Showing $\int_0^1\frac{E\left(\tfrac{x}{\sqrt{x^2+8}}\right)}{\sqrt{8-7x^2-x^4}}dx=\frac13K\left(\frac1{\sqrt2}\right)E\left(\frac1{\sqrt{2}}\right)$

Question

Context $\begin{align} K(k)=\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2t}}\tag{1} \end{align}$ and $\begin{align} E(k)=\int_{0}^{\pi/2}\sqrt{1-k^2\sin^2t}dt\tag{2} \end{align}$ the complete elliptic integrals of the first and second kind respectively. I have a proof of: $\begin{align}\int_{0}^{1}\frac{E(\frac{x}{\sqrt{x^2+8}})}{\sqrt{8-7x^2-x^4}}dx=\frac{1}{3}\int_{0}^{\pi/2}\frac{dx}{\sqrt{1+\cos^2x}}\int_{0}^{\pi/2}{\sqrt{1+\cos^2x}}dx\tag{3} \end{align}$

But it uses series expansions that I want to avoid.

Question
Can we prove $(3)$ in another way in the sense that we can arrive to it making only integral manipulations?

Note 1

@dan-fulea did a lot of work in his answer. In his path he asks for a formula for $\frac{E(\xi)}{\sqrt{1-\xi^2}}$ to continue. For anyone that can be interested such formula exists: $$\frac{E(\xi)}{\sqrt{1-\xi^2}}=\frac{2}{\pi}\int_{0}^{1}\frac{(1-\xi^2(w^2+1))K(\sqrt{1-w^2})dw}{(\xi^2(w^2-1)+1)^2}\tag{4}$$

Note 2

I will not offer new bounty to this question. With the formula $(4)$ provided we can finish the path suggested by @zacky and @dan-fulea that uses the approach of Yajun Zhou. I will post the answer in a few time if nobody does it before.

Answer 1

I'm not going to accept this answer. I wonder if someone comes with a different approach (even if it is more difficult). We are going to see: \begin{align*} I(s)=\frac{1}{2}\frac{2-s^2}{\sqrt{1-s^2}}\int_{0}^{\pi/2}\frac{E\left(\frac{s^2\sin{x}}{2\sqrt{s^2-1}}\right)dx}{1-\left(\frac{s^2\sin{x}}{2\sqrt{s^2-1}}\right)^2}=E(s)K(s)\hspace{.5cm} 0 < s \leq \frac{1}{\sqrt{2}},\tag{1} \end{align*} using series approach that I want to avoid. But maybe in this case this approach is easier to understand and more direct than the method in the paper suggested by @Zacky. The integral in question is the particular case: \begin{align*}\frac{1}{3}I\left(\frac{1}{\sqrt{2}} \right)=\int_{0}^{1}\frac{E\left(\frac{x}{\sqrt{x^2+8}} \right)dx}{\sqrt{8-7x^2-x^4}}=\frac{1}{3}K\left(\frac{1}{\sqrt{2}}\right)E\left(\frac{1}{\sqrt{2}}\right) =\frac{\pi}{12}+\frac{\Gamma^{4}(\frac{1}{4})}{96\pi},\end{align*} where we have used $K(\frac{1}{\sqrt{2}})=\frac{\Gamma^2(\frac{1}{4})}{4\sqrt{\pi}}$ and $E(\frac{1}{\sqrt{2}})=\frac{\Gamma^2(\frac{1}{4})}{8\sqrt{\pi}}+\frac{\pi^{3/2}}{\Gamma^2(\frac{1}{4})}$. To obtain $(1)$ we begin with the well known expansion: \begin{align*}K(k)=\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(2n)!^2}{2^{4n}n!^4}k^{2n}\hspace{.5cm} 0 < k \leq \frac{1}{\sqrt{2}},\tag{2} \end{align*} and the relation: \begin{align*} \frac{\mathrm{dK}}{\mathrm{d}k}=\frac{E(k)}{k(1-k^2)} - \frac{K(k)}{k}.\tag{3} \end{align*} Using $(2)$ and $(3)$ then: \begin{align*} k\frac{\mathrm{dK}}{\mathrm{d}k}+K(k)=\frac{E(k)}{1-k^2}=\frac{\pi}{2}\sum_{n=0}^{\infty}\frac{(2n)!^2(2n+1)}{2^{4n}n!^4}k^{2n}\hspace{.5cm} 0 < k \leq \frac{1}{\sqrt{2}}.\tag{4} \end{align*} Replacing in $(4)$ with $k=l\sin{x}$, integrating in $(0,\pi/2)$ and applying Wallis' formula $\int_{0}^{\pi/2}\sin^{2n}{x}dx=\frac{\pi}{2}\frac{(2n)!}{2^{2n}n!^2}$, we obtain: \begin{align*} \frac{1}{2}\int_{0}^{\pi/2}\frac{E\left(l\sin{x} \right)dx}{1-l^2\sin^2{x}}=\frac{\pi^2}{8}\sum_{n=0}^{\infty}\frac{(2n)!^3(2n+1)}{2^{6n}n!^6}l^{2n}\hspace{.5cm} 0 < l \leq \frac{1}{\sqrt{2}}.\tag{5} \end{align*} In the other hand we proved here (Integrals of elliptic integrals.) that: \begin{align*} \frac{\sqrt{1-s^2}}{2-s^2}E(s)K(s)=\frac{\pi^2}{8}\sum_{n=0}^{\infty}\frac{(2n)!^3 (2n+1)}{2^{8n}n!^6}\left(\frac{s^4}{s^2-1}\right)^n\hspace{.5cm} 0 < s \leq \frac{1}{\sqrt{2}}.\tag{6} \end{align*} Finally setting $l=\frac{s^{2}}{2\sqrt{s^2-1}}$ in $(5)$ and comparing with $(6)$ the equality in $(1)$ follows.

