}{1-a^2} da = \frac{7}4\zeta(3)$$ $$I = \frac8{\pi^3}J(1)= \frac8{\pi^3}\int_0^1 J’(a)da =\frac4{\pi^3}\int_0^1 \frac{\ln^2a}{1-a^2} da =\frac{7 \zeta(3)}{\pi^3}$$ • Very nice change of variable in the derivative. The trick will be more aesthetic if you start from$\displaystyle \int_0^\infty \dfrac{\arctan x\ln x}{1+x^2}dx– FDP Mar 29, 2022 at 20:11 Following Frank W., \begin{align} J&=\int_0^1 \frac{x^2-x}{\sin(\pi x)}\,dx\\ &=\frac 8{\pi^3}\int\limits_0^{\pi/2}x\log\cot x \,dx\end{align} Perform the change of variable $$y=\tan x$$, \begin{align}J&=-\frac 8{\pi^3}\int_0^{\infty}\frac{\ln x\arctan x}{1+x^2} \,dx\\ &=-\frac 8{\pi^3}\int_0^{1}\frac{\ln x\arctan x}{1+x^2} \,dx-\frac 8{\pi^3}\int_1^{\infty}\frac{\ln x\arctan x}{1+x^2} \,dx \\\end{align} In the second integral perform the change of variable $$y=\dfrac{1}{x}$$, \begin{align}J&=-\frac 8{\pi^3}\int_0^{1}\frac{\ln x\arctan x}{1+x^2} \,dx+\frac 8{\pi^3}\int_0^{1}\frac{\ln x\arctan\left( \frac{1}{x}\right)}{1+x^2} \,dx\\ &=-\frac 8{\pi^3}\int_0^{1}\frac{\ln x\arctan x}{1+x^2} \,dx+\frac 8{\pi^3}\int_0^{1}\frac{\left(\frac{\pi}{2}-\arctan x\right)\ln x}{1+x^2} \,dx\\ &-\frac {16}{\pi^3}\int_0^{1}\frac{\ln x\arctan x}{1+x^2} \,dx-\frac{4}{\pi^2}\text{G} \end{align} $$\text{G}$$ is the Catalan constant. Let, $$\displaystyle K=\int_0^{1}\frac{\ln x\arctan x}{1+x^2} \,dx$$ Define for $$x\in [0;1]$$, \begin{align}R(x)&=\int_0^x\frac{\ln t}{1+t^2}\,dt\\ &=\int_0^1 \frac{x\ln(xt)}{1+t^2x^2}\,dt \end{align} Observe that, $$\displaystyle R(0)=0,R(1)=-\text{G}$$. Perform integration by parts, \begin{align}K&=\Big[R(x)\arctan x\Big]_0^1-\int_0^1 \frac{R(x)}{1+x^2}\,dx\\ &=-\frac{\pi}{4}\text{G}-\int_0^1 \int_0^1 \frac{x\ln(xt)}{(1+t^2x^2)(1+x^2)}\,dt\,dx \\ &=-\frac{\pi}{4}\text{G}-\int_0^1 \int_0^1 \frac{x\ln t}{(1+t^2x^2)(1+x^2)}\,dt\,dx-\int_0^1 \int_0^1 \frac{x\ln x}{(1+t^2x^2)(1+x^2)}\,dt\,dx\\ &=-\frac{\pi}{4}\text{G}-\frac{1}{2}\int_0^1 \left[\frac{\ln t}{1-t^2}\times \ln\left(\frac{1+x^2}{1+t^2x^2}\right)\right]_{x=0}^{x=1}\,dt-\int_0^1 \Big[\frac{\arctan(tx)\ln x}{1+x^2}\Big]_{t=0}^{t=1} \,dx\\ &=-\frac{\pi}{4}\text{G}-\frac{1}{2}\int_0^1 \frac{\ln\left( \frac{2}{1+t^2}\right)\ln t}{1-t^2}\,dt-K \end{align} Therefore, \begin{align} K&=-\frac{\pi}{8}\text{G}-\frac{1}{4}\int_0^1 \frac{\ln\left( \frac{2}{1+t^2}\right)\ln t}{1-t^2}\,dt\\ &=-\frac{\pi}{8}\text{G}-\frac{\ln 2}{4}\int_0^1 \frac{\ln t}{1-t^2}\,dt+\frac{1}{4}\int_0^1 \frac{\ln(1+t^2)\ln t}{1-t^2}\,dt \end{align} Let, $$\displaystyle L=\int_0^1 \frac{\ln(1+t^2)\ln t}{1-t^2}\,dt$$ For $$x\in [0;1]$$ define, \begin{align}S(x)&=\int_0^x\frac{\ln t}{1-t^2}\,dt\\ &=\int_0^1 \frac{x\ln(tx)}{1-t^2x^2}\,dt \end{align} Perform integration by parts, \begin{align}L&=\Big[S(x)\ln(1+x^2)\Big]_0^1 -\int_0^1 \int_0^1\frac{2x^2\ln(tx)}{(1+x^2)(1-t^2x^2)}\,dt\,dx\\ &=S(1)\ln 2-\int_0^1 \int_0^1 \frac{2x^2\ln t}{(1+x^2)(1-t^2x^2)}\,dt\,dx-\int_0^1 \int_0^1\frac{2x^2\ln x}{(1+x^2)(1-t^2x^2)}\,dt\,dx\\ &=S(1)\ln 2-\int_0^1 \left[-\frac{t\ln t}{1+t^2}\ln\left(\frac{1+tx}{1-tx}\right)+\frac{\ln t}{t}\ln\left(\frac{1+tx}{1-tx}\right)-\frac{2\arctan x \ln t}{1+t^2}\right]_{x=0}^{x=1}\,dt-\\ &\int_0^1 \left[\frac{x\ln x}{1+x^2}\ln\left(\frac{1+tx}{1-tx}\right)\right]_{t=0}^{t=1}\,dx\\ &=S(1)\ln 2-\int_0^1 \left[\frac{\ln t}{t}\ln\left(\frac{1+tx}{1-tx}\right)-\frac{2\arctan x \ln t}{1+t^2}\right]_{x=0}^{x=1}\,dt\\ &=S(1)\ln 2-\int_0^1 \frac{\ln t}{t}\ln\left(\frac{1+t}{1-t}\right)\,dt+\frac{\pi}{2}\int_0^1 \frac{\ln t}{1+t^2}\,dt\\ &=S(1)\ln 2-\int_0^1 \frac{\ln t}{t}\ln\left(\frac{1+t}{1-t}\right)\,dt-\frac{1}{2}\pi\text{G} \end{align} Let, $$\displaystyle M=\int_0^1 \frac{\ln t}{t}\ln\left(\frac{1+t}{1-t}\right)\,dt$$ Perform integration by parts, \begin{align}M&=\Big[\frac{1}{2}\ln^2 t \ln\left(\frac{1+t}{1-t}\right)\Big]_0^1-\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt\\ &=-\int_0^1 \frac{\ln^2 t}{1-t^2}\,dt\\ \end{align} Using Taylor expansion, $$\displaystyle M=-\frac{7}{4}\zeta(3)$$ Therefore, $$\displaystyle L=S(1)\ln 2+\frac{7}{4}\zeta(3)-\frac{1}{2}\pi\text{G}$$ Therefore, \begin{align}K&=-\frac{\pi}{8}\text{G}-\frac{\ln 2}{4}S(1) +\frac{1}{4}L\\ &=-\frac{\pi}{8}\text{G}-\frac{\ln 2}{4}S(1) +\frac{1}{4}\left(S(1)\ln 2+\frac{7}{4}\zeta(3)-\frac{1}{2}\pi\text{G}\right)\\ &=\frac{7}{16}\zeta(3)-\frac{1}{4}\pi\text{G} \end{align} Therefore, \begin{align}J&=-\frac {16}{\pi^3}K-\frac{4}{\pi^2}\text{G}\\ &=-\frac {16}{\pi^3}\left(\frac{7}{16}\zeta(3)-\frac{1}{4}\pi\text{G}\right)-\frac{4}{\pi^2}\text{G}\\ &=\boxed{-\frac{7\zeta(3)}{\pi^3}} \end{align} Addendum April 2022 