Interesting closed form for $\int_0^{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{3}+\sin^2{\theta}\right)^{\frac{1}{3}}}\;d\theta$

Question

Some time ago I used a formal approach to derive the following identity:

$$\int_0^{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{3}+\sin^2{\theta}\right)^{\frac{1}{3}}}\;d\theta=\frac{3^{\frac{1}{12}}\pi\sqrt{2}}{AGM(1+\sqrt{3},\sqrt{8})}\tag{1}$$

where $AGM$ is the arithmetic-geometric mean. Wolfram Alpha does not tell me whether this is correct, but it does appear to be accurate to many decimal places. I have three questions:

1. Can anyone verify whether $(1)$ is in fact correct?
2. Is there a way of generalizing $(1)$ to integrals of the form $\int_0^{\frac{\pi}{2}}\left(a+\sin^2{\theta}\right)^{-\frac{1}{3}}\;d\theta$ or is this integral more special? My derivation (see below) appears to only work for $a=\frac{1}{3}$.
3. There is a superficial similarity between $(1)$ and elliptic integrals (e.g. the $AGM$ evaluation); is there a way to transform this integral into an elliptic integral that I have missed, or is it merely a coincidence that an integral of this form is the reciprocal of an $AGM$?

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Derivation: I have put this here in case it helps to see where I am coming from; I apologize for its length. I began by using a multiple integration trick of squaring the integral and converting to polar coordinates to evaluate $\int_0^\infty e^{-x^6}dx=\frac{1}{6}\Gamma(\frac{1}{6})$ as follows:

$$\left[\int_0^\infty e^{-x^6}\;dx\right]^2=\int_0^\infty\int_0^{\frac{\pi}{2}}re^{-r^6(\cos^6\theta\;+\;\sin^6\theta)}\;d\theta\;dx={\int_0^\infty re^{-r^6}\int_0^{\frac{\pi}{2}}e^{3r^6\cos^2\theta\sin^2\theta}\;d\theta\;dx}$$

$$=\int_0^\infty re^{-r^6}\int_0^{\frac{\pi}{2}}e^{\frac{3r^6}{4}\sin^22\theta}\;d\theta\;dx={\int_0^\infty re^{-r^6}\int_0^{\frac{\pi}{2}}e^{\frac{3r^6}{4}\cos^2\theta}\;d\theta\;dx}$$

I then made use of the following formula (see here):

$$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}x^n=\frac{2}{\pi}\int_0^{\frac{\pi}{2}}e^{4x\cos^2\theta}\;d\theta\tag{2}$$

Using $(2)$ and formally interchanging integration and summation we get:

$$\frac{\Gamma(\frac{1}{6})^2}{36}=\int_0^\infty re^{-r^6}\int_0^{\frac{\pi}{2}}e^{4\left(\frac{3r^6}{16}\right)\cos^2\theta}\;d\theta\;dx=\frac{\pi}{2}\int_0^\infty re^{-r^6}\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}\left(\frac{3r^6}{16}\right)^n\;dx$$

$$=\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}\left(\frac{3}{16}\right)^n \int_0^\infty r^{6n+1}e^{-r^6}\;dx=\frac{\pi}{12}\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}\left(\frac{3}{16}\right)^n \Gamma\left(n+\frac{1}{3}\right)$$

I then used Laplace transform identities and $(2)$, freely interchanging integrals and sums, to write:

$$\sum_{n=0}^{\infty}\frac{(2n)!}{(n!)^3}\frac{\Gamma\left(n+\frac{1}{3}\right)}{s^{n+\frac{1}{3}}}=L\left[\sum_{n=0}^\infty \frac{(2n)!}{(n!)^3}t^{n-\frac{2}{3}}\right](s)={\frac{2}{\pi}L\left[t^{-\frac{2}{3}}\int_0^\frac{\pi}{2}e^{4t\cos^2\theta}\;d\theta\right](s)}={\frac{2}{\pi}\int_0^\frac{\pi}{2}L\left[t^{-\frac{2}{3}}e^{4t\cos^2\theta}\right](s)\;d\theta}={\frac{2}{\pi}\int_0^\frac{\pi}{2}\frac{\Gamma(\frac{1}{3})}{(s-4\cos^2\theta)^{\frac{1}{3}}}\;d\theta}$$

