[Largeint_0^{large frac{pi}{2}} ln^2left( anfrac{x}{2}
ight){ _3F_2}left(frac12,1,1;frac32,frac32;sin^2 x
ight)mathrm{d}x = frac{5 pi^5}{384}
]
(Largemathbf{Proof:})
The ({;}_3F_2) is just a deception.
[{;}_3F_2left(frac{1}{2},1,1;frac{3}{2},frac{3}{2};sin^2(x)
ight) =sum_{n=0}^inftyfrac{(2sin x)^{2n}}{(2n+1)^2dbinom{2n}{n}}= frac{1}{sin x}int_0^xfrac{ heta}{sin heta}mathrm{d} heta
]
This is obtained by the simple observation
[frac{left(dfrac{1}{2}
ight)_n (1)_n^2}{left(dfrac{3}{2}
ight)_n^2 n!}=frac{4^n}{dbinom{2n}{n} (2n+1)^2}
]
Also, note that (displaystyle frac{mathrm{d}}{mathrm{d}x}lnleft( an frac{x}{2} ight)=frac{1}{sin x}). Therefore, by repeated integration by parts:
[egin{align*}
&int_0^{large frac{pi}{2}} ln^2left( anfrac{x}{2}
ight){ _3F_2}left(frac12,1,1;frac32,frac32;sin^2 x
ight)mathrm{d}x=frac{1}{12}int_0^{frac{pi}{2}}ln^4left( anfrac{x}{2}
ight)mathrm{d}x\
&=frac{1}{6}int_0^1frac{ln^4 x}{1+x^2}mathrm{d}x=frac{1}{6}sum_{n=0}^{infty }left ( -1
ight )^{n}int_{0}^{1}x^{2n}ln^{4}xmathrm{d}x=frac{1}{6} cdot 24sum_{n=0}^{infty }frac{left ( -1
ight )^{n}}{left ( 2n+1
ight )^{5}}\&=frac{1}{6}cdot 24cdot eta left ( 5
ight )=frac{1}{6}cdot 24cdot frac{5pi ^{5}}{1536}=Largeoxed{color{blue}{dfrac{5pi ^{5}}{384}}}
end{align*}]