On the path of Quanto. A simpler computation. \begin{align}Z&=\int_0^\infty\frac{\arctan x\ln x}{1+x^2}dx\\ &\overset{\text{IBP}}=\underbrace{\left[\arctan x\int_0^x \frac{\ln t}{1+t^2}dt\right]_0^\infty}_{=0}-\int_0^\infty \left(\int_0^1 \frac{x\ln(tx)}{(1+x^2)(1+t^2x^2}\right)dx\\ &=-\int_0^1 \left(\int_0^\infty \frac{x\ln(t)}{(1+x^2)(1+t^2x^2}\right)dx-\int_0^\infty \left(\int_0^1 \frac{x\ln(x)}{(1+x^2)(1+t^2x^2}\right)dx\\ &=-\frac{1}{4}\int_0^1 \left[\frac{\ln\left(\frac{1+x^2}{1+t^2x^2}\right)}{1-t^2}\right]_0^\infty \ln t dt-Z\\ &=\frac{1}{2}\int_0^1 \frac{\ln^2 t}{1-t^2}dt=\frac{1}{2}\int_0^1 \frac{\ln^2 t}{1-t}dt-\frac{1}{2}\underbrace{\int_0^1 \frac{t\ln^2 t}{1-t^2}dt}_{u=t^2}\\ &=\frac{1}{2}\int_0^1 \frac{\ln^2 t}{1-t}dt-\frac{1}{16}\int_0^1 \frac{\ln^2 u}{1-u}du=\frac{7}{16}\int_0^1 \frac{\ln^2 t}{1-t}dt\\ &=\frac{7}{16}\times 2\zeta(3)=\boxed{\frac{7}{8}\zeta(3)} \end{align} $$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$$ $$\ds{\int_{0}^{1}{x - x^{2} \over \sin\pars{\pi x}} \,\dd x = {7\zeta\pars{3} \over \pi^{3}}:\ {\LARGE ?}}$$. \begin{align} &\bbox[10px,#ffd]{\int_{0}^{1}{x - x^{2} \over \sin\pars{\pi x}}\,\dd x} \,\,\,\stackrel{x\ \mapsto\ x + 1/2}{=}\,\,\, \int_{-1/2}^{1/2}{1/4 - x^{2} \over \cos\pars{\pi x}}\,\dd x \\[5mm] = &\ {1 \over 2}\int_{0}^{1/2}{1 - 4x^{2} \over \cos\pars{\pi x}}\,\dd x \,\,\,\stackrel{\pi x\ \mapsto\ x}{=}\,\,\, {1 \over 2\pi^{3}}\int_{0}^{\pi/2}{\pi^{2} - 4x^{2} \over \cos\pars{x}}\,\dd x \\[5mm] = &\ \left. {1 \over 2\pi^{3}}\Re\int_{x\ =\ 0}^{x\ =\ \pi/2}{\pi^{2} - 4\bracks{-\ic\ln\pars{z}}^{2} \over \pars{z + 1/z}/2}\,{\dd z \over \ic z} \,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ \left. {1 \over \pi^{3}}\,\Im\int_{x\ =\ 0}^{x\ =\ \pi/2}{\pi^{2} + 4\ln^{2}\pars{z} \over 1 + z^{2}}\,\dd z\,\right\vert_{\ z\ =\ \exp\pars{\ic x}} \\[5mm] = &\ -\,{1 \over \pi^{3}}\,\Im\int_{1}^{0}{\pi^{2} + 4\bracks{\ln\pars{y} + \ic\pi/2}^{\, 2} \over 1 + \pars{\ic y}^{2}}\,\ic\,\dd y \\[5mm] = &\ {4 \over \pi^{3}}\ \underbrace{\int_{0}^{1} {\ln^{2}\pars{y} \over 1 - y^{2}}\,\dd y} _{\ds{7\zeta\pars{3} \over 4}} = \bbx{7\zeta\pars{3} \over \pi^{3}} \approx 0.2714 \end{align} • Why is the last identity true? Dec 22, 2019 at 1:11 •\int_0^1 x^n\ln^2(x)dx =2/(n+1)^3\$. Apr 8, 2022 at 20:06

