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$$\int_0^1\frac{3x^3-2x}{(1+x)\sqrt{1-x}}K\left(\frac{2x}{1+x}\right)\,dx\stackrel ?=\frac\pi{5\sqrt2}$$

The integral above comes from the evaluation of the integral $A=\int_0^{\pi/2}\frac{f(\theta)}\pi d\theta$, where

$$f(\theta)=\int_0^\pi\frac{(3\sin^2\theta-2)\sin\theta\,d\phi}{\sqrt{2+2\sin\theta\cos\phi}}=\frac{\sqrt2(3\sin^2\theta-2)}{\csc\theta\sqrt{1+\sin\theta}}\int_0^{\pi/2}\left(1-\frac{2\sin\theta}{1+\sin\theta}\sin^2\gamma\right)^{-1/2}d\gamma,$$

where $\gamma=\phi/2$, and the right side integral is $K\big(\!\frac{2\sin\theta}{1+\sin\theta}\!\big)$ by definition. After substitutions $x=\sin\theta$ and $y=\frac{2x}{1+x}$ we get

$$\begin{align}A&=\frac{\sqrt2}\pi\int_0^1\frac{3x^3-2x}{(1+x)\sqrt{1-x}}K\left(\frac{2x}{1+x}\right)\,dx\\ &=\frac1\pi\int_0^1\frac{y^3+8y^2-8y}{(2-y)^3\sqrt{y^2-3y+2}}K(y)\,dy\stackrel ?=\frac15,\end{align}$$

where the integral has been numerically evaluated to suggest the analytic result on the right to several thousand digits. Is there any way to prove this equality, and are there any generalizations of this conjecture to include other parameters?

P.S. Not that it's overly relevant to the question, but for the interested reader, the original integral comes from the following physics problem:

[Morin Intro to Classical Mechanics, Ex. 10.12] The earth bulges slightly at the equator, due to the centrifugal force in the earth’s rotating frame. The goal of this exercise is to find the shape of the earth, first incorrectly, and then correctly.

(a) The common incorrect method is to assume that the gravitation force from the slightly nonspherical earth points toward the center, and to then calculate the equipotential surface (incorporating both the gravitational and centrifugal forces). Show that this method leads to a surface whose height (relative to a spherical earth of the same volume) is given by $h(\theta)=R\big(\!\frac{R\omega^2}{6g}\!\big)(3\sin^2⁡\theta-2)$, where $\theta$ is the polar angle (the angle down from the north pole), and $R$ is the radius of the earth.

(b) The above method is incorrect, because the slight distortion of the earth causes the gravitational force to not point toward the center of the earth (except at the equator and the poles). This tilt in the force direction then changes the slope of the equipotential surface, and it turns out (although this is by no means obvious) that this effect is of the same order as the slope of the surface found in part (a). Your task: Assuming that the density of the earth is constant, and that the correct height takes the form of some constant factor $f$ times the result found in part (a), show that $f=5/2$. Do this by demanding that the potential at a pole equals the potential at the equator.

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  • $\begingroup$ Can you just expand out the expression for $K(x)$ or $K(y)$ and reverse the order of integration? That would at least offer a first step and the (newly) inner integral should at least be tractable, if ugly... $\endgroup$ – Steven Stadnicki Oct 30 '13 at 0:11
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    $\begingroup$ Based on numerical evaluations, I conjecture that \begin{align} I_1=\frac1\pi\int^1_0\frac{y^3}{(2-y)^3\sqrt{y^2-3y+2}}K(y)dy&=\frac{11}{15}\\ I_2=\frac1\pi\int^1_0\frac{y(1-y)}{(2-y)^3\sqrt{y^2-3y+2}}K(y)dy&=\frac{1}{15} \end{align} $\endgroup$ – Chen Wang Feb 7 '14 at 16:39
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    $\begingroup$ Looks like more is happening. If we write $ I_n=\frac1{\pi}\int^1_0\left(\frac{y}{2-y}\right)^{2n+1}\frac{K(y)dy}{\sqrt{y^2-3y+2}}$ for each $n\in\mathbb{N}$, Then each $I_n$ looks like a rational number. The integral in the OP is equal to $I_0-12I_1$. The conjectured value of the first $I_n$s are: \begin{align} I_0&\stackrel ?=1\\ I_1&\stackrel ?=\frac{11}{15}\\ I_2&\stackrel ?=\frac{13}{21}\\ I_3&\stackrel ?=\frac{1181}{2145}\\ I_4&\stackrel ?=\frac{385397}{765765}. \end{align} $\endgroup$ – Chen Wang Feb 7 '14 at 17:11
  • $\begingroup$ @ChenWang How to you relate the $I_2$ of your first comment (which I will rename to $J_2$) to the $I_0$ of your second? I can see that $I_1=J_1$ and $A=J_1-8J_2$, but I'm not seeing $A=I_0-12I_1$. $\endgroup$ – Mario Carneiro Feb 7 '14 at 17:21
  • $\begingroup$ Because $I_0=J_1+4J_2$; the numertor of integrand of $J_1+4J_2$ is $y^3+4y(1-y)=y(2-y)^2$. $\endgroup$ – Chen Wang Feb 7 '14 at 17:24
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Edit@3.30: Found another way to expand the elliptic integral.

