Which closed form expression for this series involving Catalan numbers : $\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4^nn^2}\binom{2n}{n}$

Question

Obtain a closed-form for the series: $$\mathcal{S}=\sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{4^nn^2}\binom{2n}{n}$$

From here https://en.wikipedia.org/wiki/List_of_m ... cal_series we know that for $\displaystyle{\left| z \right| \le \frac{1}{4}}$ holds $$\sum\limits_{n = 0}^\infty \binom{2n}{n} z^n = \frac{1}{\sqrt {1 - 4z} }$$

Then \begin{align} &\sum\limits_{n = 0}^\infty \binom{2n}{n} \frac{{{z^n}}}{{{4^n}}} = \frac{1}{\sqrt {1 - z} } \\ &\Rightarrow \sum_{n = 1}^\infty \binom{2n}{n} \frac{z^n}{4^n} = \frac{1}{\sqrt {1 - z} } - 1 \\ &\Rightarrow \sum_{n = 1}^\infty \binom{2n}{n} \frac{z^{n - 1}}{4^n} = \frac{1 - \sqrt {1 - z} }{z \sqrt {1 - z} } \\ &\Rightarrow \sum\limits_{n = 1}^\infty \binom{2n}{n} \frac{y^n}{n 4^n} = \int_0^y {\frac{{1 - \sqrt {1 - z} }} {{z \sqrt {1 - z} }}dz} = 2\left( \log 2 - \log \left(1 + \sqrt {1 - y} \right) \right) \\ &\Rightarrow \sum_{n = 1}^\infty \binom{2n}{n} \frac{y^{n - 1}}{n 4^n} = 2\left( {\frac{{\log 2 - \log \left( {1 + \sqrt {1 - y} } \right)}}{y}} \right) \\ &\Rightarrow \sum_{n = 1}^\infty \binom{2n}{n} \frac{x^n}{n^2 4^n} = 2 \int_0^x {\frac{{\log 2 - \log \left( {1 + \sqrt {1 - y} } \right)}}{y}dy} \end{align}

Answer 1

With $S$ the constant to be computed, we have the following explicit yellow expression: $$ \begin{aligned} S&=\sum\frac{(-1)^{n-1}}{4^n\: n^2}\binom{2n}n\\ &=2\int_0^{-1}(\log2 -\log(1+\sqrt{1-y}))\;\frac{dy}y\qquad\text{ (OP formula)} \\ &=2\int_0^1\log\underbrace{\left(\frac{1+\sqrt{1+y}}2\right)}_{=:t}\;\frac{dy}y \\ & =2\int_1^{(1+\sqrt 2)/2=:a}\frac{2t-1}{t(t-1)}\;\log t\;dt = 2\int_1^a\left(\frac{t}{t(t-1)}+\frac{t-1}{t(t-1)}\right) \;\log t\;dt \\ &= 2\int_1^a\frac{\log t}{t-1}\; dt+ 2\int_1^a\frac{\log t}t\; dt \\ &= -2\Bigg[\operatorname{Li}_2(1-t)\Bigg]_1^a+ \Bigg[\log^2 t\Bigg]_1^a \\ &=-2\operatorname{Li}_2(1-a)+\log^2 a =\bbox[yellow]{-2\operatorname{Li}_2\left(\frac{1-\sqrt2}2\right)+\log^2\left(\frac{1+\sqrt2}2\right)}\ . \end{aligned} $$

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Numerical checks, here pari/gp

? sum(n=1, 10000, (-1)^(n-1)/4^n/n^2 * gamma(2*n+1)/gamma(n+1)^2)
%40 = 0.42996693557993137159907473218062445342
? 
? 2 * intnum(y=0, 1, log( (1 + sqrt(1+y))/2 ) / y) 
%41 = 0.42996693560813697203703367869938799117

? a = (1 + sqrt(2))/2;
 
? 2 * intnum(t=1, a, (2*t - 1)/t/(t - 1) * log(t))
%43 = 0.42996693560813697203703367869938799116
 
? -2*dilog( (1-sqrt(2))/2 ) + log( (1+sqrt(2))/2 )^2
%44 = 0.42996693560813697203703367869938799117

Answer 2

\begin{align*} S &= \sum\limits_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^2 \cdot 4^n} \binom{2n}{n} \ &= -2 \int_0^{-1} \frac{\log 2 - \log (1 + \sqrt{1 - y})}{y} dy \quad (Integration by parts) \ &= 2 \int_{-1}^0 \frac{\log 2 - \log (1 + \sqrt{1 + x})}{-x} dx \quad (y = -x \text{ substitution}) \ &\stackrel{y = 1 + \sqrt{1 + x}}{=} -2 \int_1^0 \frac{\log y}{x} dx \quad (u \text{ substitution}) \ &= 2 \int_0^1 \frac{\log y - \log 2}{x} dx \ &\stackrel{u = 2y}{=} 4 \int_2^{1 + \sqrt{2}} \frac{\log (u/2)}{u (u - 2)} (u - 1) du \quad (u/2 \text{ substitution}) \ &\stackrel{v = y/(1 - y)}{=} 2 \int_1^{(1 + \sqrt{2}) / 2} \frac{\log v}{v (v - 1)} (2v - 1) dv \quad (v \text{ substitution}) \ &= 2 \int_1^{(1 + \sqrt{2}) / 2} \log v \left( \frac{1}{v} + \frac{1}{v - 1} \right) dv \ &= \left[ \log^2 v \right]1^{(1 + \sqrt{2}) / 2} + 2 \int_1^{(1 + \sqrt{2}) / 2} \frac{\log v}{v - 1} dv \ &= \log^2 \left( \frac{1 + \sqrt{2}}{2} \right) + 2 \int_1^{(1 + \sqrt{2}) / 2} \frac{\log v}{v - 1} dv \ &\stackrel{a = 1 + \sqrt{2}}{=} 2 \int_1^a \frac{\log v}{v (1 - v^{-1})} dv \quad (a \text{ substitution}) \ &= 2 \sum{n = 0}^\infty \int_1^a \frac{\log v}{v^{n + 1}} dv \quad (Geometric series expansion) \ &= 2 \int_1^a \frac{\log v}{v} dy + 2 \sum_{n = 1}^\infty \int_1^a \frac{\log v}{v^{n + 1}} dy \ &= \log^2 a + 2 \sum_{n = 1}^\infty \left( \frac{1}{n^2} - \frac{1}{n^2 a^n} - \frac{\log a}{n \cdot a^n} \right) \ &= \log^2 a + \frac{\pi^2}{3} - 2 \text{Li}_2 \left( \frac{2}{1 + \sqrt{2}} \right) \ &\quad + 2 \log \left( \frac{1 + \sqrt{2}}{2} \right) \cdot \log \left( \frac{\sqrt{2} - 1}{\sqrt{2} + 1} \right) \end{align*}