I will present an evaluation that makes use of the following two Euler sums:

$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3) \qquad \text{and} \qquad \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^2} = \frac{23}{16} \zeta (3) - \pi \mathbf{G}.$$

Here $$\mathbf{G}$$ is Catalan's constant. For a proof of the first, see either here or Eq. (646) in this link. For a proof of the second, see Eq. (659) in this link.

Writing $$I = \int_0^1 \frac{x(1 - x)}{\sin (\pi x)} \, dx,$$ following on from what others have already recognised, we can write the integral as $$I = -\frac{8}{\pi^3} \int_0^{\frac{\pi}{2}} x \ln (\cot x) \, dx = \frac{8}{\pi^3} \int_0^{\frac{\pi}{2}} x \ln (\tan x) \, dx.$$ Enforcing a substitution of $$x \mapsto \arctan x$$ gives \begin{align} I &= \frac{8}{\pi^3} \int_0^\infty \frac{\ln x \arctan x}{1 + x^2} \, dx\\ &= \frac{8}{\pi^3} \int_0^1 \frac{\ln x \arctan x}{1 + x^2} \, dx + \frac{8}{\pi^3} \int_1^\infty \frac{\ln x \arctan x}{1 + x^2} \, dx\\ &= \frac{8}{\pi^3} \int_0^1 \frac{\ln x \arctan x}{1 + x^2} \, dx - \frac{8}{\pi^3} \int_0^1 \frac{\ln x \left (\frac{\pi}{2} - \arctan x \right )}{1 + x^2} \, dx\\ &= \frac{16}{\pi^3} \int_0^1 \frac{\ln x \arctan x}{1 + x^2} \, dx - \frac{4}{\pi^2} \int_1^\infty \frac{\ln x}{1 + x^2} \, dx\\ &= \frac{16}{\pi^3} I_1 - \frac{4}{\pi^2} I_2.\tag1 \end{align}

For the first integral $$I_1$$

Applying the Cauchy product to the product between the Macluarin series for $$\arctan x$$ and $$\frac{1}{1 + x^2}$$, one readily finds $$\frac{\arctan x}{1 + x^2} = \sum_{n = 0}^\infty (-1)^n \left (H_{2n + 1} - \frac{1}{2} H_n \right ) x^{2n + 1}, \qquad |x| < 1.$$ Here $$H_n$$ is the $$n$$th Harmonic number.

Thus \begin{align} I_1 &= \sum_{n = 0}^\infty (-1)^n \left (H_{2n + 1} - \frac{1}{2} H_n \right ) \int_0^1 x^{2n + 1} \, dx\\ &= \sum_{n = 0}^\infty (-1)^n \left (H_{2n + 1} - \frac{1}{2} H_n \right ) \frac{d}{ds} \left [\int_0^1 x^{2n + s + 1} \, dx \right ]_{s = 0}\\ &= -\sum_{n = 0}^\infty (-1)^n \left (H_{2n + 1} - \frac{1}{2} H_n \right ) \frac{1}{(2n + 2)^2}\\ &= -\frac{1}{4} \underbrace{\sum_{n = 0}^\infty \frac{(-1)^n H_{2n + 1}}{(n + 1)^2}}_{n \, \mapsto \, n -1} + \frac{1}{8} \underbrace{\sum_{n = 0}^\infty \frac{(-1)^{n} H_n}{(n + 1)^2}}_{n \, \mapsto \, n - 1}\\ &= -\frac{1}{4} \sum_{n = 1}^\infty \frac{(-1)^{n-1} H_{2n - 1}}{n^2} + \frac{1}{8} \sum_{n = 1}^\infty \frac{(-1)^{n-1} H_{n-1}}{n^2}\\ &= \frac{1}{4} \sum_{n = 1}^\infty \frac{(-1)^n}{n^2} \left (H_{2n} - \frac{1}{2n} \right ) - \frac{1}{8} \sum_{n = 1}^\infty \frac{(-1)^n}{n^2} \left (H_{n} - \frac{1}{n} \right )\\ &= \frac{1}{4} \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^2} - \frac{1}{8} \sum_{n = 1}^\infty \frac{(-1)^n}{n^3} - \frac{1}{8} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} + \frac{1}{8} \sum_{n = 1}^\infty \frac{(-1)^n}{n^3}\\ &= \frac{1}{4} \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^2} - \frac{1}{8} \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2}\\ &= \frac{1}{4} \left (\frac{23}{16} \zeta (3) - \pi \mathbf{G} \right ) - \frac{1}{8} \left (-\frac{5}{8} \zeta (3) \right )\\ &= \frac{7}{16} \zeta (3) - \frac{\pi}{4} \mathbf{G}.\tag2 \end{align}

For the second integral $$I_2$$

\begin{align} I_2 &= \int_0^1 \frac{\ln x}{1 + x^2} \, dx\\ &= \sum_{n = 0}^\infty (-1)^n x^{2n} \ln x \, dx\\ &= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [ \int_0^1 x^{2n + s} \right ]_{s = 0}\\ &= \sum_{n = 0}^\infty (-1)^n \frac{d}{ds} \left [\frac{1}{2n + s + 1} \right ]_{s = 0}\\ &= -\sum_{n = 0}^\infty \frac{(-1)^n}{(2n + 1)^2}\\ &= - \mathbf{G}\tag3 \end{align}

