Proving that $\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}=\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$ Trying to show using a different approach that $\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx =\frac{\pi^2\sqrt 3}{9}-\frac{8}{3}G\, $  I have stumbled upon this series:   $$\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}$$ The linked proof relies upon this trigamma identity. Now by rewriting the integral as:
$$I=\int_0^1 \frac{\sqrt{x}\ln x}{x^2-2\cos\left(\frac{\pi}{3}\right)x+1}dx$$ 
And using that: $$\frac{\sin t}{x^2-2x\cos t+1}=\frac{1}{2i}\left(\frac{e^{it}}{1-xe^{it}}-\frac{e^{-it}}{1-xe^{-it}}\right)=\Im \left(\frac{e^{it}}{1-xe^{it}}\right)=$$
$$=\sum_{n=0}^{\infty} \Im\left(x^n e^{i(n+1)t}\right)=\sum_{n=0}^\infty x^n\sin((n+1)t)$$
$$I=\frac{1}{\sin \left(\frac{\pi}{3}\right)}\sum_{n=0}^\infty  \sin\left(\frac{\pi}{3} (n+1) \right)\int_0^1 x^{n+1/2} \ln x dx$$
$$\text{Since} \  \int_0^1 x^p \ln x dx= -\frac{1}{(p+1)^2}$$
$$I=-\frac{2}{\sqrt 3} \sum_{n=0}^\infty \frac{\sin\left((n+1)\frac{\pi}{3}\right)}{(n+1+1/2)^2}=-\frac{8}{\sqrt 3}\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2} $$
And well by using the previous link we can deduce that the series equals to  $\frac{G}{\sqrt 3} -\frac{\pi^2}{24}$, where $G$ is Catalan's constant.
I thought this might be a coefficient of some Fourier series, or taking the imaginary part of $\left(\sum_{n=1}^\infty \frac{e^{i\frac{n\pi}{3}}}{(2n+1)^2}\right)$, but I was not that lucky afterwards.
Is there a way to show the result without relying on that trigamma identity? Another approach to the integral would of course be enough.
 A: This is a 
major revision 
of the last few lines.
The conclusion is that
$\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}
=\frac{\sqrt{3}}{72}\psi^{(1)}(\frac56)-\frac{\sqrt{3}\pi^2}{144}
$
where
$\psi^{(1)}$
is a polygamma function
(reference below).
$\begin{array}\\
\sum_{n=1}^\infty \frac{\sin\left(n\frac{\pi}{3}\right)}{(2n+1)^2}
&=\sum_{n=1}^\infty \left(\frac{\sin\left((3n-2)\frac{\pi}{3}\right)}{(2(3n-2)+1)^2}+\frac{\sin\left((3n-1)\frac{\pi}{3}\right)}{(2(3n-1)+1)^2}+\frac{\sin\left(3n\frac{\pi}{3}\right)}{(2(3n)+1)^2}\right)\\
&=\sum_{n=1}^\infty \left(\frac{\sin\left(-2\frac{\pi}{3}\right)}{(6n-3)^2}+\frac{\sin\left(-\frac{\pi}{3}\right)}{(6n-1)^2}+\frac{\sin\left(n\pi\right)}{(6n+1)^2}\right)\\
&=\sum_{n=1}^\infty \left(-\frac{\sqrt{3}/2}{(6n-3)^2}+\frac{\sqrt{3}/2}{(6n-1)^2}\right)\\
&=\frac{\sqrt{3}}{2}\sum_{n=1}^\infty \left(-\frac1{(6n-3)^2}+\frac1{(6n-1)^2}\right)\\
&=\frac{\sqrt{3}}{2}\sum_{n=1}^\infty \left(\frac1{(6n-1)^2}-\frac1{(6n-3)^2}\right)\\
&=\frac{\sqrt{3}}{2}\left(\sum_{n=1}^\infty \frac1{(6n-1)^2}-\frac19\sum_{n=1}^\infty\frac1{(2n-1)^2}\right)\\
&=\frac{\sqrt{3}}{2}\left(\frac1{36}\sum_{n=1}^\infty \frac1{(n-1/6)^2}-\frac19(1-\frac14)\zeta(2)\right)\\
&=\frac{\sqrt{3}}{2}\left(\frac1{36}\sum_{n=0}^\infty \frac1{(n+5/6)^2}-\frac19\frac34\frac{\pi^2}{6}\right)\\
&=\frac{\sqrt{3}}{72}\psi^{(1)}(\frac56)-\frac{\sqrt{3}\pi^2}{144}\\
\end{array}
$
$\psi^{(1)}(z)$
is a polygamma function:
https://en.wikipedia.org/wiki/Polygamma_function
$\psi^{(m)}(z)
=(-1)^{m+1}m!\sum_{n=0}^{\infty} \dfrac1{(z+n)^{m+1}}
$
so that
$\psi^{(1)}(z)
=\sum_{n=0}^{\infty} \dfrac1{(z+n)^2}
$
Also
$\begin{array}\\
\zeta(m)
&=\sum_{n=1}^{\infty} \dfrac1{n^m}\\
&=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}+\sum_{n=1}^{\infty} \dfrac1{(2n)^m}\\
&=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}+\dfrac1{2^m}\sum_{n=1}^{\infty} \dfrac1{n^m}\\
&=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}+\dfrac1{2^m}\zeta(m)\\
\end{array}
$
so
$\zeta(m)(1-2^{-m})
=\sum_{n=1}^{\infty} \dfrac1{(2n-1)^m}
$.
