<s>[INST] 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.
 [/INST] $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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$\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
$\ds{\color{red}{\Psi\, '\pars{11 \over 12} -
\Psi\, '\pars{5 \over 12} \color{black}{=} 4\root{3}\pi^{2} - 80G}}$
</s>