Main integral

On returning to the main integral, substituting (2) and (3) into (1) one sees that $$I = \frac{16}{\pi^3} \left (\frac{7}{16} \zeta (3) - \frac{\pi \mathbf{G}}{4} \right ) + \frac{4 \mathbf{G}}{\pi^2},$$ or $$\int_0^1 \frac{x(1 - x)}{\sin (\pi x)} \, dx = \frac{7}{\pi^3} \zeta (3),$$ as required.

Basically The Same As @FrankW.

First, a warmup integral.

Let $$S(x)=\int\frac{dx}{\sin\pi x}\overset{t=\pi x}=\frac1\pi \int\frac{dt}{\sin t}.$$ The sub $$u=\tan(t/2)$$ provides $$S(x)=\frac1\pi\int\frac{1}{\frac{2u}{1+u^2}}\frac{2du}{1+u^2}=\frac1\pi\ln\tan\frac{\pi x}{2}\ .$$ So we have that the integral in question is $$I=\frac1\pi\int_0^1(x-x^2)\left(\ln\tan\tfrac{\pi x}{2}\right)'\ dx.$$ Thus \begin{align} \pi I&=\underbrace{(x-x^2)\ln\tan\tfrac{\pi x}{2}\bigg |_0^1}_{=0}+\int_0^1(2x-1)\ln\tan\tfrac{\pi x}{2}\ dx\\ &=2\int_0^1 x\ln\tan\tfrac{\pi x}{2}\ dx-\underbrace{\int_0^1\ln\tan\tfrac{\pi x}{2}\ dx}_{=0}\\ &=\frac2{\pi^2}\int_0^\pi x\ln\tan\tfrac{x}{2}\ dx\ . \end{align} Then recall the definition of the Clausen function of order $$2$$: $$\mathrm{Cl}_2(x)=-\int_0^x \ln\left|2\sin\tfrac{t}{2}\right|\ dt.$$ Then using the Fourier series given by @FrankW. one can show that $$\mathrm{Cl}_2(x)=\sum_{k\ge1}\frac{\sin kx}{k^2}.$$ So we have \begin{align} \int_0^x \ln\tan\tfrac{t}{2}\ dt&=\int_0^x \ln\left(2\sin\tfrac{t}{2}\right)\ dt-\int_0^x \ln\left(2\cos\tfrac{t}{2}\right)\ dt\\ &=-\mathrm{Cl}_2(x)-\int_0^x \ln\left(2\cos\tfrac{t}{2}\right)\ dt\\ &=-\mathrm{Cl}_2(x)-\mathrm{Cl}_2(\pi-x). \end{align} Integration by parts again: \begin{align} \frac{\pi^3}{2}I&=\left[-x(\mathrm{Cl}_2(x)+\mathrm{Cl}_2(\pi-x))\right]_0^\pi+\int_0^\pi\mathrm{Cl}_2(x)dx+\int_0^\pi \mathrm{Cl}_2(\pi-x)dx\\ &=\int_0^\pi\mathrm{Cl}_2(x)dx+\int_0^\pi \mathrm{Cl}_2(\pi-x)dx\\ &=2\int_0^\pi\mathrm{Cl}_2(x)dx. \end{align} Next, we recall the definition of the $$n$$-th order Clausen function: $$\mathrm{Cl}_n(x)=\sum_{k\ge1}\frac{p_n(kx)}{k^n}$$ where $$p_n(x)=\Bigg\{{{\cos x\qquad n \text{ odd}}\atop{\sin x\qquad n\text{ even}}}$$ so that $$\int \mathrm{Cl}_n(x)dx=(-1)^{n+1}\mathrm{Cl}_{n+1}(x).$$ So at long last, $$I=\frac4{\pi^3}\left(\mathrm{Cl}_3(0)-\mathrm{Cl}_3(\pi)\right).$$ Since $$p_3(0)=1$$ and $$p_3(\pi k)=(-1)^k$$ we have that $$I=\frac{4}{\pi^3}\sum_{k\ge1}\frac1{k^3}[1-(-1)^k]=\frac{7\zeta(3)}{\pi^3}.$$