Nice hypergeometric series involving the silver ratio $\delta_S$
Today we show the proof of this result involving hypergeometric series and the silver ratio $\delta_S= 1+\sqrt{2}$ proposed by @mamekebi \[ {}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] = \int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = \frac{\pi^2}{8} -\frac{\ln^2(1+\sqrt{2})}{2}\] The proof relies on some properties of the rising factorial and some identities of the dilogarithm function.
Proof
Recall the recursion formula for the argument and the degree of the rising factorial: \[(x)_{n+1} = (n+x)(x)_{n} = x(x+1)_{n}\quad \tag{1}\] and the duplication formula \[ (x)_{2n} = 2^{2n} \left(\frac{x}{2}\right)_n\left(\frac{x+1}{2}\right)_n \tag{2} \] Hence \begin{align*} {}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] = &\sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}\left(\frac{1}{2}\right)_{j}}{\left(\frac{3}{2}\right)_{j}\left(\frac{3}{2}\right)_{j}} \frac{(-1)^j}{j!}\\ =&\frac{1}{2}\sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}\left(j+\frac{1}{2}\right)} \frac{(-1)^j}{j!} \quad \textrm{from (1)}\\ =&\sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}\left(2j+1\right)} \frac{(-1)^j}{j!}\\ \end{align*} We previously showed the proof of this series expansion: \begin{align*}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} =& \sum_{j=0}^{\infty} \frac{2^{2j}}{\binom{2j}{j}(2j+1)}(-1)^jx^{2j} = \sum_{j=0}^{\infty} \frac{2^{2j}j!^2}{(2j)!(2j+1)}(-1)^jx^{2j} \\ =& \sum_{j=0}^{\infty} \frac{2^{2j}(1)_{j}(1)_{j}}{(1)_{2j}(2j+1)}(-1)^jx^{2j} \quad \left( (1)_{j} = j!\right)\\ =& \sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}}\frac{(-1)^jx^{2j}}{j!} \quad \textrm{ from (1) and (2)} \end{align*} Integrating from $0$ to $1$ \[\int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = \sum_{j=0}^{\infty} \frac{(1)_{j}(1)_{j}}{\left(\frac{3}{2}\right)_{j}(2j+1)}\frac{(-1)^j}{j!} ={}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] \] \begin{align*} I = \int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = &\int_{0}^{1}\frac{\ln\left(\sqrt{x^2+1}+x\right)}{x\sqrt{1+x^2}} dx\\ =& 2\int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{(w+1)(w-1)}dw \quad \left(w \mapsto \sqrt{x^2+1}+x\right) \\ =& -\int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{1-w}dw - \int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{w+1}dw\\ =& \int_{0}^{-\sqrt{2}}\frac{\ln(1-t)}{t}dw - \int_{1}^{1+\sqrt{2}}\frac{\ln(w)}{w+1}dw \quad \left(t\mapsto 1-w\right) \\ \stackrel{IBP}{=} & \int_{0}^{-\sqrt{2}}\frac{\ln(1-t)}{t}dw - \ln(1+\sqrt{2})\ln(2+\sqrt{2})-\int_{-1-\sqrt{2}}^{-1}\frac{\ln(1-w)}{w}dw\\ =& -\int_{-\sqrt{2}}^{0}\frac{\ln(1-t)}{t}dw - \ln(1+\sqrt{2})\ln(2+\sqrt{2})-\int_{-1-\sqrt{2}}^{0}\frac{\ln(1-w)}{w}dw+\int_{-1}^{0}\frac{\ln(1-w)}{w}dw\\ =& -\operatorname{Li}_{2}(-\sqrt{2})- \ln(1+\sqrt{2})\ln(2+\sqrt{2})-\operatorname{Li}_{2}(-1-\sqrt{2})+\operatorname{Li}_{2}(-1) \end{align*} From the identity \[ \operatorname{Li}_2(-z)-\operatorname{Li}_2(1-z)+\frac{1}{2}\operatorname{Li}_2(1-z^2)=-\frac{{\pi}^2}{12}-\ln z \cdot \ln(z+1)\] If we put $z = \sqrt{2}$ \[ \operatorname{Li}_2(-\sqrt{2})=\operatorname{Li}_2(1-\sqrt{2})-\frac{1}{2}\operatorname{Li}_2(-1)-\frac{{\pi}^2}{12}-\ln \sqrt{2} \cdot \ln(\sqrt{2}+1) \quad \quad (3)\] From the identity \[\operatorname{Li}_2(z) +\operatorname{Li}_2\left(\frac{1}{z}\right) = - \frac{\pi^2}{6} - \frac{\ln^2(-z)}{2}\] If we put $z = 1-\sqrt{2}$
Therefore \[\operatorname{Li}_2\left(-1-\sqrt{2}\right) = -\operatorname{Li}_2(1-\sqrt{2}) - \frac{\pi^2}{6} - \frac{\ln^2(\sqrt{2}-1)}{2} \quad \quad (4)\] Using the fact that \[\operatorname{Li}_{2}(-1)= -\frac{\pi^2}{12} \] and from $(3)$ and $(4)$ we have: \begin{align*} I =& \frac{\pi^2}{8} +\ln \sqrt{2} \cdot \ln(\sqrt{2}+1)- \ln(1+\sqrt{2})\ln(2+\sqrt{2}) + \frac{\ln^2(\sqrt{2}-1)}{2}\\ =& \frac{\pi^2}{8} -\frac{\ln^2(1+\sqrt{2})}{2} \end{align*} \[ \boxed{ {}_{3}F_{2}\left[{1,1,\frac{1}{2} \atop \frac{3}{2}, \frac{3}{2} }, -1\right] = \int_{0}^{1}\frac{\operatorname{arcsinh}(x)}{x\sqrt{1+x^2}} dx = \frac{\pi^2}{8} -\frac{\ln^2(1+\sqrt{2})}{2}}\]
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