Accordingly, since $\frac{4}{3}-\cos^2\theta=\frac{1}{3}+\sin^2{\theta}$ we can deduce that:

$$\frac{\Gamma(\frac{1}{6})^2}{36}=\frac{\Gamma(\frac{1}{3})}{6}\left(\frac{4}{3}\right)^\frac{1}{3}\int_0^\frac{\pi}{2}\frac{1}{(\frac{1}{3}+\sin^2\theta)^{\frac{1}{3}}}\;d\theta$$

Reflection and duplication give $\Gamma(\frac{1}{6})=2^{-\frac{1}{3}}\sqrt{\frac{3}{\pi}}\Gamma(\frac{1}{3})^2$ and hence we have the following identity:

$$\int_0^{\frac{\pi}{2}}\frac{1}{\left(\frac{1}{3}+\sin^2{\theta}\right)^{\frac{1}{3}}}\;d\theta=\frac{3^\frac{1}{3}\Gamma(\frac{1}{3})^3}{2^\frac{7}{3}\pi}\tag{3}$$

while $(1)$ may be obtained by using the following identity (see here):

$$\Gamma\left(\frac{1}{6}\right)=\frac{2^\frac{14}{9}3^\frac{1}{3}\pi^\frac{5}{6}}{AGM(1+\sqrt{3},\sqrt{8})^\frac{2}{3}}$$

This completes the derivation; I cannot see how a method like this (especially with the conversion to polar coordinates) could be used to give results more general than $(1)$ and $(3)$.

Answer 1

This is answer to question 3.

Expanding into the powers of $\frac{\cos^2x}{s+4}$ and integrating termwise one obtains $$ \int_0^{\pi/2}\frac{1}{\sqrt[3]{s+4 \sin ^2x}}dx=\int_0^{\pi/2}\frac{1}{\sqrt[3]{s+4 -4\cos ^2x}}dx=\frac{\pi \, _2F_1\left(\frac{1}{2},\frac{1}{3};1;\frac{4}{s+4}\right)}{2 \sqrt[3]{s+4}}. $$ Use Pfaff's transformation to write $$ {}_2F_1\left(\frac{1}{2},\frac{1}{3};1;\frac{4}{s+4}\right)=\sqrt{\frac{s+4}{s}} \, _2F_1\left(\frac{1}{2},\frac{2}{3};1;-\frac{4}{s}\right). $$ By transformation 2.11(5) from Erdelyi, Higher transcendental functions $$ _2F_1\left(\frac{1}{2},\frac{2}{3};1;-\frac{4}{s}\right)=\left(\frac{2}{\sqrt{\frac{4}{s}+1}+1}\right)^{4/3} \, _2F_1\left(\frac{2}{3},\frac{2}{3};1;\left(\frac{1-\sqrt{1+\frac{4}{s}}}{\sqrt{1+\frac{4}{s}}+1}\right)^2\right) $$ Now apply Pfaff's transformation $$ {}_2F_1\left(\frac{2}{3},\frac{2}{3};1;\left(\tfrac{1-\sqrt{1+\frac{4}{s}}}{\sqrt{1+\frac{4}{s}}+1}\right)^2\right)=\left(\tfrac{4 \sqrt{\frac{4}{s}+1}}{\left(\sqrt{\frac{4}{s}+1}+1\right)^2}\right)^{-2/3} \, _2F_1\left(\frac{1}{3},\frac{2}{3};1;-\tfrac{\left(\sqrt{\frac{s+4}{s}}-1\right)^2}{4 \sqrt{\frac{s+4}{s}}}\right) $$ It is known that this last hypergeometric function is elliptic integral. Introduce the following parametrization from Ramanujan's Notebooks, PART V, formula 5.17 $$ s=-2+\frac{2i}{3 \sqrt{3}}\frac{\left(2+2p-p^2\right) \left(1+4p+p^2\right) \left(1-2p-2 p^2\right)}{p \left(1-p^2\right) (p+2) (2 p+1)},\tag{1} $$ then $$ \sqrt{2 p+1} \, _2F_1\left(\frac{1}{3},\frac{2}{3};1;-\frac{\left(\sqrt{\frac{s+4}{s}}-1\right)^2}{4 \sqrt{\frac{s+4}{s}}}\right)=\left(p^2+p+1\right) \, _2F_1\left(\frac{1}{2},\frac{1}{2};1;\frac{p^3 (p+2)}{2 p+1}\right), $$ and after combining all the formulas

