# Simpler closed form for $\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}$

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!

#### Solutions Collecting From Web of "Simpler closed form for $\sum_{n=1}^\infty\frac{\Gamma\left(n+\frac{1}{2}\right)}{(2n+1)^4\,4^n\,n!}$"

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

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

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.

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