# Closed-forms for $\int_0^\infty\frac{dx}{\sqrt{55+\cosh x}}$ and $\int_0^\infty\frac{dx}{\sqrt{45\big(23+4\sqrt{33}\big)+\cosh x}}$

(This summarizes results for cube roots from here and here. The fourth root version is this post.)

Define $\beta= \tfrac{\Gamma\big(\tfrac56\big)}{\Gamma\big(\tfrac13\big)\sqrt{\pi}}=\frac1{B\big(\tfrac{1}{3},\tfrac{1}{2}\big)}$ with beta function $B(a,b)$. Then we have the nice evaluations,

\begin{aligned}\frac{3}{5^{5/6}} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-4\big)\\ &=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^2+4x^3}}\\[1.7mm] &=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt{\color{blue}{9+4\sqrt{5}}\,x}}\\[1.7mm] &=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt{9+\cosh x}} \end{aligned}\tag1 and, \begin{aligned}\frac{4}{7} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-27\big)\\ &=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^2+27x^3}}\\[1.7mm] &=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt{\color{blue}{55+12\sqrt{21}}\,x}}\\[1.7mm] &=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt{55+\cosh x}} \end{aligned}\tag2 Note the powers of fundamental units, $$U_{5}^6 = \big(\tfrac{1+\sqrt{5}}{2}\big)^6=\color{blue}{9+4\sqrt{5}}$$ $$U_{21}^3 = \big(\tfrac{5+\sqrt{21}}{2}\big)^3=\color{blue}{55+12\sqrt{21}}$$ Those two instances can't be coincidence.

Question: Is it true this observation can be explained by, let $b=2a+1$, then, $$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt{x^2+ax^3}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt{b+\sqrt{b^2-1}\,x}}=2^{1/3}\int_0^\infty\frac{dx}{\sqrt{b+\cosh x}}$$

Example: We assume it is true and use one of Noam Elkies' results as, $$\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6}; -a\big) = \frac{6}{11^{11/12}\, U_{33}^{1/4}}$$ where $a=\sqrt{11}\,(U_{33})^{3/2}$ with fundamental unit $U_{33}=23+4\sqrt{33}$. Since $b=2a+1=45\,U_{33}$, we then have the nice integral,

$$2^{1/3}\beta\,\int_0^\infty\frac{dx}{\sqrt{45\big(23+4\sqrt{33}\big)+\cosh x}}=\frac{6}{11^{11/12}\,U_{33}^{1/4}}=0.255802\dots$$

where $\beta= \tfrac{\Gamma\big(\tfrac56\big)}{\Gamma\big(\tfrac13\big)\sqrt{\pi}}.\,$ So is it true in general?

• I observe that Mathematica gives easily this one: $\int_0^\infty \frac{dx}{\sqrt{1+\cosh x}}=\frac{2^{2/3}\,\sqrt{3\pi}\, \Gamma\big(\tfrac76\big)}{\Gamma\big(\tfrac23\big)}.$ – Olivier Oloa Dec 4 '16 at 9:12
• Possibly relevant: math.stackexchange.com/questions/1379472/… and math.stackexchange.com/questions/1326557/… – nospoon Dec 4 '16 at 10:21
• @OlivierOloa: If $b=1$, then $a=\color{blue}0$. Let $\beta$ be defined as above. Then $2^{1/3} \beta \int_0^\infty \frac{dx}{\sqrt{1+\cosh x}} = 1$ since $\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};\color{blue}0 \big)=1$. – Tito Piezas III Dec 4 '16 at 15:09
• @nospoon: I've summarized some related results in this question. – Tito Piezas III Dec 5 '16 at 3:36

I've noticed that for $$I = \int_0^\infty \frac{dx}{\sqrt{b + \cosh(x)}}$$ if we take a Mellin transform with respect to $b$, we get $$\mathcal{M}_b[I](s) = \frac{\Gamma(\frac{1}{3}-s)\Gamma(s)}{\Gamma(\frac{1}{3})}\int_0^\infty \frac{\text{sech}^{-s}{x}}{\cosh^{1/3}(x)}\;dx$$ $$\mathcal{M}_a[I](s) = \frac{\sqrt{\pi}\Gamma(\frac{1}{3}-s)\Gamma(\frac{1}{6}-\frac{s}{2})\Gamma(s)}{2\Gamma(\frac{1}{3})\Gamma(\frac{2}{3}-\frac{s}{2})}, \; \Re(s)<\frac{1}{3}$$ $$I(s) = \mathcal{M}^{-1}_s\left[\frac{\sqrt{\pi}\Gamma(\frac{1}{3}-s)\Gamma(\frac{1}{6}-\frac{s}{2})\Gamma(s)}{2\Gamma(\frac{1}{3})\Gamma(\frac{2}{3}-\frac{s}{2})}\right](b)$$ $$I(s) = \frac{\Gamma(\frac{1}{6})^2}{2^{5/3}\Gamma(\frac{1}{3})}\;_2F_1\left(\frac{1}{6},\frac{1}{6},\frac{1}{2},b^2\right) - \frac{b \Gamma(\frac{2}{3})^2}{2^{2/3}\Gamma(\frac{1}{3})}\;_2F_1\left(\frac{2}{3},\frac{2}{3},\frac{3}{2},b^2\right)$$ $$2^{1/3}\beta I(s) = \frac{\sqrt{3}}{2^{2/3}}\;_2F_1\left(\frac{1}{6},\frac{1}{6},\frac{1}{2},b^2\right) - \frac{2\sqrt{\pi}b \Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6})^2}\;_2F_1\left(\frac{2}{3},\frac{2}{3},\frac{3}{2},b^2\right)$$ so you are looking for $a$ and $b$ such that satisfy $$\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-a\big) = \frac{\sqrt{3}}{2^{2/3}}\;_2F_1\left(\frac{1}{6},\frac{1}{6},\frac{1}{2},b^2\right) - \frac{2\sqrt{\pi}b \Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6})^2}\;_2F_1\left(\frac{2}{3},\frac{2}{3},\frac{3}{2},b^2\right)$$ i.e. $a=4,b=9$ and $a=27,b=55$, and would need to prove $$\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-a\big) = \frac{\sqrt{3}}{2^{2/3}}\;_2F_1\left(\frac{1}{6},\frac{1}{6},\frac{1}{2},(2a+1)^2\right) - \frac{2\sqrt{\pi}(2a+1) \Gamma(\frac{5}{6})}{\Gamma(\frac{1}{6})^2}\;_2F_1\left(\frac{2}{3},\frac{2}{3},\frac{3}{2},(2a+1)^2\right)$$ The residues if you subtract the two and plot are around machine epsilon. I hope this is helpful.