A: Integrate by parts
\begin{align}
\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx
=& \int_0^1 \ln x \>d\left( \tan^{-1} \frac{\sqrt x} {1-x}
-\frac1{\sqrt3 } \tanh^{-1} \frac{\sqrt {3x}} {1+x} \right)\\
=&\frac1{\sqrt3 }I_1 - I_2\tag1
\end{align}
where
\begin{align} 
I_1&=\int_0^1 \frac{\tanh^{-1} \frac{\sqrt {3x}} {1+x}}x dx\\
I_2 & = \int_0^1 \frac{\tan^{-1} \frac{\sqrt {x}} {1-x}}x dx 
= \underset{\sqrt x\to x}{\int_0^1 \frac{\tan^{-1} \sqrt x}x dx }
 + \underset{\sqrt {x^3}\to x}{\int_0^1 \frac{\tan^{-1} \sqrt {x^3}}x }dx \\
&= \left(2+ \frac23 \right) \int_0^1 \frac{\tan^{-1} x}x dx = \frac83G\tag2
\end{align}
Evaluate $I_1$ with $J(a) = \int_0^1 \frac{\tanh^{-1} \frac{2a\sqrt {x}} {1+x}}x dx$
$$J’(a) =\int_0^1 \frac{2(\frac1{\sqrt x}+\sqrt x)dx}{(x+1)^2-(2a\sqrt x)^2}
=\frac{2 \tan^{-1}\frac{\sqrt{(1-a^2)x}}{1-x}\bigg|_0^1} {\sqrt{1-a^2}}=\frac\pi{\sqrt{1-a^2}}
$$
Then
$$I_1 = J(\frac{\sqrt3}2)=\int_0^{ \frac{\sqrt3}2}J’(a)da
=\int_0^{ \frac{\sqrt3}2} \frac\pi{\sqrt{1-a^2}} da =\frac{\pi^2}3\tag3
$$
Plug (2) and (3) into (1) to obtain
$$\int_0^1 \frac{\sqrt x \ln x}{x^2-x+1}dx = \frac{\pi^2}{3\sqrt3}-\frac83G$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\ds{\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} =
{G \over \root{3}} - {\pi^{2} \over 24}} \approx 0.1176:\
{\Large ?}}$. $\ds{G = 0.9159\ldots}$ is the
Catalan Constant.

\begin{align}
&\bbox[5px,#ffd]{\sum_{n = 1}^{\infty}
{\sin\pars{n\pi/3} \over \pars{2n + 1}^{2}}} =
\sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} +
\sum_{n = 6}^{11}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} + \sum_{n = 12}^{17}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} + \cdots
\\[5mm] = &\
\sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over
\pars{2n + 1}^{2}} +
\sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over
\bracks{2n + 2\pars{6} + 1}^{\, 2}} + \sum_{n = 0}^{5}{\sin\pars{n\pi/3} \over
\bracks{2n + 4\pars{6} + 1}^{\, 2}} + \cdots
\\[5mm] = &\
\sum_{n = 0}^{5}\sin\pars{n\,{\pi \over 3}}
\sum_{k = 0}^{\infty}{1 \over \pars{2n + 12k + 1}^{2}} =
{1 \over 144}\sum_{n = 0}^{5}\sin\pars{n\,{\pi \over 3}}\,
\Psi\, '\pars{2n + 1 \over 12}
\\[5mm] = &\
{G \over \root{3}}\ -\
\underbrace{\bracks{80G + \Psi\, '\pars{11 \over 12} -
\Psi\, '\pars{5 \over 12}}{\root{3} \over 288}}
_{\ds{=\ {\pi^{2} \over 24}}}
\end{align}
I'll $\ds{\underline{\mbox{have}}}$ to prove that
$\ds{\color{red}{\Psi\, '\pars{11 \over 12} -
\Psi\, '\pars{5 \over 12} \color{black}{=} 4\root{3}\pi^{2} - 80G}}$