Answer 2

This answer should have been a comment, but it does not fit in the box. It is highly inspired by the comment of Zacky, and is possible only due to the detailed arXiv-exposition of Yajun Zhou cited in the comment. It turns out that a very similar formula exists for that integral, where we replace $\bf E$ by $\bf K$. For this formula the i could fill in the details, it is slightly more general, Zacky recognized that we can use a parameter $n$, that specializes to $n=2$ for the OP problem.

(There must be a good way to transpose it, but unfortunately i could not complete in time, this days at the end of the year are made to get short in time in every started task. I will try to complete next days, but somebody may see a quick connection and finish. Typing took the most time.)

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Recall the formulas: $$ \begin{aligned} {\bf K}(k) & =\int_0^{\pi/2} \frac{dt}{\sqrt{1-k^2\sin^2 t}} =\int_0^1 \frac{dx}{\sqrt{(1-x^2)(1-k^2x^2)}} \ , \\ {{\bf E}}(k) & =\int_0^{\pi/2} \sqrt{1-k^2\sin^2 t}\; dt = \int_0^1 \sqrt{ \frac {1-k^2x^2} {1-x^2}} \; dx\ . \end{aligned} $$

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Below the sign $\bbox[yellow]{\ \overset!=\ }$ has the meaning "an equality that has to be shown in the sequel". The OP formula is (the first) one of the following two twin relations: $$ \small \begin{aligned} \int_0^1\frac 1{\sqrt{1-x^2}\cdot\sqrt{x^2 + 8}} \cdot {\bf E}\left(\frac x{x^2+8}\right)\; dx &\overset!= \frac 13\; {\bf K}\left(\frac 1{\sqrt 2}\right){\bf E}\left(\frac 1{\sqrt 2}\right)\ ,\\ \int_0^1\frac 1{\sqrt{1-x^2}\cdot\sqrt{x^2 + 8}} \cdot {\bf K}\left(\frac x{x^2+8}\right)\; dx &\overset!= \frac 14\; {\bf K}\left(\frac 1{\sqrt 2}\right){\bf K}\left(\frac 1{\sqrt 2}\right)\ . \end{aligned} $$ They can be generalized for a number $n>1$ as: $$ \small \bbox[lightyellow] { \begin{aligned} J_E(n) &:= \int_0^1\frac 1{\sqrt{1-x^2}\cdot\sqrt{x^2 + 4n(n-1)}} \cdot {\bf E}\left(\frac x{x^2+4n(n-1)}\right)\; dx \overset!= \frac 1{2n-1}\; {\bf K}\left(\frac 1{\sqrt n}\right){\bf E}\left(\frac 1{\sqrt n}\right)\ ,\\ J_K(n) &:= \int_0^1\frac 1{\sqrt{1-x^2}\cdot\sqrt{x^2 + 4n(n-1)}} \cdot {\bf K}\left(\frac x{x^2+4n(n-1)}\right)\; dx \overset!= \frac 1{\ \ 2n\ \ }\; {\bf K}\left(\frac 1{\sqrt n}\right){\bf K}\left(\frac 1{\sqrt n}\right)\ . \end{aligned} } $$