Using DLMF 15.8.13 with $a=b=\tfrac12$ and $z=\frac{2x}{1+x}$, we conclude that $$K\left(\frac{2x}{1+x}\right)=\frac{\pi}{2}\sqrt{1+x}{~_2F_1}(\tfrac14,\tfrac34;1;x^2).$$

Therefore, we have $$ I_n=\frac{1}{\sqrt2}\int^1_0\frac{x^{2n+1}}{\sqrt{1-x^2}}{~_2F_1}(\tfrac14,\tfrac34;1;x^2)dx\\ =\frac{1}{\sqrt2}\int^1_0\frac{x^{2n+1}}{\sqrt{1-x^2}}\sum_{m=0}^{\infty}\frac{\Gamma(m+\tfrac14)\Gamma(m+\tfrac34)x^{2m}}{\Gamma(\tfrac14)\Gamma(\tfrac34)\Gamma(m+1)^2}dx\\ =\frac{1}{\sqrt2}\sum_{m=0}^{\infty}\frac{\Gamma(m+\tfrac14)\Gamma(m+\tfrac34)}{\Gamma(\tfrac14)\Gamma(\tfrac34)\Gamma(m+1)^2}\int^1_0\frac{x^{2m+2n+1}}{\sqrt{1-x^2}}dx\\ =\frac{1}{\sqrt2}\sum_{m=0}^{\infty}\frac{\Gamma(m+\tfrac14)\Gamma(m+\tfrac34)}{\Gamma(\tfrac14)\Gamma(\tfrac34)\Gamma(m+1)^2}\frac{\sqrt{\pi}\Gamma(m+n+1)}{2\Gamma(m+n+\tfrac32)}\\ =\frac{\sqrt{\pi}\Gamma(n+1)}{2\sqrt2\Gamma(n+\tfrac32)}{~_3F_2}\left(\begin{array}c\tfrac14,\tfrac34,n+1\\1,n+\tfrac32\end{array}\middle|1\right).\\ $$

In particular, we have $$ I_0=\frac{\sqrt{\pi}}{2\sqrt2\Gamma(\tfrac32)}\frac{\Gamma(\tfrac32)\Gamma(\tfrac12)}{\Gamma(\tfrac34)\Gamma(\tfrac54)}=1\\ I_1=\frac{\sqrt{2}}{3}{~_3F_2}\left(\begin{array}c\tfrac14,\tfrac34,2\\1,\tfrac52\end{array}\middle|1\right)=\frac{11}{15}. $$

The evaluation of the last ${~_3F_2}$ function is due to Mathematica.

Therefore, OP's original integral $A=3I_1-2I_0=\frac15$.

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  • $\begingroup$ An observation: I notice that unlike with $A$, Mathematica can do the $dx$ integrals for the first few $I_n$ a la Steven Stadnicki's suggestion, although you get a whole mess of elliptic functions, and still have the $dt$ integrals to do. $\endgroup$ – Mario Carneiro Feb 7 '14 at 20:57
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This is a collection of musings that may or may not constitute an answer, but are too big to append to the OP while keeping it restricted to the problem statement.

As suggested by Steven Stadnicki, we can use the integral definition for $K(x)$ and interchange the integrals (I have switched $y\mapsto x$ from the convention in the OP):

$$\begin{align}A&=\frac1\pi\int_0^1\frac{dt}{\sqrt{1-t^2}}\int_0^1dx\frac{x^3+8x^2-8x}{(2-x)^3}\big[(x-1)(x-2)(1-t^2x)\big]^{-1/2}\end{align}$$

This integrand is an elliptic integral (w.r.t. $dx$), since it is a rational function of $x$ and $s$, where $s^2=(x-1)(x-2)(1-t^2x)$ is a cubic function of $x$. (Not sure how to reduce... will add more later.)


Thanks to the work of Chen Wang in the comments, it seems worthwhile to extend the conjecture to the one-parameter family

$$I_n=\frac1\pi\int_0^1 \frac{x^{2n+1}K(x)\,dx}{(2-x)^{2n+3/2}\sqrt{1-x}},$$

for which $A=3I_1-2I_0$. Each $I_n$ is apparently rational, with $$I_0\stackrel?=1,\quad I_1\stackrel?=\frac{11}{15},\quad I_2\stackrel?=\frac{13}{21},\quad I_3\stackrel?=\frac{1181}{2145},\quad I_4\stackrel?=\frac{385397}{765765}.$$

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