1. Introduction
Significant progress has been made on the calculations of the
\(\varepsilon\)-expansion of multiloop
Feynman diagram in the past quarter of a century. It turns out that
special values of multiple polylogarithms have played indispensable
roles in these computations.
Many experimental work emerged around the beginning of this century,
for instance, see [1-4], in which a special class of
series emerges. These infinite sums are often called Apéry-type
series (or Apéry-like sums) because the simplest cases
were used by Apéry in his celebrated proof of irrationality of \(\zeta(3)\) in 1979. In particular, he
showed that \[\label{Apery}
\zeta(3)=\dfrac{5}{2}\sum_{n=1}^{\infty}\dfrac{\left(-1\right)^{n-1}}{n^3\binom{2n}{n}}.\]
In his new book [5],
Zhi-Wei Sun listed 820 mathematical conjectures, including several
concerning the Apéry-like sums involving ordinary and hyper-harmonic
numbers, some of which had appeared in his previous paper [6].
The main aim of this breif article is develop an elementary aproach
to prove the following six conjectures of Sun [6,Conjecture 3.1 and 3.2]. Our approach throughout is completely
analytic and the proofs rely on simple yet interesting logarithmic
integral evaluations.
Conjecture 1 ([6,Conjecture 3.1 and 3.2]). The following identities
hold \[\sum_{n=1}^{\infty}\dfrac{1}{n^2\binom{2n}{n}}\left(H_{2n}+\dfrac{2}{3n}\right)=
\zeta(3),\] \[\sum_{n=1}^{\infty}\dfrac{1}{n^2\binom{2n}{n}}\left(H_{2n}+2H_n\right)=
\dfrac{5}{3}\,\zeta(3),\] \[\sum_{n=1}^{\infty}\dfrac{H_{2n}+17H_{n}}{n^2\binom{2n}{n}}=\dfrac{5\pi}{6\sqrt{3}}\left(\psi^{\left(1\right)}\left(\dfrac{1}{3}\right)
– \psi^{\left(1\right)}\left(\dfrac{2}{3}\right)\right),\] \[\sum_{n=1}^{\infty}\dfrac{2^n}{n^2\binom{2n}{n}}\left(2H_{2n}-3H_n+\dfrac{2}{n}\right)=\dfrac{7}{4}\,\zeta(3),\]
\[\sum_{n=1}^{\infty}\dfrac{2^n}{n^2\binom{2n}{n}}\left(6H_{2n}-11H_n+\dfrac{8}{n}\right)=2\pi
G,\] \[\sum_{n=1}^{\infty}\dfrac{2^n}{n^2\binom{2n}{n}}\left(7H_{n}-2H_{2n}-\dfrac{2}{n}\right)=\dfrac{\pi^2}{2}\log2,\]
where \(H_n=\sum_{k=1}^{n}k^{-1}\)
denotes the classical harmonic number, \(G=\sum_{n=0}^{\infty}\left(-1\right)^n\left(2n+1\right)^{-2}\)
denotes the Catalans constant, \(\psi^{\left(1\right)}(z)=\sum
_{{n=0}}^{{\infty}}{(z+n)^{-2}}\) represents the trigamma
function, and \(\zeta(n)=\sum_{k=1}^{\infty}k^{-n}\)
represents the Riemann zeta function.