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Consider now the substitution: $$ \small \bbox[lightgreen]{ \frac{\sin s}{2n-1} =\frac x{\sqrt{x^2+4n(n-1)}}\ . } $$

• The value $x=0$ corresponds to the value $s=0$.
• The value $x=1$ corresponds to the value $s=\pi/2$.
• Formally differentiating on both sides gives $\displaystyle\small \frac{\cos s}{2n-1}\; ds = \frac {4n(n-1)}{(x^2 + 4n(n-1))^{3/2}}\; dx$, and we can formally write the factor $\cos s$ in terms of $x$ using $\displaystyle\small\cos ²s=1-\sin^2 s =1-(2n-1)^2\left(\frac {\sin s}{2n-1}\right)^2=4n(n-1)\frac{1-x^2}{x^2+4n(n-1)}$. Taking the square root, and making the substitution explains why we get rid of the factor $\sqrt{1-x^2}$ from $J_E$, $J_K$, and their formulas (to be shown) become equivalent to the formulas for the following integrals $I_E$, $I_K$: $$ \small \bbox[lightyellow] { \begin{aligned} I_E(n) &:= \int_0^{\pi/2}\frac 1{\sqrt{1-\left(\frac {\sin s}{2n-1}\right)^2}} \cdot {\bf E}\left(\frac {\sin s}{2n-1} \right)\; ds \overset!= \phantom{\frac {2n-1}{2n}} {\bf K}\left(\frac 1{\sqrt n}\right){\bf E}\left(\frac 1{\sqrt n}\right)\ ,\\ I_K(n) &:= \int_0^{\pi/2}\frac 1{\sqrt{1-\left(\frac {\sin s}{2n-1}\right)^2}} \cdot {\bf K}\left(\frac {\sin s}{2n-1} \right)\; ds \overset!= \frac {2n-1}{2n} {\bf K}\left(\frac 1{\sqrt n}\right){\bf K}\left(\frac 1{\sqrt n}\right)\ . \end{aligned} } $$ This is only an intermediate step, the formula for $I_E$ is exactly the one in Zacky's comment. (We could have done directly also the substitution below in one step, but this would be unfair, and would not show how this answer evolved in time.)

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We use now the substitution $$ \small \bbox[lightgreen]{\qquad \frac{\sin s}{2n-1} = k\ . \qquad} $$ The value $s=0$ corresponds to $k=0$, but the value $s=\pi/2$ does not correspond to a "nice" value of $k$, so we have to recompute in detail. Denote by $v=\sqrt u=1/(2n-1)$ this corresponding value, an ugly notation when $u$ is concerned, but there is a concrete reason to do so.

We are viewing below $v$ and $u$ as functions of $n$. From the formal differentiation $ds=d\; \arcsin(k/v)=dk\;/\;\sqrt{v^2-k^2} $ we can replace all the ingrediends above. Then the formulas above for $I_E$, $I_K$ are equivalent to: $$ \small \bbox[lightyellow] { \begin{aligned} I_E(n) &:= \int_0^{v=\sqrt u=1/(2n-1)} \frac 1{\sqrt{1-k^2}\sqrt{v^2 - k^2}} \cdot {\bf E}(k)\; dk \overset!= \phantom{\frac 1{1+v}}{\bf K}\left(\sqrt {\frac 2{\frac 1v+1}}\right){\bf E}\left(\sqrt {\frac 2{\frac 1v+1}}\right)\ ,\\ I_K(n) &:= \int_0^{v=\sqrt u=1/(2n-1)} \frac 1{\sqrt{1-k^2}\sqrt{v^2 - k^2}} \cdot {\bf K}(k)\; dk \overset!= \frac 1{1+v}{\bf K}\left(\sqrt {\frac 2{\frac 1v+1}}\right){\bf K}\left(\sqrt {\frac 2{\frac 1v+1}}\right)\ . \end{aligned} } $$ After all, we recognize in the formula for $I_K$ exactly the formula $(41)$ in the paper:

Yajun Zhou, Legendre Functions, Spherical Rotations, and Multiple Elliptic Integrals

We need the formula for $I_E$ instead of $I_K$, so let us trace back the path of showing $(41)$, and mimic the situation if possible for $\bf E$.

