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I'm trying to find a closed form of this sum: $$S=\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}.\tag{1}$$ WolframAlpha gives a large expressions containing multiple generalized hypergeometric functions, that is quite difficult to handle. After some simplification it looks as follows: $$S=\frac{\pi^{3/2}}{3}-\sqrt{\pi}-\frac{\sqrt{\pi}}{324}\left[9\,_3F_2\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)\\+3\,_4F_3\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)+\,_5F_4\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)\right].\tag{2}$$ I wonder if there is a simpler form. Elementary functions and simpler special funtions (like Bessel, gamma, zeta, polylogarithm, polygamma, error function etc) are okay, but not hypergeometric functions.

Could you help me with it? Thanks!

X.C.
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    $\displaystyle\sum_{n=0}^\infty\frac{\Big(n-\frac12\Big)!}{n!}\cdot\frac{x^{2n+1}}{2n+1}~=~\sqrt\pi\cdot\arcsin x.~$ Now, by repeatedly dividing and integrating with regard to x three times, and letting $x=\dfrac12$, we arrive at an alternate expression for S. – Lucian Aug 31 '14 at 17:54
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    This is direct if we eliminate the very first term of $$\small , _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)=\frac{\pi \zeta (3)}{12}+\frac{\psi ^{(3)}\left(\frac{1}{3}\right)}{6912 \sqrt{3}}-\frac{\psi ^{(3)}\left(\frac{2}{3}\right)}{6912 \sqrt{3}}+\frac{\psi ^{(3)}\left(\frac{1}{6}\right)}{6912 \sqrt{3}}-\frac{\psi ^{(3)}\left(\frac{5}{6}\right)}{6912 \sqrt{3}}$$ – Infiniticism Oct 08 '20 at 15:24
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    Generalizations $$\small , _7F_6\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)=\frac{7 \pi ^3 \zeta (3)}{864}+\frac{\pi \zeta (5)}{16}+\frac{\psi ^{(5)}\left(\frac{1}{3}\right)}{6635520 \sqrt{3}}-\frac{\psi ^{(5)}\left(\frac{2}{3}\right)}{6635520 \sqrt{3}}+\frac{\psi ^{(5)}\left(\frac{1}{6}\right)}{6635520 \sqrt{3}}-\frac{\psi ^{(5)}\left(\frac{5}{6}\right)}{6635520 \sqrt{3}}$$ – Infiniticism Oct 08 '20 at 15:26
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    We can also express $S$ simply as, $$S = \sqrt{\pi}\left(-1+\frac{\pi}{12}\zeta(3)+\frac{27}{32}\operatorname{Cl}_4\big(\tfrac23\pi\big)\right)$$ where $\operatorname{Cl}_n(x)$ is the Clausen function. For derivation, see fourth answer below. – Tito Piezas III Nov 03 '23 at 03:22
  • Since $$16\psi^{(3)}\big(\tfrac13\big)-\psi^{(3)}\big(\tfrac23\big) =\psi^{(3)}\big(\tfrac16\big)\ \psi^{(3)}\big(\tfrac13\big)+16\psi^{(3)}\big(\tfrac23\big) =\psi^{(3)}\big(\tfrac56\big)$$ and similar relations for higher levels, then the $\big(\tfrac16\big)$ and $\big(\tfrac56\big)$ are really not needed. – Tito Piezas III Nov 04 '23 at 08:22

4 Answers4

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First, in view of Legrende's duplication formula, $$S=\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}=2\sqrt{\pi}\sum_{n=1}^\infty\frac{\Gamma(2n)}{\Gamma(n)\,n!\,(2n+1)^4\, 16^n} \\=-\frac{\sqrt{\pi}}{3}\int_0^1 \ln^3(x)\sum_{n=1}^{\infty}\frac{\Gamma(2n)}{\Gamma(n)\,n!}\left(\frac{x^2}{16}\right)^ndx\\ =-\sqrt{\pi}-\frac{\sqrt{\pi}}{6}\int_0^1\frac{\ln^3(x)}{\sqrt{1-x^2/4}}dx=-\sqrt{\pi}-\frac{\sqrt{\pi}}{3}\int_0^{\frac{\pi}{6}}\ln^3(2\sin x)dx$$

Claim: for $0<a\leq \frac{\pi}{2}$,

$$\int_0^a \ln^3\left(\frac{\sin x}{\sin a}\right)dx\tag{0}=\frac{4a-3\pi}{2}a^2\ln(2\sin a)-\frac{3\pi}{4}\zeta(3)+3\left(\frac{\pi}{2}-a\right)\Re\left(\frac12 \operatorname{Li}_3(e^{2ia})+\operatorname{Li}_3(1-e^{2ia})\right)+3\Im\left(\frac14\operatorname{Li}_4(e^{2ia})+\operatorname{Li}_4(1-e^{2ia})\right) $$