Our main result in this manuscript is as follows:
Theorem 2. The following identities hold \[\ label{I1}\sum_{n=1}^{\infty}\dfrac{H_n}{n^2\binom{2n}{n}}
=-
\dfrac{1}{9}\,\zeta(3)+ \dfrac{\pi}{18\sqrt{3}}\left(\psi^{\left(1\right)}\left(\dfrac{1}{3}\right)
– \psi^{\left(1\right)}\left(\dfrac{2}{3}\right)\right), \tag{1}\] \[
\sum_{n=1}^{\infty}\dfrac{2^nH_n}{n^2\binom{2n}{n}} =
\dfrac{7}{16}\,\zeta(3) + \dfrac{3}{4}\,\zeta(2)\log2, \tag{2}\] \[\
sum_{n=1}^{\infty}\dfrac{2^nH_{2n}}{n^2\binom{2n}{n}}
= \dfrac{119}{32}\,\zeta(3) + \dfrac{3}{8}\,\zeta(2)\log2 -\pi
G, \tag{3}\] \[\
sum_{n=1}^{\infty}\dfrac{H_{2n}}{n^2\binom{2n}{n}}
=\dfrac{17}{9}\,\zeta(3)- \dfrac{\pi}{9\sqrt{3}}\left(\psi^{\left(1\right)}\left(\dfrac{1}{3}\right)
–
\psi^{\left(1\right)}\left(\dfrac{2}{3}\right)\right), \tag{4}\]
Proof of Theorem 2 is provided in Section 2. Combining Theorem
2
with the remarkable identities proved by I. Zucker [9,Eqn.2.9 and 2.10] in 1985, that is \[\sum_{n=1}^{\infty}\dfrac{1}{n^3\binom{2n}{n}}=\dfrac{\pi}{6\sqrt{3}}\left(\psi^{\left(1\right)}\left(\dfrac{1}{3}\right)
– \psi^{\left(1\right)}\left(\dfrac{2}{3}\right)\right),\] and
\[\sum_{n=1}^{\infty}\dfrac{2^n}{n^3\binom{2n}{n}}=\dfrac{\pi^2}{8}\log2
+ \pi G – \dfrac{35}{16}\,\zeta(3),\] completes the proof of
Conjecture 1.
Beyond this specific case, we hope that the tools used in this study
will be useful to the readers in their endeavour of evaluating more
interesting Apéry-like series and integrals, and they will be an
integral tool in proving remaining unsolved similar conjectures of
Sun.
Proof of Theorem 2
2.1. Proof of identity 1
Start with the following series expansion: \[\
dfrac{\arcsin
z}{\sqrt{1-z^2}}=\dfrac{1}{2}\sum_{n=1}^{\infty}\dfrac{2^{2n}}{n\binom{2n}{n}}\,z^{2n-1},\quad\left|z\right|
< 1. \tag{5}\] Some routine manipulations produce \[\sum_{n=1}^{\infty}\dfrac{H_n}{n^2\binom{2n}{n}} =
– 4\int_0^1 z\log\left(1-4\sin^2 z\right)\mathrm{d}z.\] Next, we
replace \(z \mapsto \arctan z\) to get
\[\
sum_{n=1}^{\infty}\dfrac{H_n}{n^2\binom{2n}{n}} =
4\left(\int_0^{\frac{1}{\sqrt{3}}}\dfrac{\arctan\left(z\right)
\log\left(1+z^2\right)}{1+z^2}\,\mathrm{d}z –
\int_0^{\frac{1}{\sqrt{3}}}\dfrac{\arctan\left(z\right)
\log\left(1-3z^2\right)}{1+z^2}\,\mathrm{d}z\right)
\nonumber\\ = 4\left(\mathfrak{T}_1 –
\mathfrak{T}_2\right).\label{FI}, \tag{6}\] The change of variable \(z=\tan x\) produces \[\
mathfrak{T}_1 =-2
\int_0^{\pi/6}x\log\left(\cos\left(x\right)\right)\mathrm{d}x\nonumber\\= \dfrac{11}{72}\,\zeta(3)
+ \dfrac{1}{6}\log\left(2\right)\zeta(2) –
\dfrac{\pi}{36\sqrt{3}}\left(\psi^{(1)}\left(\dfrac{1}{3}\right) –
\psi^{(1)}\left(\dfrac{2}{3}\right)\right), \tag{7}\] where we have used the Fourier series expansion
of \(\log\left(\cos x\right)\). To
evaluate \(\mathfrak{T}_2\), let \[\mathfrak{F}(\omega)=\int_0^{\omega}\dfrac{\arctan\left(z\right)\log\left(1-\omega^{-2}z^2\right)}{1+z^2}\,\mathrm{d}z,
\quad \omega \in \mathbb{R},\] and observe that \(\mathfrak{T}_2=\mathfrak{F}(1/\sqrt{3})\).
