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


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 $AGM$?

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):


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:


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:


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


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 $(3)$.

  • 2
    $\begingroup$ Testing your result numerically in Maple, it's correct to 100 decimal places. $\endgroup$
    – quasi
    Jan 21, 2017 at 6:04
  • 1
    $\begingroup$ A partial result: your integral is equivalent to $$3\sqrt[3]\frac34\int_{\sqrt[3]\frac43}^{\infty }\frac{\mathrm d w}{\sqrt[3]{w^3-1} \sqrt{3 w^3-4}}$$ $\endgroup$ Jan 21, 2017 at 6:47
  • 1
    $\begingroup$ $$\int_{0}^{+\infty}\frac{dx}{\sqrt{(x^2+a^2)(x^2+b^2)}}=\frac{\pi}{2\,\text{AGM}(a,b)}.$$ $\endgroup$ Jan 21, 2017 at 12:30
  • 4
    $\begingroup$ It and others can be expressed as,$$\int_0^{\pi/2}\frac1{\sqrt[3]{\tfrac13+\sin^2 x}}dx=3^{1/12}\,K(k_3)$$ $$\int_0^{\pi/2}\frac1{\sqrt[3]{4+\sin^2 x}}dx=\frac{3^{3/4}}{5^{5/6}}\,K(k_3)$$ $$\int_0^{\pi/2}\frac1{\sqrt[3]{27+\sin^2 x}}dx=\frac{3^{3/4}}7\,K(k_3)$$ with elliptic integral singular value $\displaystyle K(k_3) = \frac{3^{1/4}\,\Gamma^3\big(\tfrac13\big)}{2^{7/3}\,\pi}$. See answer below. $\endgroup$ Jan 21, 2017 at 16:12
  • 2
    $\begingroup$ @J.M.isn'tamathematician: It turns out this integral is connected to another integral I asked about before. $\endgroup$ Jan 21, 2017 at 16:15

3 Answers 3


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.

  • $\begingroup$ @Nemo I understood his questions as seeking both a generalization, and a verifciation of this specific case. $\endgroup$ Jan 21, 2017 at 10:17
  • $\begingroup$ Thank you for this answer +1. I am interested to see that the integral can be converted to a similar representation to that which Jack D'Aurizio used to derive the $\Gamma(\frac{1}{6})$ identity I quoted in my answer. $\endgroup$
    – Anon
    Jan 22, 2017 at 23:07

(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,

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

  • $\begingroup$ +1. Thank you for this nice answer. This was the sort of nice connection to existing theory I was looking for. It seems $\alpha=\frac{1}{3}$ is not alone in having a nice answer of this form. I'll take a while to go through everything here! $\endgroup$
    – Anon
    Jan 22, 2017 at 23:11
  • 2
    $\begingroup$ This is a really good answer. I had no idea what I was getting into when I posted my question (I hadn't studied the hypergeometric function much) but there is some very beautiful theory here and in the linked post. $\endgroup$
    – Anon
    Jan 23, 2017 at 1:17

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 $\alpha=\frac{1}{3}$.

  • $\begingroup$ The ratio of the integral of this post and the other post, if $\alpha$ is used, turns out to be a simple algebraic form dependent on $\tau$. $\endgroup$ Jan 21, 2017 at 16:44
  • $\begingroup$ @TitoPiezasIII , yes, it is no surprize because both are proportional to the same elliptic integral $K\Big(\sqrt{\tfrac{p^3 (2+p)}{1+2 p}}\Big)$. $\endgroup$
    – Nemo
    Jan 21, 2017 at 16:56
  • 1
    $\begingroup$ For the conclusion, I added some equivalent forms as Jacobi thetas and $_2F_1$. I hope it's ok. By the way, I like how the form $r=\alpha(1+\alpha)$ always turns up. $\endgroup$ Jan 22, 2017 at 4:41
  • $\begingroup$ Many thanks for this awesome answer to my third question; the elliptic integral transformation certainly doesn't seem simple. I'm not sure which answer to mark correct since my questions have been answered in different answers. I'm certainly very grateful for the work that's gone into the answers. $\endgroup$
    – Anon
    Jan 23, 2017 at 1:12

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