# 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!

• $\displaystyle\sum_{n=0}^\infty\frac{\Big(n-\frac12\Big)!}{n!}\cdot\frac{x^{2n+1}}{2n+1}~=~\sqrt\pi\cdot\arcsin x.~$ Now, by repeatedly dividing and integrating with regard to x three times, and letting $x=\dfrac12$, we arrive at an alternate expression for S. Aug 31 '14 at 17:54
• This is direct if we eliminate the very first term of $$\small \, _5F_4\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)=\frac{\pi \zeta (3)}{12}+\frac{\psi ^{(3)}\left(\frac{1}{3}\right)}{6912 \sqrt{3}}-\frac{\psi ^{(3)}\left(\frac{2}{3}\right)}{6912 \sqrt{3}}+\frac{\psi ^{(3)}\left(\frac{1}{6}\right)}{6912 \sqrt{3}}-\frac{\psi ^{(3)}\left(\frac{5}{6}\right)}{6912 \sqrt{3}}$$ Oct 8 '20 at 15:24
• Generalizations $$\small \, _7F_6\left(\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2};\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2},\frac{3}{2};\frac{1}{4}\right)=\frac{7 \pi ^3 \zeta (3)}{864}+\frac{\pi \zeta (5)}{16}+\frac{\psi ^{(5)}\left(\frac{1}{3}\right)}{6635520 \sqrt{3}}-\frac{\psi ^{(5)}\left(\frac{2}{3}\right)}{6635520 \sqrt{3}}+\frac{\psi ^{(5)}\left(\frac{1}{6}\right)}{6635520 \sqrt{3}}-\frac{\psi ^{(5)}\left(\frac{5}{6}\right)}{6635520 \sqrt{3}}$$ Oct 8 '20 at 15:26

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,

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

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

• I've also worked on generalizations. Another interesting way to generalize: writing $(2n+1)^{1/a}$ instead of $(2n+1)^4$. Oct 10 '15 at 9:33
• 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. 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? May 28 '18 at 9:39

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. $$S = \sqrt{\pi}\left(\frac{\pi}{12}\zeta(3)+\frac{1}{192\sqrt3}\psi^{(3)}\left(\tfrac13\right)-\frac{\pi^4}{72\sqrt3}-1\right).$$

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

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

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

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

and $$\small{\,_2F_1\left(\begin{array}{c}\tfrac12,\tfrac12\\ \tfrac32\end{array}\middle|\tfrac14\right)} = \frac{\pi}3$$, then,

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