For the reader’s convenience, we will use and restate computation of
\(\mathfrak{F}(\omega)\) from [10]: \[\mathfrak{F}(\omega) = \dfrac{1}{8}\zeta{(3)} –
\dfrac{1}{8}Cl_3(4\alpha) –
\dfrac{1}{2}\,\alpha^{2}\log{\left(\sin^{2}{\alpha}\right)}
-\alpha\operatorname{Cl}_{2}(2\alpha), \tag{8}\] where \(\alpha = \arctan\omega\), \(Cl_n(\vartheta)\) represents the Clausen
function of order \(n\), defined by the
Fourier series \[Cl_{{2m}}(\vartheta)=\sum
_{{k=1}}^{\infty }{\frac {\sin \left(k\vartheta\right)}{k^{{2m}}}},
\quad Cl_{{2m+1}}(\vartheta)=\sum
_{{k=1}}^{\infty}{\frac {\cos\left(k\theta\right)}{k^{{2m+1}}}}.\]
The change of variable \(\arctan z =
\tau\) with \(\omega=\tan\alpha\) produces \[\begin{aligned}
\mathfrak{F}(\omega)&=\int_0^{\omega}\dfrac{\arctan\left(z\right)\log\left(1-\omega^{-2}z^2\right)}{1+z^2}\,\mathrm{d}z
\\&=
\int_{0}^{\tan{\alpha}}\,\frac{\arctan{\left(z\right)}\log{\left(1-z^2\cot^{2}\left(\alpha\right)\right)}}{1+z^{2}}\,\mathrm{d}z
\\&=\int_{0}^{\alpha}\tau\log{\left(1-\cot^{2}{\left(\alpha\right)}\tan^{2}{\left(\tau\right)}\right)}\mathrm{d}\tau\\&=\int_{0}^{\alpha}\tau\log{\left(\frac{\sin^{2}{\left(\alpha\right)}\cos^{2}{\left(\tau\right)}-\cos^{2}{\left(\alpha\right)}\sin^{2}{\left(\tau\right)}}{\sin^{2}{\left(\alpha\right)}\cos^{2}{\left(\tau\right)}}\right)}\mathrm{d}\tau
\\&=\int_{0}^{\alpha}\tau\log{\left(\csc^{2}{\left(\alpha\right)}\sec^{2}{\left(\tau\right)}\sin{\left(\alpha-\tau\right)}\sin{\left(\alpha+\tau\right)}\right)}
\mathrm{d}\tau
\\&
=-\int_{0}^{\alpha}\tau\left(\log{\left(\sin^{2}{\alpha}\right)}+\log{\left(\cos^{2}{\tau}\right)}-\log{\left(\sin{\left(\alpha-\tau\right)}\right)}-\log{\left(\sin{\left(\alpha+\tau\right)}\right)}\right)\mathrm{d}\tau
\\&
=-\int_{0}^{\alpha}\tau\log{\left(\sin^{2}{\alpha}\right)}\mathrm{d}\tau-\int_{0}^{\alpha}\tau\left(2\log{\left(\cos{\tau}\right)}-\log{\left(\sin{\left(\alpha-\tau\right)}\right)}-\log{\left(\sin{\left(\alpha+\tau\right)}\right)}\right)\mathrm{d}\tau
\\& =
-\dfrac{1}{2}\,\alpha^{2}\log{\left(\sin^{2}{\alpha}\right)}-\int_{0}^{\alpha}\int_{0}^{\tau}\left(2\log{\left(\cos{\tau}\right)}-\log{\left(\sin{\left(\alpha-\tau\right)}\right)}-\log{\left(\sin{\left(\alpha+\tau\right)}\right)}\right)\mathrm{d}\vartheta\,\mathrm{d}\tau
\\& =
-\dfrac{1}{2}\,\alpha^{2}\log{\left(\sin^{2}{\alpha}\right)}-\int_{0}^{\alpha}\int_{\vartheta}^{\alpha}\left(2\log{\left(\cos{\tau}\right)}-\log{\left(\sin{\left(\alpha-\tau\right)}\right)}-\log{\left(\sin{\left(\alpha+\tau\right)}\right)}\right)\mathrm{d}\tau\,\mathrm{d}\vartheta
\\&=-\dfrac{1}{2}\,\alpha^{2}\log{\left(\sin^{2}{\alpha}\right)}-\mathfrak{T}_3.