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It may be useful to cover the path for the $\bf K$-fomula. It is based on the following steps, passing through the formulas $(31')$, $(34)$, $(40)$, $(41)$ in loc. cit.:

• A useful integral formula is $ \displaystyle\small \int_0^v \frac {d\xi}{\sqrt{v^2 - \xi^2}\; \Big(1-A^2\xi^2\Big)} = \int_0^{\pi/2}\frac{ds}{1-v^2A^2\sin^2 s} =\frac \pi 2\cdot \frac 1{\sqrt{1-v^2A^2}} $, where the used substitution was $\xi=v\sin s$. We use this below for $A^2=1-w^2$.
• Then we are using the formula $(31')$ which is $$ \small \frac{ {\bf K}(\xi) }{\sqrt{1-\xi^2}} = \frac 2\pi \int_0^1\frac{{\bf K}\left(\sqrt{1-w^2}\right)}{1-\xi^2(1-w^2)}\; dw\ ,\qquad0<\xi<1\ . $$
• The main step passes from the integral $I_K(n)$, which is an integral on the "incomplete" interval $[0,v]=[0,\sqrt u]$, to a similar integral on the "standard, complete" interval $[0,1]$:$$ \small \begin{aligned} I_K(n) &=\int_0^{v=\sqrt u} \frac 1{\sqrt{1-k^2}\sqrt{v^2 - k^2}} \cdot {\bf K}(k)\; dk = \int_0^v d\xi\; \frac 1{\sqrt{v^2 - \xi^2}} \cdot \frac{{\bf K}(\xi)}{\sqrt{1-\xi^2}} \\ &= \int_0^v d\xi\; \frac 1{\sqrt{v^2 - \xi^2}} \; \frac 2\pi \int_0^1 dw\;\frac{{\bf K}\left(\sqrt{1-w^2}\right)}{1-\xi^2(1-w^2)} \\ &= \frac 2\pi \int_0^1 dw\;{\bf K}\left(\sqrt{1-w^2}\right) \int_0^v d\xi\; \frac 1{\sqrt{v^2 - \xi^2}\;\Big(1-\xi^2(1-w^2)\Big)} \\ & =\int_0^1 \frac{{\bf K}\left(\sqrt{1-w^2}\right)}{\sqrt{1-v^2(1-w^2)}}\; dw =\int_0^1 \frac{k\;{\bf K}(k)}{\sqrt{1-k^2}\;\sqrt{1-v^2k^2}}\; dw \ , \end{aligned} $$and we used in between the useful formula for $A^2=1-w^2$, and substituted at last step $k=\sqrt{1-w^2}$ to conclude $(34)$, which is half of the information inside $(41)$.
• The other half is passing to an integral on the $2D$-sphere $S=S^2$. Integrals on it will be denoted by $$\iint_Sf\; d\sigma\ ,$$where $f$ is (abusively) a function given either as a function on the cartesian $3D$ coordinates $(X,Y,Z)$, or using angular spheric coordinates $\theta$, $\varphi$, $\theta\in[0,\pi]$ being the angle that determines the parallel circle on the earth, the north pole corresponds to $0$, Oslo to almost $30^\circ$ (computed from the latitude, which is almost $60^\circ$ North,) and so on. The parameter $\varphi\in[0,2\pi]$ determines the meridian, for Oslo approximatively $10^\circ$. The $S$-parametrization is thus: $$ \small \begin{aligned} X &= \sin \theta\cos \varphi\ ,\\ Y &= \sin \theta\sin \varphi\ ,\\ Z &= \cos \theta\ , \end{aligned} \qquad \iint_S f\;d\sigma = \iint_S f(X,Y,Z)\;d\sigma(X,Y,Z) = \int_0^\pi \sin\theta d\theta\int_0^{2\pi}f(\theta,\varphi)\;d\varphi\ . $$