$$ \int_0^{\pi/2}\frac{1}{\sqrt[3]{s+4 \sin ^2x}}dx=\frac{\pi \left(p^2+p+1\right) \, _2F_1\left(\frac{1}{2},\frac{1}{2};1;\frac{p^3 (p+2)}{2 p+1}\right)}{2\sqrt{2 p+1} \sqrt[6]{s (s+4)}} $$ with parametrization (1).

As an example $p=-\frac{1}{2}+\frac{3}{\sqrt[3]{2}}-\frac{3}{2^{2/3}}+\frac{i}{2} \sqrt{3 \left(13-4 \sqrt[3]{2}-5\cdot 2^{2/3}\right)}$, then $s=\frac43$ and $$ \int_0^{\pi/2}\frac{1}{\sqrt[3]{\frac13+\sin ^2x}}dx=\frac{3^{1/3} \pi \left(p^2+p+1\right) \, _2F_1\left(\frac{1}{2},\frac{1}{2};1;\frac{p^3 (p+2)}{2 p+1}\right)}{2^{4/3}\sqrt{2 p+1} } $$

A more convenient parametrization in terms of eta-quotients can be obtained similar to this question:

Define $\tau=\frac{1+n\sqrt{-3}}{2}$,$~w=-\frac{1}{\tau+1}$,$~\alpha=\frac{1}{4\sqrt{27}}\left(e^{\frac{\pi i}{4}} \sqrt{27} \left(\frac{\eta (3 w)}{\eta (w)}\right)^3-e^{-\frac{\pi i}{4} }\left(\frac{\eta (3 w)}{\eta (w)}\right)^{-3}\right)^2$, and $~r=\alpha(\alpha+1),$ then $$ \begin{aligned}\int_0^{\pi/2}\frac{1}{\sqrt[3]{\alpha+\sin ^2x}}dx &=-\frac{\pi w\sqrt{3} ~i }{2\sqrt[6]{r}}\left(\frac{\eta (2 w) \eta^6 (3 w)}{\eta^2 (w) \eta^3 (6 w)}+4\frac{\eta (w)\eta^6 (6 w) }{\eta^2 (2 w) \eta^3 (3 w)}\right)\\[2.5mm] &=-\frac{\pi w\sqrt{3} ~i }{2\sqrt[6]{r}}\left(\frac{\big(\vartheta_4(0,q^6)\big)^3}{\vartheta_4(0,q^2)}+\frac{\big(\vartheta_2(0,q^3)\big)^3}{\vartheta_2(0,q)}\right)\\[2.5mm] &=-\frac{\pi w\sqrt{3} ~i }{2\sqrt[6]{r}}\;_2F_1\large\left(\tfrac13,\tfrac23;1;\tfrac{2r\,+\,(1+2\alpha)\sqrt{r}}{4r}\right)\end{aligned} $$ with Jacobi theta functions $\vartheta_n(0,q)$ and nome $q=e^{\pi i\, w}$.

For example $n=3$ gives $\alpha=\frac{1}{3}$.