\end{aligned}\] Next, we find that \[\begin{aligned}
\mathfrak{T}_3&=\int_{0}^{\alpha}\int_{\vartheta}^{\alpha}\left(2\log{\left(2\cos{\tau}\right)}-\log{\left(2\sin{\left(\alpha-\tau\right)}\right)}-\log{\left(2\sin{\left(\alpha+\tau\right)}\right)}\right)\mathrm{d}\tau\,\mathrm{d}\vartheta
\nonumber\\&=\int_{0}^{\alpha}\,\bigg{[}2\int_{\vartheta}^{\alpha}\log{\left(2\cos{\tau}\right)}\,\mathrm{d}\tau-\int_{\vartheta}^{\alpha}\log{\left(2\sin{\left(\alpha-\tau\right)}\right)}\,\mathrm{d}\tau-\int_{\vartheta}^{\alpha}\log{\left(2\sin{\left(\alpha+\tau\right)}\right)}\,\mathrm{d}\tau\bigg{]}\mathrm{d}\vartheta
\nonumber \\&=\int_{0}^{\alpha}\,\bigg{[}2\int_{\pi/2-\alpha}^{\pi/2-\vartheta}\log{\left(2\sin{\tau}\right)}\,\mathrm{d}\tau-\int_{\vartheta}^{\alpha
–
\vartheta}\log{\left(2\sin{\tau}\right)}\,\mathrm{d}\tau-\int_{\alpha+\vartheta}^{2\alpha}\log{\left(2\sin{\tau}\right)}\,\mathrm{d}\tau\bigg{]}\mathrm{d}\vartheta
\nonumber \\&
=\dfrac{1}{2}\int_{0}^{\alpha}\,\bigg{[}2\int_{\pi-2\alpha}^{\pi-2\vartheta}\log{\left(2\sin{\dfrac{\tau}{2}}\right)}\mathrm{d}\tau-\int_{0}^{2\alpha
–
2\vartheta}\log{\left(2\sin{\dfrac{\tau}{2}}\right)}\mathrm{d}\tau-\int_{2\alpha+2\vartheta}^{4\alpha}\log{\left(2\sin{\dfrac{\tau}{2}}\right)}\mathrm{d}\tau\bigg{]}\mathrm{d}\vartheta
\nonumber \\&=\dfrac{1}{2}\int_{0}^{\alpha}\bigg{[}\operatorname{Cl}_{2}{\left(2\alpha-2\vartheta\right)}+\operatorname{Cl}_{2}{\left(4\alpha\right)}-\operatorname{Cl}_{2}{\left(2\alpha+2\vartheta\right)}+2\operatorname{Cl}_{2}{\left(\pi-2\alpha\right)}-2\operatorname{Cl}_{2}{\left(\pi-2\vartheta\right)}\bigg{]}\mathrm{d}\vartheta
\nonumber\\&
=\alpha\operatorname{Cl}_{2}{\left(2\alpha\right)}
+\dfrac{1}{2}\int_{0}^{\alpha}\operatorname{Cl}_{2}\left(2\alpha-2\vartheta\right)\mathrm{d}\vartheta-\dfrac{1}{2}\int_{0}^{\alpha}\operatorname{Cl}_{2}\left(2\alpha+2\vartheta\right)\mathrm{d}\vartheta-\int_{0}^{\alpha}\operatorname{Cl}_{2}\left(\pi-2\vartheta\right)\mathrm{d}\vartheta,\nonumber
\end{aligned}\] where we have used the integral representation of
Clausen function of order 2, that is \[\operatorname{Cl}_{2}{\left(\vartheta\right)}=-\int_{0}^{\vartheta}\log{\left(\left|2\sin{\left(\frac{\tau}{2}\right)}\right|\right)}\,\mathrm{d}\tau.\]
Replacing \(\vartheta \mapsto
\dfrac{\vartheta}{2}\) produces \[\begin{aligned}
&\dfrac{1}{2}\int_{0}^{\alpha}\operatorname{Cl}_{2}\left(2\alpha-2\vartheta\right)\mathrm{d}\vartheta-\dfrac{1}{2}\int_{0}^{\alpha}\operatorname{Cl}_{2}\left(2\alpha+2\vartheta\right)\mathrm{d}\vartheta-\int_{0}^{\alpha}\operatorname{Cl}_{2}\left(\pi-2\vartheta\right)\mathrm{d}\vartheta\nonumber
\\&