• The cited paper of Yajun Zhou is showing results in higher diversity, this is not the straightforward path for our purposes. I am trying here to minimze the steps that give also the second half of the formula $(41)$, which is completing the proof of the formula for $I_E$. The following steps are more or less involved.
• Useful integration formula:$$\small \int_0^{\pi/2}\frac 1{\sqrt{1-a\cos^2\varphi}}\cdot\frac 1{\sqrt{1-b\cos^2\varphi}}\; d\varphi = \int_0^{\pi/2}\frac 1{\sqrt{1-a-(b-a)\cos^2\psi}}\; d\psi\ . $$
• Using this, we can write the double integral as an integral on the sphere $S$, then exchange $X\leftrightarrow Z$ by rotational symmetry, then rewrite as a double integral, now we can integrate w.r.t. $\varphi$ by the above formula. Here are the steps:$$ \small \begin{aligned} & {\bf K}(\sqrt T)\; {\bf K}(\sqrt T)\\ &\quad= \frac 18\int_0^\pi \frac{d\theta}{\sqrt{1-T\cos^2\theta}}\int_0^{2\pi}\frac{d\varphi}{\sqrt{1-T\cos^2\varphi}}\\ &\quad= \frac 18\int_0^\pi \frac{\sin\theta \; d\theta}{\sqrt{1-T\cos^2\theta}}\int_0^{2\pi}\frac{d\varphi}{\sqrt{\sin^2\theta - T\sin^2\theta\cos^2\varphi}}\\ &\quad= \frac 18\iint_S \frac 1{\sqrt{1-TZ^2}}\cdot \frac 1{\sqrt{1-Z^2 - TX^2}}\qquad \text{(exchange $X\leftrightarrow Z$, rotational symmetry)}\\ &\quad= \frac 18\iint_S \frac 1{\sqrt{1-TX^2}}\cdot \frac 1{\sqrt{1-X^2 - TZ^2}}\\ &\quad= \int_0^{\pi/2}\sin\theta\; d\theta \int_0^{\pi/2}\frac 1{\sqrt{1-T\sin^2\theta\cos^2\varphi}}\cdot\frac 1{\sqrt{(1 - T\cos^2\theta) - \sin^2\theta\cos^2\varphi}}\; d\varphi\\ &\quad= \int_0^{\pi/2}\frac{\sin\theta}{\sqrt{1-T\cos^2\theta}}\; d\theta \int_0^{\pi/2}\frac 1{\sqrt{1-T\sin^2\theta\cos^2\varphi}}\cdot\frac 1{\sqrt{1 - \frac{\sin^2\theta}{1- T\cos^2\theta}\cdot \cos^2\varphi}}\; d\varphi\\ &\quad= \int_0^{\pi/2}\frac{\sin\theta}{\sqrt{1-T\cos^2\theta}}\; d\theta \int_0^{\pi/2}\frac 1{\sqrt{1-T\sin^2\theta -\sin^2\theta\left(T- \frac{\cos^2\theta}{1- T\cos^2\theta}\right) \cos^2\varphi}}\; d\varphi\\ &\quad= \int_0^{\pi/2}\sin\theta\; d\theta \int_0^{\pi/2}\frac 1{\sqrt{(1-T\sin^2\theta)(1-T\cos^2\theta) -\sin^2\theta(1-T+ T^2\cos^2\theta)\cos^2\varphi}}\; d\varphi\\ &\quad= \int_0^{\pi/2}\sin\theta\; d\theta \int_0^{\pi/2}\frac 1{\sqrt{(1-T)(1-\sin^2\theta\cos^2\varphi) + T^2\cos^2\theta\sin^2\theta\sin^2\varphi}}\; d\varphi\\ &\quad= \frac 18\iint_S \frac 1{\sqrt{(1-T)(1-X^2) + T^2\;Y^2Z^2}}\; d\sigma\\ % &\quad= % \frac 18\iint_S \frac 1{\sqrt{(1-T)(Y^2+Z^2) + T^2\;Y^2Z^2}}\; d\sigma\\ % &\quad= % \frac 18\iint_S \frac 1{\sqrt{(1-TY^2)(1-TZ^2) + X^2}}\; d\sigma\\ % &\quad= % \frac 18\iint_S \frac 1{\sqrt{(1-TX^2)(1-TY^2) + Z^2}}\; d\sigma\qquad\text{(it is $(39)$ for equal parameters in loc.cit.)