Answer 2

Making the substitution $\displaystyle \,\, \cos x = \operatorname{sech} t,$ the integral is transformed into $$I=\int_0^{\pi/2} \left(\frac{4}{3}-\cos^2 x\right)^{-1/3} \, dx = \int_0^\infty \left(\frac{4}{3}-\operatorname{sech}^2 t\right)^{-1/3} \, \operatorname{sech} t \, dt.$$ Now, using the triple-angle formula $\displaystyle \, \, \cosh(3 z) =4 \, \cosh^3 z -3 \, \cosh z,$ we can simplify it to $$I= \int_0^\infty \left( \dfrac{\cosh(3 t)}{3 \, \cosh^3 t} \right)^{-1/3} \, \operatorname{sech} t \, dt \\\\= 3^{1/3} \int_0^{\infty} (\operatorname{sech} 3 t)^{1/3} \, dt.$$

See for example here how we can easily show that $$\int_0^{\infty} (\operatorname{sech} t)^{s} dt = \frac12 B\left(\frac{s}{2},\frac12\right).$$

This gives your integral as $$ I = 3^{-2/3} \cdot \frac12 B\left(\frac16,\frac12\right)$$

which you can bring to your equivalent representation using the functional equations of the Gamma function.

Answer 3

(Too long for a comment.)

Expanding on a concluding remark by Nemo, there is an interesting connection between the OP's, $$\int_0^{\pi/2}\frac1{\sqrt[3]{\color{blue}{\alpha}+\sin^2 x}}dx=\frac{\pi}{2\,\sqrt[3]{\alpha+1}}\,_2F_1\Big(\frac12,\frac13;1;\,\frac1{\color{blue}{\alpha}+1}\Big)\tag1$$ and the integral involved in this post, $$\int_0^{\infty}\frac1{\sqrt[3]{\color{blue}{1+2\alpha}+\cosh x}}dx=2^{2/3}\,3^{1/4}K(k_3)\;_2F_1\Big(\frac13,\frac13;\frac56;-\color{blue}{\alpha}\Big)\tag2$$ Equations $(1),(2)$ are valid for arbitrary $\alpha$. But if it is chosen such that, $$\color{blue}{\alpha} =\alpha(\tau)= \frac1{4\sqrt{27}}\big(\lambda^3-\sqrt{27}\lambda^{-3}\big)^2$$ where, $$\lambda =\frac{\eta\big(\frac{\tau+1}3\big)}{\eta(\tau)},\quad \tau=\frac{1+N\sqrt{-3}}2$$ then there is a beautifully simple relationship between $(1),(2)$ as,

$$\frac{N+1}{2^{1/3}\,3^{1/2}}\,\int_0^{\pi/2}\frac1{\sqrt[3]{\color{blue}{\alpha}+\sin^2 x}}dx=\int_0^{\infty}\frac1{\sqrt[3]{\color{blue}{1+2\alpha}+\cosh x}}dx\tag3$$

The closed-form for the RHS (via its hypergeometric equivalent) is already given in this answer, so that automatically leads to the form for the LHS as well. Some rational values are,

$$\alpha\big(\tfrac{1+3\sqrt{-3}}2\big)=\large\tfrac13$$ $$\alpha\big(\tfrac{1+5\sqrt{-3}}2\big)=4$$ $$\alpha\big(\tfrac{1+7\sqrt{-3}}2\big)=27$$

which will be algebraic for integer $N>1$. The first one was re-discovered by the OP and lead to the nice evaluations,

$$\int_0^{\pi/2}\frac1{\sqrt[3]{\tfrac13+\sin^2 x}}dx=3^{1/12}\,K(k_3)$$ $$\int_0^{\pi/2}\frac1{\sqrt[3]{4+\sin^2 x}}dx=\frac{3^{3/4}}{5^{5/6}}\,K(k_3)$$ $$\int_0^{\pi/2}\frac1{\sqrt[3]{27+\sin^2 x}}dx=\frac{3^{3/4}}7\,K(k_3)$$