=\dfrac{1}{4}\int_{0}^{2\alpha}\operatorname{Cl}_{2}{\left(2\alpha-\vartheta\right)}\,\mathrm{d}\vartheta-\dfrac{1}{4}\int_{0}^{2\alpha}\operatorname{Cl}_{2}{\left(2\alpha+\vartheta\right)}\,\mathrm{d}\vartheta-\dfrac{1}{2}\int_{0}^{2\alpha}\operatorname{Cl}_{2}{\left(\pi-\vartheta\right)}\,\mathrm{d}\vartheta
\nonumber\\&=\dfrac{1}{4}\int_{0}^{2\alpha}\operatorname{Cl}_{2}{\left(\vartheta\right)}\,\mathrm{d}\vartheta-\dfrac{1}{4}\int_{2\alpha}^{4\alpha}\operatorname{Cl}_{2}{\left(\vartheta\right)}\,\mathrm{d}\vartheta-\dfrac{1}{2}\int_{\pi
–
2\alpha}^{\pi}\operatorname{Cl}_{2}{\left(\vartheta\right)}\,\mathrm{d}\vartheta
\nonumber\\&=\dfrac{1}{4}\operatorname{Cl}_{3}{\left(0\right)}-\dfrac{1}{4}\operatorname{Cl}_{3}{\left(2\alpha\right)}+\dfrac{1}{4}\operatorname{Cl}_{3}{\left(4\alpha\right)}-\dfrac{1}{4}\operatorname{Cl}_{3}{\left(2\alpha\right)}+\dfrac{1}{2}\operatorname{Cl}_{3}{\left(\pi\right)}-\dfrac{1}{2}\operatorname{Cl}_{3}{\left(\pi-2\alpha\right)}
\nonumber\\&=\dfrac{1}{4}\operatorname{Cl}_{3}{\left(0\right)}+\dfrac{1}{2}\operatorname{Cl}_{3}{\left(\pi\right)}+\dfrac{1}{4}\operatorname{Cl}_{3}{\left(4\alpha\right)}-\dfrac{1}{2}\operatorname{Cl}_{3}{\left(2\alpha\right)}-\dfrac{1}{2}\operatorname{Cl}_{3}{\left(\pi-2\alpha\right)}
= \dfrac{1}{8}\operatorname{Cl}_{3}{\left(4\alpha\right)}
-\dfrac{1}{8}\zeta{\left(3\right)}.\label{claus2} \nonumber
\end{aligned}\] Putting all things together produces identity 8.
Plugging in \(\omega=\frac{1}{\sqrt{3}}\) in identity 8
produces \[\alpha =
\arctan\left(\dfrac{1}{\sqrt{3}}\right) = \dfrac{\pi}{6},\] and
thus \[
\mathfrak{T}_2 =\dfrac{1}{8}\,\zeta(3) + \dfrac{1}{6}\,\zeta(2)\log
2-\dfrac{\pi}{6}\,Cl_2\left(\dfrac{\pi}{3}\right)-\dfrac{1}{8}\,Cl_3\left(\dfrac{2\pi}{3}\right)
,\tag{9}\]
\[\ dfrac{13}{72}\,\zeta(3) + \dfrac{1}{6}\,\zeta(2)\log2 –
\dfrac{\pi}{24\sqrt{3}}\left(\psi^{\left(1\right)}\left(\dfrac{1}{3}\right)
– \psi^{\left(1\right)}\left(\dfrac{2}{3}\right)\right), \tag{10}\] where we have simply used the fact that \[Cl_2\left(\dfrac{\pi}{3}\right)=
\dfrac{1}{4\sqrt{3}}\left(\psi^{\left(1\right)}\left(\dfrac{1}{3}\right)
– \psi^{\left(1\right)}\left(\dfrac{2}{3}\right)\right).\]
Combining equations (6),(7), and (10) finally gives
us the desired result.
2.2. Proof of identity 2
Using the series expansion 5, we find that \[\sum_{n=1}^{\infty}\dfrac{2^nH_n}{n^2\binom{2n}{n}}
=
-4\int_{0}^{\frac{1}{\sqrt{2}}}\dfrac{\log\left(1-2z^2\right)\arcsin\left(z\right)}{\sqrt{1-z^2}}\,\mathrm{d}z.\]
The change of variable \(z = \sin z\)
produces \[\begin{aligned}
\int_{0}^{\frac{1}{\sqrt{2}}}\dfrac{\arcsin\left(z\right)\log\left(1-2z^2\right)}{\sqrt{1-z^2}}\,\mathrm{d}z
&=
\int_{0}^{\pi/4}z\log\left(1-2\sin^2(z)\right)\mathrm{d}z=\int_{0}^{\pi/4}z\log\left(\cos\left(2z\right)\right)\mathrm{d}z.