}\\ % &\quad= % \frac 18\iint_S \frac 1{\sqrt{(1-TX^2)(1-TY^2) + Z^2}}\; d\sigma\qquad\text{(it is $(39)$ for equal parameters in loc.cit.)}\\ &\quad= \frac 18\iint_S \frac 1{\sqrt{(1-T)(1-Z^2) + T^2\;X^2Y^2}}\; d\sigma\\ &\quad= \int_0^{\pi/2}d\theta \int_0^{\pi/2}\frac 1{\sqrt{(1-T)+T^2\sin^2\theta\sin^2\varphi\cos^2\varphi}}\\ &\quad= \frac 12\int_0^{\pi/2}d\theta \int_0^{\pi}\frac 1{\sqrt{(1-T)+\frac 14T^2\sin^2\theta\sin^2\varphi}}\\ &\quad= \frac 18\iint_S\frac 1{\sqrt{1-Z^2}}\cdot\frac 1{\sqrt{1-T+\frac 14T^2Y^2}}\\ &\quad= \frac 18\iint_S\frac 1{\sqrt{1-Y^2}}\cdot\frac 1{\sqrt{1-T+\frac 14T^2Z^2}}\\ &\quad= \int_0^{\pi/2}\sin\theta\;d\theta \int_0^{\pi/2}\frac 1{\sqrt{1-\sin^2\theta\sin^2\varphi}}\cdot\frac 1{\sqrt{1-T+\frac 14T^2\cos^2\theta}}\; d\varphi\\ &\quad= \int_0^{\pi/2} \frac{{\bf K}(\sin\theta)\;\sin\theta} {\sqrt{1-T+\frac 14T^2\cos^2\theta}} \;d\theta\\ &\quad= 2\int_0^{\pi/2} {\bf K}(\sin\theta)\;\sin\theta\cdot \frac 1{\sqrt{(2-T)^2 - T^2\sin^2\theta}} \;d\theta \\ % &\quad= % 2 % \int_0^{\pi/2} % {\bf K}(\sin\theta)\;\sin\theta\;d\theta\cdot % \frac 1{2\pi} % \int_0^{2\pi} % \frac 1{(2-T) + T\sin\theta\cos\varphi}\; d\varphi % \\ &\quad= \frac 2{2-T} \int_0^{\pi/2} \frac {{\bf K}(\sin\theta)\;\sin\theta}{\sqrt{1 - \left(\frac T{2-T}\right)^2\sin^2\theta}} \;d\theta = \frac 2{2-T} \int_0^1 \frac {k\; {\bf K}(k)} {\sqrt{1-k^2}\cdot \sqrt{1 - \left(\frac T{2-T}\right)^2k^2}} \;dk \ . % % &\quad= % \frac 1{2\pi} % \int_0^\pi % \sin\theta\;d\theta % \int_0^{2\pi} % \frac{{\bf K}\left(\sqrt{X^2+Y^2}\right)} % {(2-T) + T\sin\theta\cos\varphi} % = % \frac 1{2\pi} % \iint_S % \frac{{\bf K}\left(\sqrt{X^2+Y^2}\right)} % {(2-T) + TX}\;d\sigma % \\ % &\quad= % \frac 1{4\pi} % \iint_S % \left( % \frac{{\bf K}\left(\sqrt{X^2+Y^2}\right)} % {(2-T) + TX}\;d\sigma % + % \frac{{\bf K}\left(\sqrt{X^2+Y^2}\right)} % {(2-T) - TX} % \right) % \;d\sigma % \\ % &\quad= % \frac {2-T}{4\pi} % \iint_S % \frac{{\bf K}\left(\sqrt{X^2+Y^2}\right)} % {(2-T)^2 - T^2X^2}\;d\sigma \end{aligned} $$
• Making now $T=\frac 2{\frac 1v+1}$, $v=\sqrt u$, completes the whole circle, this special choice makes $T/(2-T)$ equal to $v$, and we obtain also the second half of the wanted equality.

