Closed form for $\sum_{n=0}^\infty\frac{\Gamma\left(n+\tfrac14\right)}{2^n\,(4n+1)^2\,n!}$

Question

I was experimenting with hypergeometric-like series and discovered the following conjecture (so far confirmed by more than $5000$ decimal digits): $$\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\tfrac14\right)}{2^n\,(4n+1)^2\,n!}\stackrel{\color{gray}?}=$$ $$\frac{\Gamma\!\left(\tfrac14\right)\sqrt[4]2}{192}\left[\vphantom{\huge|}6\sqrt{2}\left(2\pi\ln2-\ln^22-8\operatorname{Li}_2\left(\tfrac1{\sqrt2}\right)\right)+3\psi^{(1)}\!\left(\tfrac18\right)-48G+\left(\vphantom{\large|}7\sqrt2-6\right)\pi^2\right]$$ where $G$ is the Catalan constant, $\operatorname{Li}_2(x)$ is the dilogarithm and $\psi^{(1)}(x)$ is the trigamma function. Could you suggest any ideas how to prove it?

---

To see what approach I use to find conjectures like this, see my another question.

---

Update: I've found a generalization of this conjecture. See the corresponding Mathematica expression here. Hopefully, it can be simplified.

Answer 1

First note that $$\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\tfrac14\right)}{2^n\,(4n+1)^2\,n!}=\Gamma\left(\frac14\right){}_3F_2\biggl[ \begin{array}{c}\frac14,\frac14,\frac14 \\ \frac54,\frac54\end{array};\frac12\biggr]$$

We will obtain an "elementary" expression for ${}_3F_2\biggl[ \begin{array}{c}\frac14,\frac14,\frac14 \\ \frac54,\frac54\end{array};z\biggr]$ with arbitrary $z$ using as a main tool the differentiation formula \begin{align}\left(z\frac{d}{dz}+\beta_k-1\right){}_pF_q\biggl[ \begin{array}{c}\alpha_1,\ldots,\alpha_p \\ \beta_1,\ldots,\beta_k,\ldots,\beta_q\end{array};z\biggr]&=\\ =\left(\beta_k-1\right) {}_pF_q\biggl[ \begin{array}{c}\alpha_1,\ldots,\alpha_p \\ \beta_1,\ldots,\beta_k-1,\ldots,\beta_q\end{array};z\biggr]&. \tag{$\spadesuit$}\end{align}

Lemma 1. We have $${}_2F_1\biggl[ \begin{array}{c}\frac14,\frac14 \\ \frac54\end{array};z\biggr]= \frac{z^{-\frac14}}4\left[e^{\frac{\pi i}4}\ln\frac{1+e^{-\frac{\pi i}4}t(z)}{1-e^{-\frac{\pi i}4}t(z)}+e^{-\frac{\pi i}4}\ln\frac{1+e^{\frac{\pi i}4}t(z)}{1-e^{-\frac{\pi i}4}t(z)}\right], \tag{$\clubsuit$}$$ where $t(z)=\left(\frac{z}{1-z}\right)^{\frac14}$.

Proof. Setting in ($\spadesuit$) $p=2$, $q=1$, $\alpha_1=\alpha_2=\frac14$, $\beta_1=\frac54$, we get $$\left(z\frac{d}{dz}+\frac14\right){}_2F_1\biggl[ \begin{array}{c}\frac14,\frac14 \\ \frac54\end{array};z\biggr]=\frac14{}_2F_1\biggl[ \begin{array}{c}\frac14,\frac14 \\ \frac14\end{array};z\biggr]=\frac{\left(1-z\right)^{-\frac14}}{4}.$$ The resulting 1st order ODE $zy'+\frac y4=\frac{\left(1-z\right)^{-\frac14}}{4}$ can be integrated by variation of integration constant: setting $y(z)=C(z)z^{-\frac14}$, one obtains $$C'(z)=\frac{z^{-\frac34}\left(1-z\right)^{-\frac14}}{4}\qquad \Longrightarrow\quad C(z)=\frac14\int z^{-\frac34}\left(1-z\right)^{-\frac14}dz.$$ The antiderivative can be computed in terms of elementary functions: setting $$z=\frac{t^4}{1+t^4},\qquad 1-z=\frac{1}{1+t^4},\qquad dz=\frac{4t^3dt}{(1+t^4)^2},\qquad t=\left(\frac{z}{1-z}\right)^{\frac14},$$ we get $$C(z)=\int\frac{dt}{1+t^4}=\frac14\left[e^{\frac{\pi i}4}\ln\frac{1+e^{-\frac{\pi i}4}t}{1-e^{-\frac{\pi i}4}t}+e^{-\frac{\pi i}4}\ln\frac{1+e^{\frac{\pi i}4}t}{1-e^{-\frac{\pi i}4}t}\right]+\operatorname{const}.$$ Fixing the integration constant via the condition ${}_2F_1\biggl[ \begin{array}{c}\frac14,\frac14 \\ \frac54\end{array};0\biggr]=1$ (in fact it suffices to know that ${}_2F_1$ is regular as $z\to 0$), we arrive at the representation ($\clubsuit$). In Prudnikov et al there is a formula (7.3.2.66, Vol. III) from which one should in principle be able to derive the same result after transformation of parameters. $\square$