\end{aligned}\] Therefore we have \[-4\int_{0}^{\frac{1}{\sqrt{2}}}\dfrac{\arcsin\left(z\right)\log\left(1-2z^2\right)}{\sqrt{1-z^2}}\,\mathrm{d}z=-\int_{0}^{\pi/2}z\log\left(\cos
z\right)\mathrm{d}z.\] Next, we use the Fourier series expansion
of \(\log\left(\cos z\right)\) to
finally deduce that \[\int_{0}^{\pi/2}z\log\left(\cos
z\right)\mathrm{d}z = -\dfrac{7}{16}\,\zeta(3) –
\dfrac{3}{4}\,\zeta(2)\log2.\] Putting all things together gives
us the desired result.
2.3. Proof of identity 3
Following similar steps as above gives \[\label{2nH2n1}
\sum_{n=1}^{\infty}\dfrac{2^nH_{2n}}{n^2\binom{2n}{n}}=-\,4\int_0^{\pi/4}z\log\left(1-\sqrt{2}\sin z\right)\mathrm{d}z.\]
Now some routine trigonometric substitutions and simple manipulations
yield \[\begin{aligned}
&\int_0^{\pi/4}z\log\left(1-\sqrt{2}\sin z\right)\mathrm{d}z\\&=\int_0^{\pi/4}z\log\left(\cos
2z\right)\mathrm{d}z-\int_0^{\pi/4}z\log\left(1+\sqrt{2}\sin
z\right)\mathrm{d}z
\\&=\dfrac{1}{2}\int_0^{\pi/4}z\log\left(\cos
2z\right)\mathrm{d}z+\dfrac{1}{2}\int_{-\pi/4}^{\pi/4}z\log\left(1-\sqrt{2}\sin
z\right)\mathrm{d}z
\\&=\int_0^{\pi/4}\left[\left(\dfrac{\pi}{4} –
2z\right)\log\left(2\sqrt{2}\right) + \dfrac{z}{2}\log\left(\cos
2z\right)-\dfrac{1}{4}\,\pi\log\left(\cot z\right) – 2z\log\left(\tan
z\right)\right]\mathrm{d}z
\\&=\dfrac{1}{2}\int_0^{\pi/4}z\log\left(\cos
2z\right)\mathrm{d}z – \dfrac{\pi}{4}\int_0^{\pi/4}\log\left(\cot
z\right)\mathrm{d}z – 2\int_0^{\pi/4}z\log\left(\tan z\right)\mathrm{d}z
\\&=\dfrac{1}{8}\int_0^{\pi/2}z\log\left(\cos
z\right)\mathrm{d}z – \dfrac{\pi}{4}\int_0^{\pi/4}\log\left(\cot
z\right)\mathrm{d}z – 2\int_0^{\pi/4}z\log\left(\tan z\right)\mathrm{d}z
\\&=-\dfrac{7}{128}\,\zeta(3)-\dfrac{7}{8}\,\zeta(3)-\dfrac{3}{32}\,\zeta(2)\log2+\dfrac{\pi
G}{2}-\dfrac{\pi G}{4}
=-\dfrac{119}{128}\,\zeta(3)-\dfrac{3}{32}\,\zeta(2)\log2+\dfrac{\pi
G}{4},
\end{aligned}\] where we have simply used the fact that \[\int_0^{\pi/4}z\log\left(\tan z\right)\mathrm{d}z
= \dfrac{7}{16}\,\zeta(3)-\dfrac{\pi G}{4}, \quad
\int_0^{\pi/2}z\log\left(\cos z\right)\mathrm{d}z =
-\dfrac{7}{16}\,\zeta(3) -\dfrac{3}{4}\,\zeta(2)\log2,\] and the
integral representation of Catalan’s constant, that is \[G=\int_0^{\pi/4}\log\left(\cot
z\right)\mathrm{d}z.\] Therefore, putting all things together
gives us the desired result.
2.4. Proof of identity 4
Using the series exansion 5, we get \[\sum_{n=1}^{\infty}\dfrac{H_{2n}}{n^2\binom{2n}{n}}=-4\int_0^{\pi/6}z\log\left(1-2\sin\left(z\right)\right)\mathrm{d}z.\]
Now following the same steps as we did above for a similar integral
yields \[\int_0^{\pi/6}z\log\left(1-2\sin
z\right)\mathrm{d}z=\dfrac{17}{36}\,\zeta(3)- \dfrac{\pi}{36\sqrt{3}}\left(\psi^{\left(1\right)}\left(\dfrac{1}{3}\right)
– \psi^{\left(1\right)}\left(\dfrac{2}{3}\right)\right).\]
Combining all things together gives us the desired result.