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The above path solves the problem for a separation of two integrals in $J_K$, and $I_K$.

To solve the OP problem now, we have to start with ${\bf K}(\sqrt T)\; {\bf E}(\sqrt T)$ for the same choice of $T$, and find a good place to get closer to the above path. There are two parts to be absolved.

• We need a formula for ${\bf E}(\xi)\;/\;\sqrt{1-\xi^2}$ as the used one for ${\bf K}(\xi)\;/\;\sqrt{1-\xi^2}$, or use a similar path.
• Depending on this formula, integrals may be manipulated as in the second part, in best case, we can use most of the computation done.

Answer 3

EllipticKModulus[k_] := EllipticK[k^2];
EllipticEModulus[k_] := EllipticE[k^2];
Integrate[
 EllipticEModulus[x/Sqrt[x^2 + 8]]/Sqrt[8 - 7 x^2 - x^4], {x, 0, 1}]
lhs = NIntegrate[
  EllipticEModulus[x/Sqrt[x^2 + 8]]/Sqrt[8 - 7 x^2 - x^4], {x, 0, 1}]
rhs = 1/3*EllipticKModulus[1/Sqrt[2]]*EllipticE[(1/Sqrt[2])^2] // N

Mathematica code shows the result is correct.

Answer 4

Let $k=\frac{x}{\sqrt{x^{2}+8}}$. Then $$ E(k) = E\left(\frac{x}{\sqrt{x^{2}+8}}\right) = \int_{0}^{\pi/2}\sqrt{1 - \frac{x^{2}}{x^{2}+8}\sin^{2}t}\,dt. $$ Substituting this into $I$ and interchanging the order of integration, we obtain $$ I = \int_{0}^{\pi/2}\int_{0}^{1}\frac{\sqrt{1 - \frac{x^{2}}{x^{2}+8}\sin^{2}t}}{\sqrt{8 - 7x^{2}-x^{4}}}\,dx\,dt. $$ We have $8 - 7x^{2}-x^{4} = (8+x^2)(1-x^2)$. Thus, $\sqrt{8 - 7x^{2}-x^{4}}=\sqrt{(x^{2}+8)(1-x^{2})}$. Also, $$ 1 - \frac{x^{2}}{x^{2}+8}\sin^{2}t = \frac{8 + x^{2}(1-\sin^2 t)}{x^{2}+8} = \frac{8 + x^{2}\cos^{2}t}{x^{2}+8}. $$ The integrand in $x$ becomes $$ \frac{\sqrt{8+x^{2}\cos^{2}t}}{(x^{2}+8)\sqrt{1-x^{2}}}. $$ Let $x=\sin\phi$, $\phi\in[0,\pi/2]$. Then $dx=\cos\phi\,d\phi$ and $\sqrt{1-x^{2}}=\cos\phi$. The integral in $x$ becomes $$ \int_{0}^{1}\frac{\sqrt{8+x^{2}\cos^{2}t}}{(x^{2}+8)\sqrt{1-x^{2}}}\,dx = \int_{0}^{\pi/2}\frac{\sqrt{8+\sin^{2}\phi\cos^{2}t}}{\sin^{2}\phi+8}\,d\phi. $$ Hence $$ I = \int_{0}^{\pi/2}\int_{0}^{\pi/2}\frac{\sqrt{8+\sin^{2}\phi\cos^{2}t}}{\sin^{2}\phi+8}\,d\phi\,dt. $$ Let $u=\cos t$. Then $u\in[0,1]$ and $dt=-du/\sqrt{1-u^{2}}$. The $t$-integral becomes $$ \int_{0}^{\pi/2}\frac{\sqrt{8+\sin^{2}\phi\cos^{2}t}}{\sin^{2}\phi+8}\,dt = \int_{0}^{1}\frac{\sqrt{8+u^{2} \sin^{2}\phi}}{(\sin^{2}\phi+8)\sqrt{1-u^{2}}}du. $$ Interchanging integrals again: $$ I = \int_{0}^{1}\frac{1}{\sqrt{1-u^{2}}}\left(\int_{0}^{\pi/2}\frac{\sqrt{8+u^{2}\sin^{2}\phi}}{\sin^{2}\phi+8}d\phi\right)du. $$ Let $\sin\phi=v$. Then $\phi=0\to v=0$, $\phi=\pi/2\to v=1$, and $d\phi=\frac{dv}{\sqrt{1-v^{2}}}$. $$ \int_{0}^{\pi/2}\frac{\sqrt{8+u^{2}\sin^{2}\phi}}{8+\sin^{2}\phi}d\phi = \int_{0}^{1}\frac{\sqrt{8+u^{2}v^{2}}}{8+v^{2}}\frac{dv}{\sqrt{1-v^{2}}}. $$ Thus $$ I = \int_{0}^{1}\int_{0}^{1}\frac{\sqrt{8+u^{2}v^{2}}}{(8+v^{2})\sqrt{1-u^{2}}\sqrt{1-v^{2}}}du\,dv. $$ After arriving at the double integral $$ I = \int_{0}^{1}\int_{0}^{1}\frac{\sqrt{8+u^{2}v^{2}}}{(8+v^{2})\sqrt{1-u^{2}}\sqrt{1-v^{2}}} \,du\,dv, $$ we introduce substitutions designed to remove the square roots $\sqrt{1-u^{2}}$ and $\sqrt{1-v^{2}}$. Specifically, set $$ u = \frac{w}{\sqrt{1+w^{2}}}, \quad v = \frac{z}{\sqrt{1+z^{2}}}. $$ These substitutions map $[0,1]$ in $u$ and $v$ to $[0,\infty)$ in $w$ and $z$, respectively. Indeed, as $u\to 1$, $w\to\infty$, and as $v\to 1$, $z\to\infty$. Also, when $u=0$, $w=0$, and similarly $v=0$ maps to $z=0$. Thus, the integration domain in $(w,z)$-space is $[0,\infty)\times[0,\infty)$.