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: $$\ln(2\sin x)=\ln(1-e^{2ix})+i\left(\frac{\pi}{2}-x\right) \tag{1}$$ $$\small\int\frac{\ln^3(1-x)}{x}dx=\ln^3(1-x)\ln(x)+3\ln^2(1-x)\text{Li}_2(1-x)-6\ln(1-x)\text{Li}_3(1-x)+6\text{Li}_4(1-x) \tag{2}$$ $$\int_0^a x\ln(2\sin x)dx=-\frac{a}{2}\text{Cl}_2(2a)-\frac14\Re\text{Li}_3(e^{2ia})+\frac{\zeta(3)}{4}\tag{3}$$ $$\int_0^a x^2\ln(2\sin x)dx=-\frac{a^2}{2}\text{Cl}_2(2a)-\frac{a}{2}\Re\text{Li}_3(e^{2ia})+\frac14\Im\text{Li}_4(e^{2ia})\tag{4}$$ $$\int_0^a \ln(\sin x)dx=-a\ln2-\frac12 \text{Cl}_2(2a)\tag{5}$$ $$\int_0^a \ln^2(\sin x)dx=\frac{a^3}{3}+a\ln^2 2-a\ln^2(2\sin a)-\ln(\sin a)\text{Cl}_2(2a)-\Im\text{Li}_3(1-e^{2ia})\tag{6}$$

$(1)$ 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 $\,\,\ln(2\sin x)=-\sum_{n\geq1}\frac{\cos(2xn)}{n}$.

It is obvious that since we have $(5)$ and $(6)$, the claim $(0)$ depends on a closed form for $\displaystyle\int_0^a \ln^3(\sin x)dx$, and the latter may be evaluated in terms of $\displaystyle\int_0^a \ln^3(2\sin x)dx$.

But, with the help of $(1)$, $$\int_0^a \ln^3(2\sin x)dx=\Re\int_0^a \ln^3(1-e^{2ix})dx+3\int_0^a \ln(2\sin x)\left(\frac{\pi}{2}-x\right)^2dx\\ =\frac12\Im\int_1^{e^{2ia}}\frac{\ln^3(1-x)}{x} dx+3\int_0^a \ln(2\sin x)\left(\frac{\pi}{2}-x\right)^2dx$$

(Same idea @RandomVariable had in this answer.)

Now we employ $(2),(3),(4),$ and $(5)$. Some expressions cancel and claim follows.$\square $

This result, together with the fact that $e^{i\pi/3}$ and $1-e^{i\pi/3}$ are conjugates, yields $\displaystyle \int_0^{\frac{\pi}{6}} \ln^3(2\sin x)dx=-\frac{\pi}{4}\zeta(3)-\frac94\Im\text{Li}_4(e^{i\pi/3})$, and

$$S=\sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{9}{12}\Im\text{Li}_4(e^{i\pi/3})-1\right)$$

This form is equivalent to @user153012's form, as $$\frac{2}{\sqrt{3}}\Im\text{Li}_4(e^{i\pi/3})=\sum_{n\geq 0}\frac{(-1)^n}{(3n+1)^4}+\sum_{n\geq 0}\frac{(-1)^n}{(3n+2)^4} \\=\frac{\psi^{(3)}\left(\frac13\right)}{216}-\frac{\pi^4}{81}$$

Also, as noted in the comments in the linked question, this may be used to write a closed form for a certain hypergeometric function.

This serves as a generalisation for the series, because $\displaystyle \sum_{n=1}^{\infty} \frac{\Gamma(n+1/2)}{(2n+1)^4 n!}a^{2n}=-\sqrt{\pi}\left(1+\frac1{6a}\int_0^{\sin^{-1} a}\ln^3\left(\frac{\sin x}{a}\right)dx\right)$

As an example, using closed forms for trilogarithms displayed in this post, we have $$\int_0^{\frac{\pi}{4}}\ln^3(\sqrt{2}\sin x)dx=-\frac{\pi^3}{128}\ln2-\frac{3\pi}{8}\zeta(3)+\frac34\beta(4)+3\Im\text{Li}_4(1-i)$$

where $\beta(4)=\Im\text{Li}_4(i)$ is a value of Dirichlet's beta function.

