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(This summarizes scattered results from here, here, here and elsewhere. See also this older post.)

I. Cubic

Define $\beta= \tfrac{\Gamma\big(\tfrac56\big)}{\Gamma\big(\tfrac13\big)\sqrt{\pi}}= \frac{1}{48^{1/4}\,K(k_3)}$. Then we have the nice evaluations,

$$\begin{aligned}\frac{3}{5^{5/6}} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-4\big)\\ &=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+4x^3}}\\[1.7mm] &=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{\color{blue}{9+4\sqrt{5}}\,x}}\\[1.7mm] &=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt[3]{9+\cosh x}} \end{aligned}\tag1$$ and, $$\begin{aligned}\frac{4}{7} &=\,_2F_1\big(\tfrac{1}{3},\tfrac{1}{3};\tfrac{5}{6};-27\big)\\ &=\beta\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+27x^3}}\\[1.7mm] &=\beta\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{\color{blue}{55+12\sqrt{21}}\,x}}\\[1.7mm] &=2^{1/3}\,\beta\,\int_0^\infty\frac{dx}{\sqrt[3]{55+\cosh x}} \end{aligned}\tag2$$ 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.

II. Quartic

Define $\gamma= \tfrac{\sqrt{2\pi}}{\Gamma^2\big(\tfrac14\big)}= \frac{1}{2\sqrt2\,K(k_1)}=\frac1{2L}$ with lemniscate constant $L$. Then we have the nice,

$$\begin{aligned}\frac{2}{3^{3/4}} &=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-3\big)\\ &=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+3x^4}}\\[1.7mm] &\overset{\color{red}?}=\gamma\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{\color{blue}{7+4\sqrt{3}}\,x}}\\[1.7mm] &=2^{1/4}\,\gamma\,\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}} \end{aligned}\tag3$$ and, $$\begin{aligned}\frac{3}{5}&=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-80\big)\\ &=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+80x^4}}\\[1.7mm] &\overset{\color{red}?}=\gamma\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{\color{blue}{161+72\sqrt{5}}\,x}}\\[1.7mm] &=2^{1/4}\,\gamma\,\int_0^\infty\frac{dx}{\sqrt[4]{161+\cosh x}} \end{aligned}\tag4$$

with $a=161$ given by Noam Elkies in this comment. (For $4$th roots, I just assumed the equality using the blue radicals based on the ones for cube roots.) Note again the powers of fundamental units, $$U_{3}^2 = \big(2+\sqrt3\big)^2=\color{blue}{7+4\sqrt{3}}$$ $$U_{5}^{12} = \big(\tfrac{1+\sqrt{5}}{2}\big)^{12}=\color{blue}{161+72\sqrt{5}}$$ Just like for the cube roots version, these can't be coincidence.

Questions:

Is it true these observations can be explained by, let $b=2a+1$, then,

$$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[3]{x^2+ax^3}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small2/3} \sqrt[3]{b+\sqrt{b^2-1}\,x}}=2^{1/3}\int_0^\infty\frac{dx}{\sqrt[3]{b+\cosh x}}$$

$$\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+ax^4}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{b+\sqrt{b^2-1}\,x}}=2^{1/4}\int_0^\infty\frac{dx}{\sqrt[4]{b+\cosh x}}$$

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Tito Piezas III
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    Hello, I really admire your work (especially on your blog), and I really want to someday be as much of an expert as you are. May I ask, where were you educated/where did you learn so much amazing number theory? – clathratus Jan 02 '20 at 08:11

2 Answers2

8

Starting from $$ \,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-a\big)=\gamma\,\int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+ax^4}}, $$ $$ (b+\sqrt{b^2-1})^{-1/4}\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{1}{2};\tfrac{2\sqrt{b^2-1}}{b+\sqrt{b^2-1}}\big)={\gamma}\,\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{{b+\sqrt{b^2-1}}\,x}}, $$ (with $\gamma$ defined above) and applying transformations 2.11(4), 2.10(6), 2.11(2) from Erdelyi, Higher transcendental functions, vol. I, to the second hypergeometric function one gets \begin{align} (b+\sqrt{b^2-1})^{-1/4}\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{1}{2};\tfrac{2\sqrt{b^2-1}}{b+\sqrt{b^2-1}}\big)&=b^{-1/4}\,_2F_1\big(\tfrac{1}{8},\tfrac{5}{8};\tfrac{3}{4};\tfrac{{b^2-1}}{b^2}\big)\\ &=\,_2F_1\big(\tfrac{1}{8},\tfrac{1}{8};\tfrac{3}{4};1-b^2\big)\\ &=\,_2F_1\big(\tfrac{1}{8},\tfrac{1}{8};\tfrac{3}{4};-4a(1+a)\big)\\ &=\,_2F_1\big(\tfrac{1}{4},\tfrac{1}{4};\tfrac{3}{4};-a\big), \end{align} where $b=2a+1$, thus proving that $$ \int_0^1 \frac{dx}{\sqrt{1-x}\,\sqrt[4]{x^3+ax^4}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{\small3/4} \sqrt[4]{b+\sqrt{b^2-1}\,x}}. $$

More generally application of the same series of transformations gives $$ {(b+\sqrt{b^2-1})^{-\alpha } \, _2F_1\left(\alpha ,\alpha ;2 \alpha ;\tfrac{2 \sqrt{b^2-1}}{b+\sqrt{b^2-1}}\right)}={\, _2F_1\left(\alpha ,\alpha ;\alpha +\tfrac{1}{2};-a\right)}, $$ i.e. $$ \int_0^1 \frac{dx}{\sqrt{1-x}\,x^{1-\alpha}(1+ax)^{\alpha}}=\int_{-1}^1\frac{dx}{\left(1-x^2\right)^{1-\alpha} (b+\sqrt{b^2-1}\,x)^{\alpha}}. $$ When $\alpha=1/3$ this answers the related question.

Formula 2.12(10) from Erdelyi, Higher transcendental functions, vol. I answers the second equality, namely $$ {\, _2F_1\left(\alpha ,\alpha ;\alpha +\tfrac{1}{2};-a\right)}=2^{\alpha}\frac{\Gamma(\alpha+1/2)}{\sqrt{\pi}\Gamma(\alpha)}\int_0^\infty\frac{dx}{(b+\cosh x)^\alpha}. $$

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Martin Nicholson
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Too long for a comment : In general, for strictly positive values of n we have

$$\begin{align} \sqrt[n]2\int_0^\infty\frac{dx}{\sqrt[n]{\cosh2t~+~\cosh x}} ~&=~ \int_0^1\frac{dx}{\sqrt{1-x}\cdot\sqrt[n]{x^{n-1}~+~x^n\cdot\sinh^2t}} \\\\ ~&=~ \int_{-1}^1\frac{dx}{\sqrt[n]{(1-x^2)^{n-1}}\cdot\sqrt[n]{\cosh2t~+~x\cdot\sinh2t}} \end{align}$$

Lucian
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