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I was trying to find the closed form for the series $$ S = \sum_{n = 0}^{\infty} \frac{1}{\left(2n + 1\right)^{3}\binom{n - 1/2}{n}} \approx 1.12269 $$ Although it is a bit messy, we can represent this series as a hypergeometric function $$ S = \mbox{}_{4}\!\operatorname{F}_3\left(1,1,\frac{1}{2},\frac{1}{2};\ \frac{3}{2},\frac{3}{2},\frac{3}{2};\ 1\right) $$ which allows us to see that it is an integer or half-integer away from known sums $$ \mbox{}_{4}\!\operatorname{F}_{3}\left(1,1,\frac{1}{2},\frac{3}{2};\ \frac{3}{2},\frac{3}{2},\frac{3}{2};\ 1\right) = 2{\rm G} $$ $$ \mbox{}_{4}\!\operatorname{F}_{3}\left(1,1,\frac{1}{2},\frac{1}{2};\ \frac{3}{2},\frac{3}{2},1;\ 1\right) = \frac{\pi^{2}}{8} $$ where $\rm G$ is the Catalan's Constant.

So, there's probably a way to connect this series $S$ to the value of a known one, maybe using the integral transform, but the recurrence relations I get don't quite work.

Could this be done or is there something wrong ?.

Felix Marin
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Nikitan
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    $$\sum _{n=0} ^{\infty} \frac{1}{(2n+1)^3 \binom{n-1/2}{n}}=\int_0^\infty\frac{\ln(1+x^2)}{2\sqrt{1+x^2}}\Big(\frac\pi2-\arctan x\Big)dx$$ $$=\int_0^\infty\Big(\frac\pi2-\arctan (\sinh x)\Big)\ln(\cosh x)dx=1.12269...$$ I have some doubts that the closed form exists... – Svyatoslav Jul 26 '24 at 15:49
  • That integral’s a little disheartening… But if there’s no closed form, then there’s no connection between their hypergeometric functions which seems weird, considering they differ by an integer – Nikitan Jul 26 '24 at 17:00
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    @Svyatoslav, it took me two weeks but I solved the integral! $$I = \frac{3}{16} \pi ^3 + \frac{1}{4} \pi \log ^2 2 -4 \Im \operatorname{Li} _3 (1+i) \approx 1.12269$$ – Nikitan Aug 11 '24 at 16:31
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    This is fantastic to get a closed answer! My congratulations! – Svyatoslav Aug 11 '24 at 16:41
  • @Loading-146Complete It is possible to find a closed form in terms of complex polylogarithms: ${4F_3}{\left(\frac12,\frac12,1,1;\frac32,\frac32,\frac32;1\right)}=-\frac{\pi^{3}}{32}-\frac{\pi}{8}\ln^{2}{\left(2\right)}+4,\Im{\left[\operatorname{Li}{3}{\left(\frac{1+i}{2}\right)}\right]}$ – David H Aug 31 '24 at 02:46

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Let's rewrite the series in a more standard way $$\sum_{n = 0}^{\infty} \frac{1}{\left(2n + 1\right)^m\binom{n - 1/2}{n}} = \sum_{n = 0}^{\infty} \frac{4^n}{\left(2n + 1\right)^m\binom{2n}{n}}$$ Now define $$f_m(x) = \sum_{n = 0}^{\infty} \frac{(2x)^{2n}}{\left(2n + 1\right)^m\binom{2n}{n}}$$ So we are interested in $f_3(1)$. Notice that $$f_{m+1}(x)=\frac{1}{x}\int _0 ^x f_m(t)dt$$ We'll use the fact that $$f_1(x) = \frac{\arcsin x}{x \sqrt{1-x^2}}$$ Therefore $$f_3(y)= \frac{1}{y}\int_0 ^y\frac{1}{x} \int _0 ^x \frac{\arcsin t}{t \sqrt{1-t^2}}dxdt$$ And so $$f_3(1)=\int_0 ^1\frac{1}{x} \int _0 ^x \frac{\arcsin t}{t \sqrt{1-t^2}}{dx}{dt} = \int _0 ^1 \left( \int _t ^1 \frac{1}{x} \frac{\arcsin t}{t \sqrt{1-t^2}}{dx} \right){dt}=\\ -\int _0 ^1 \frac{\arcsin t}{t \sqrt{1-t^2}} \log t ~ dt = -\int _0 ^{\pi/2} \frac{u}{\sin u}\log(\sin u) ~ {du} \quad (t \rightarrow \sin u)$$ Now we do a Weierstrass substitution $u \rightarrow 2 \arctan v$ and the integral becomes $$- 2\int _0 ^1 \frac{\arctan v}{v} \log\left( \frac{2v}{1+v^2}\right){dv} =\\ -2\log2 \int _0 ^1 \frac{\arctan v}{v}{dv} -2 \int _0 ^1 \frac{\arctan v \log v}{v}{dv} +2 \int _0 ^1 \frac{\arctan v \log (1+v^2)}{v}{dv}$$ Now I reference an article to supplant my lack of integration skill, so I apologise for that! Using the $\operatorname{QLI}$ function defined in this paper on logarithmic integrals we can rewrite them in the following way $$f_3(1) = -2\log2 \operatorname{QLI}(5;2) -2\operatorname{QLI}(25;5) + 2\operatorname{QLI}(45;5)$$ The article has a table of values for the $\operatorname{QLI}$ function, and after some simplifications we get $$\boxed{f_3(1) = \frac{3}{16} \pi ^3 + \frac{1}{4}\pi(\log 2)^2 - 4 ~ \Im \operatorname{Li} _3 (1 + i)}$$

Nikitan
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