Or equivalently, $$\sum_{n=1}^{\infty} \frac{\Gamma\left(n+\frac12\right)}{(2n+1)^4\,2^n\,n!}=-\sqrt{\pi}-\frac{\sqrt{2\pi}}{6}\left(-\frac{\pi^3}{128}\ln2-\frac{3\pi}{8}\zeta(3)+\frac34\beta(4)+3\Im\text{Li}_4(1-i)\right)$$

Noam Shalev - nospoon
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    I've also worked on generalizations. Another interesting way to generalize: writing $(2n+1)^{1/a}$ instead of $(2n+1)^4$. – user153012 Oct 10 '15 at 09:33
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    Dear nospoon, I would like to cite your last identity in a paper I am writing about hypergeometric functions and Euler sums; would you like to be mentioned by your real name? If so, please let me know. – Jack D'Aurizio May 26 '18 at 18:57
  • @JackD'Aurizio Wow, Jack, what a great honor! I would like to have my real name cited. How may I contact you in private for further details? – Noam Shalev - nospoon May 28 '18 at 09:39
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Another possible closed form of $S$ is the following. It containts also a generalized hypergeometric function, but just one.

$$S = \frac{\sqrt{\pi}}{648} {_6F_5}\left(\begin{array}c\ 1,\frac32,\frac32,\frac32,\frac32,\frac32\\2,\frac52,\frac52,\frac52,\frac52\end{array}\middle|\,\frac14\right).$$

WolframAlpha's simplification gives back your form.

user153012
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By now, I've found a closed-form by doing some integral evaluation, a lot of hypergeometric, polylogarithm and polygamma manipulation. $$ S = \sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{1}{192\sqrt3}\psi^{(3)}\left(\tfrac13\right)-\frac{\pi^4}{72\sqrt3}-1\right). $$

user153012
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    Since $$\psi^{(3)}\left(\tfrac13\right)= \tfrac83\pi^4+162\sqrt3 \operatorname{Cl}_4\big(\tfrac23\pi\big)$$ where $\operatorname{Cl}_n(x)$ is the Clausen function, then we can further simplify as, $$S = \sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{27}{32}\operatorname{Cl}_4\big(\tfrac23\pi\big)-1\right)$$ – Tito Piezas III Nov 03 '23 at 03:01
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The OP gives the evaluation

$$S=\frac{\pi^{3/2}}{3}-\sqrt{\pi}-\frac{\sqrt{\pi}}{324}\left[9\,_3F_2\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)\\+3\,_4F_3\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)+\,_5F_4\left(\begin{array}{c}\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2},\tfrac{3}{2}\\\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2},\tfrac{5}{2}\end{array}\middle|\tfrac{1}{4}\right)\right]$$

We can simplify this further. Since $\small{\,_2F_1\left(\begin{array}{c}\tfrac12,\tfrac12\\ \tfrac32\end{array}\middle|\tfrac14\right)} = \frac{\pi}3$ and,

$$\frac1{36}\,_3F_2\left(\begin{array}{c}\tfrac32,\tfrac32,\tfrac32\\ \tfrac52,\tfrac52\end{array}\middle|\tfrac14\right) = -\,_3F_2\left(\begin{array}{c}\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32\end{array}\middle|\tfrac14\right) + \,_2F_1\left(\begin{array}{c}\tfrac12,\tfrac12\\ \tfrac32\end{array}\middle|\tfrac14\right) $$

$$\frac1{108}\,_4F_3\left(\begin{array}{c}\tfrac32,\tfrac32,\tfrac32,\tfrac32\\ \tfrac52,\tfrac52,\tfrac52\end{array}\middle|\tfrac14\right) = -\,_4F_3\left(\begin{array}{c}\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac14\right) + \,_3F_2\left(\begin{array}{c}\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32\end{array}\middle|\tfrac14\right) $$

$$\frac1{324}\,_5F_4\left(\begin{array}{c}\tfrac32,\tfrac32,\tfrac32,\tfrac32,\tfrac32\\ \tfrac52,\tfrac52,\tfrac52,\tfrac52\end{array}\middle|\tfrac14\right) = -\,_5F_4\left(\begin{array}{c}\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac14\right) + \,_4F_3\left(\begin{array}{c}\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac14\right) $$

then,

$$S=\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!} = -\sqrt{\pi}+\sqrt{\pi}\,_5F_4\left(\begin{array}{c}\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac14\right) \approx 0.0028056$$

(Added 2023): Borrowing insights from user153012, we can then express the last $_pF_q$ simply as,

$$_5F_4\left(\begin{array}{c}\tfrac12,\tfrac12,\tfrac12,\tfrac12,\tfrac12\\ \tfrac32,\tfrac32,\tfrac32,\tfrac32\end{array}\middle|\tfrac14\right) = \frac{\pi}{12}\zeta(3)+\frac{27}{32}\operatorname{Cl}_4\big(\tfrac23\pi\big)$$

where $\operatorname{Cl}_n(x)$ is the Clausen function. Hence,

$$S = \sqrt{\pi}\left(-1+\frac{\pi}{12}\zeta(3)+\frac{27}{32}\operatorname{Cl}_4\big(\tfrac23\pi\big)\right)$$

Tito Piezas III
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