Lemma 2. We have \begin{align}{}_3F_2\biggl[ \begin{array}{c}\frac14,\frac14,\frac14 \\ \frac54,\frac54\end{array};z\biggr]&=\frac{z^{-\frac14}}{4}\Biggl\{\int_0^{t(z)}\frac{\ln\left(1+t^{-4}\right)dt}{1+t^4} +\Biggr.\tag{$\heartsuit$} \\ &+\Biggl.\frac{\ln z}{4}\left[e^{\frac{\pi i}4}\ln\frac{1+e^{-\frac{\pi i}4}t(z)}{1-e^{-\frac{\pi i}4}t(z)}+e^{-\frac{\pi i}4}\ln\frac{1+e^{\frac{\pi i}4}t(z)}{1-e^{-\frac{\pi i}4}t(z)}\right]\Biggr\}, \end{align} where $t(z)=\left(\frac{z}{1-z}\right)^{\frac14}$.

Proof. The same procedure: setting in ($\spadesuit$) $p=3$, $q=2$, $\alpha_1=\alpha_2=\alpha_3=\frac14$, $\beta_1=\beta_2=\frac54$, we get $$\left(z\frac{d}{dz}+\frac14\right){}_3F_2\biggl[ \begin{array}{c}\frac14,\frac14 ,\frac14\\ \frac54,\frac54\end{array};z\biggr]=\frac14{}_3F_2\biggl[ \begin{array}{c}\frac14,\frac14,\frac14 \\ \frac54,\frac14\end{array};z\biggr]=\frac14{}_2F_1\biggl[ \begin{array}{c}\frac14,\frac14\\ \frac54\end{array};z\biggr],$$ which in turn implies that $${}_3F_2\biggl[ \begin{array}{c}\frac14,\frac14 ,\frac14\\ \frac54,\frac54\end{array};z\biggr]=\frac{z^{-\frac14}}{4}\int_0^{z}s^{-\frac34}{}_2F_1\biggl[ \begin{array}{c}\frac14,\frac14\\ \frac54\end{array};s\biggr]\,ds.$$ Substituting into this expression the representation found in Lemma 1 and making the change of variables $s\to t(s)=\left(\frac{s}{1-s}\right)^{\frac14}$, we arrive at $${}_3F_2\biggl[ \begin{array}{c}\frac14,\frac14 ,\frac14\\ \frac54,\frac54\end{array};z\biggr]=\frac{z^{-\frac14}}{4}\int_0^{t(z)}\left[e^{\frac{\pi i}4}\ln\frac{1+e^{-\frac{\pi i}4}t}{1-e^{-\frac{\pi i}4}t}+e^{-\frac{\pi i}4}\ln\frac{1+e^{\frac{\pi i}4}t}{1-e^{-\frac{\pi i}4}t}\right]\frac{dt}{t\left(1+t^4\right)}.$$ This clearly can be integrated in terms of dilogarithms. It is convenient to integrate first by parts using that the derivative of the expression in the square brackets is $\frac4{1+t^4}$, which yields ($\heartsuit$). $\square$

Corollary. Hypergeometric function ${}_3F_2\biggl[ \begin{array}{c}\frac14,\frac14,\frac14 \\ \frac54,\frac54\end{array};z\biggr]$ has an explicit expression in terms of elementary functions and dilogarithms, which should give the original statement as a particular case corresponding to $z=\frac12$.

Proof. It suffices to compute in explicit form the integral from the first line of ($\heartsuit$). Since the anti-derivative (check e.g. WolframAlpha) is given by a rather long expression, it is not written here. The resulting formula, however, should simplify to linked expression from the update of @VladimirReshetnikov. $\square$

Answer 2

(Too long for a comment.) Note that,

$$\sum_{n=0}^\infty\frac{\Gamma\!\left(n+\tfrac14\right)}{2^n\,(4n+1)^2\,n!}=A=B$$ $$A=\frac{\Gamma\!\left(\tfrac14\right)\sqrt[4]2}{192}\left[\vphantom{\huge|}6\sqrt{2}\left(2\pi\ln2-\ln^22-8\operatorname{Li}_2\left(\tfrac1{\sqrt2}\right)\right)+3\psi^{(1)}\!\left(\tfrac18\right)-48G+\left(\vphantom{\large|}7\sqrt2-6\right)\pi^2\right]$$ $$B=\frac{\Gamma\!\left(\tfrac14\right)\sqrt[4]2}{192}\left[\vphantom{\huge|}6\sqrt{2}\left(2\pi\ln2-\ln^22-8\operatorname{Li}_2\left(\tfrac1{\sqrt2}\right)\right)\color{red}-3\psi^{(1)}\!\left(\color{red}{\tfrac58}\right)\color{red}+48G+\left(\vphantom{\large|}7\sqrt2\color{red}+6\right)\pi^2\right]$$

Since,

$$\psi^{(1)}\!\left(\tfrac18\right)+\psi^{(1)}\!\left(\tfrac78\right)=2\pi^2(2+\sqrt2)$$ $$\psi^{(1)}\!\left(\tfrac38\right)+\psi^{(1)}\!\left(\tfrac58\right)=2\pi^2(2-\sqrt2)$$

Or in general,

$$\psi^{(1)}\!\left(k\right)+\psi^{(1)}\!\left(1-k\right)=\pi^2\csc^2(k\pi)$$

then one can use any of the arguments $\tfrac18,\tfrac38,\tfrac58,\tfrac78$.