[Largedisplaystyle sum_{n=1}^inftyfrac{Gammaleft(n+dfrac{1}{2}
ight)}{(2n+1)^4\,4^n\,n!}
]
(Largemathbf{Solution:})
First, in view of Legrende's duplication formula,
[egin{align*}
S&=sum_{n=1}^inftyfrac{Gammaleft(n+dfrac{1}{2}
ight)}{(2n+1)^4\,4^n\,n!}\&=2sqrt{pi}sum_{n=1}^inftyfrac{Gamma(2n)}{Gamma(n)\,n!\,(2n+1)^4\, 16^n}
\&
=-frac{sqrt{pi}}{3}int_0^1 ln^3(x)sum_{n=1}^{infty}frac{Gamma(2n)}{Gamma(n)\,n!}left(frac{x^2}{16}
ight)^nmathrm{d}x\
&=-sqrt{pi}-frac{sqrt{pi}}{6}int_0^1frac{ln^3(x)}{sqrt{1-dfrac{x^2}{4}}}mathrm{d}x\&=-sqrt{pi}-frac{sqrt{pi}}{3}int_0^{frac{pi}{6}}ln^3(2sin x)mathrm{d}x
end{align*}]
Claim: for (0<aleq dfrac{pi}{2}),
[egin{align*}
int_0^a ln^3left(frac{sin x}{sin a}
ight)mathrm{d}x&=frac{4a-3pi}{2}a^2ln(2sin a)-frac{3pi}{4}zeta(3)\
&+3left(frac{pi}{2}-a
ight)Releft(frac12 operatorname{Li}_3(e^{2ia})+operatorname{Li}_3(1-e^{2ia})
ight)\
&+3Imleft(frac14operatorname{Li}_4(e^{2ia})+operatorname{Li}_4(1-e^{2ia})
ight)
end{align*}]
Proof: The idea is exactly identical to the proof displayed in this question.
The proof is rather tedious (and obviously inefficient), and ends with a somewhat of a cancellation (implying the existence of a shortcut) , so I omit the boring algebra and outline the main ideas, which can be repeated systematically to obtain closed forms for even higher powers of logsine.
Things to know:
[egin{align*}
ln(2sin x)=ln(1-e^{2ix})+ileft(frac{pi}{2}-x
ight) ag{1}
end{align*}]
[egin{align*}
intfrac{ln^3(1-x)}{x}mathrm{d}x&=ln^3(1-x)ln(x)+3ln^2(1-x) ext{Li}_2(1-x)\
&-6ln(1-x) ext{Li}_3(1-x)+6 ext{Li}_4(1-x) ag{2}
end{align*}]
[egin{align*}
int_0^a xln(2sin x)mathrm{d}x=-frac{a}{2} ext{Cl}_2(2a)-frac14Re ext{Li}_3(e^{2ia})+frac{zeta(3)}{4} ag{3}
end{align*}]
[egin{align*}
int_0^a x^2ln(2sin x)mathrm{d}x=-frac{a^2}{2} ext{Cl}_2(2a)-frac{a}{2}Re ext{Li}_3(e^{2ia})+frac14Im ext{Li}_4(e^{2ia}) ag{4}
end{align*}]
[egin{align*}
int_0^a ln(sin x)mathrm{d}x=-aln2-frac12 ext{Cl}_2(2a) ag{5}
end{align*}]
[egin{align*}
int_0^a ln^2(sin x)mathrm{d}x=frac{a^3}{3}+aln^2 2-aln^2(2sin a)-ln(sin a) ext{Cl}_2(2a)-Im ext{Li}_3(1-e^{2ia}) ag{6}
end{align*}]
(1) is trivial, (2) is not too hard to find, (5) and (6) are shown in the linked answer, and (3),(4) are easily found using (displaystyle ln(2sin x)=-sum_{ngeq1}frac{cos(2xn)}{n}).
It is obvious that since we have (5) and (6), the claim (0) depends on a closed form for (displaystyleint_0^a ln^3(sin x)mathrm{d}x), and the latter may be evaluated in terms of (displaystyleint_0^a ln^3(2sin x)mathrm{d}x).
But, with the help of (1),
[egin{align*}
int_0^a ln^3(2sin x)mathrm{d}x&=Reint_0^a ln^3(1-e^{2ix})mathrm{d}x+3int_0^a ln(2sin x)left(frac{pi}{2}-x
ight)^2mathrm{d}x\
&=frac12Imint_1^{e^{2ia}}frac{ln^3(1-x)}{x} mathrm{d}x+3int_0^a ln(2sin x)left(frac{pi}{2}-x
ight)^2mathrm{d}x
end{align*}]
(Same idea RandomVariable had in this answer.)
Now we employ (2),(3),(4), and (5). Some expressions cancel and claim follows.
This result, together with the fact that (e^{ipi/3}) and (1-e^{ipi/3}) are conjugates, yields
[displaystyle int_0^{frac{pi}{6}} ln^3(2sin x)mathrm{d}x=-frac{pi}{4}zeta(3)-frac94Im ext{Li}_4(e^{ipi/3})
]
and use
[frac{2}{sqrt{3}}Im ext{Li}_4(e^{ipi/3})=sum_{ngeq 0}frac{(-1)^n}{(3n+1)^4}+sum_{ngeq 0}frac{(-1)^n}{(3n+2)^4} =frac{psi^{(3)}left(dfrac13
ight)}{216}-frac{pi^4}{81}
]
Hence we have
[Largeoxed{displaystyle sum_{n=1}^inftyfrac{Gammaleft(n+dfrac{1}{2}
ight)}{(2n+1)^4\,4^n\,n!}= color{Blue}{sqrt{pi}left(frac{pi}{12}zeta(3)+frac{psi^{(3)}left(dfrac13
ight)}{192sqrt3}-frac{pi^4}{72sqrt3}-1
ight)}}
]
It can also be expressed in generalized hypergeometric function with the help of WolframAlpha:
[Largeoxed{displaystyle sum_{n=1}^inftyfrac{Gammaleft(n+dfrac{1}{2}
ight)}{(2n+1)^4\,4^n\,n!}=color{blue}{frac{sqrt{pi}}{648} {_6F_5}left(egin{array}c 1,dfrac32,dfrac32,dfrac32,dfrac32,dfrac32\2,dfrac52,dfrac52,dfrac52,dfrac52end{array}middle|\,frac14
ight)}}
]