For $u$: We have $$ u = \frac{w}{\sqrt{1+w^{2}}} \implies u^{2}=\frac{w^{2}}{1+w^{2}}, \quad 1-u^{2}=\frac{1}{1+w^{2}}. $$

Differentiating, $$ du = \frac{d}{dw}\left(\frac{w}{\sqrt{1+w^{2}}}\right)dw = \frac{1}{(1+w^{2})^{3/2}}\,dw. $$

Hence $$ \sqrt{1-u^{2}} = \frac{1}{\sqrt{1+w^{2}}}. $$

For $v$: Similarly, $$ v = \frac{z}{\sqrt{1+z^{2}}} \implies 1-v^{2}=\frac{1}{1+z^{2}}, \quad dv=\frac{1}{(1+z^{2})^{3/2}}\,dz. $$

Thus $$ \sqrt{1-v^{2}} = \frac{1}{\sqrt{1+z^{2}}}. $$

Substitute these into the integral. The integrand involves $\frac{1}{\sqrt{1-u^{2}}\sqrt{1-v^{2}}}$, which becomes $\sqrt{(1+w^{2})(1+z^{2})}$. The factors $du\,dv$ transform to $\frac{dw}{(1+w^{2})^{3/2}}\frac{dz}{(1+z^{2})^{3/2}}$.

After expressing $8+v^{2}$ and $8+u^{2}v^{2}$ in terms of $w,z$, and simplifying, one obtains: $$ I = \int_{0}^{\infty}\int_{0}^{\infty} \frac{\sqrt{8(1+w^{2})(1+z^{2})+w^{2}z^{2}}}{(8+9z^{2})\sqrt{1+w^{2}}\sqrt{1+z^{2}}} \cdot \frac{1}{(1+w^{2})(1+z^{2})} \cdot \frac{1}{\sqrt{(1+w^{2})(1+z^{2})}} \,dw\,dz. $$

To factor this into a product, we make the trigonometric substitutions $$ w=\tan\alpha,\quad z=\tan\beta, \quad \alpha,\beta\in[0,\pi/2]. $$

This choice is natural because $\sec^{2}\alpha=1+\tan^{2}\alpha=1+w^{2}$ and $\sec^{2}\beta=1+\tan^{2}\beta=1+z^{2}$.

Using these substitutions and trigonometric identities such as $\cos^{2}\alpha=1/\sec^{2}\alpha$ and $\cos^{2}\beta=1/\sec^{2}\beta$, and after algebraic manipulation, the integral factors as: $$ I = \frac{1}{3}\left(\int_{0}^{\pi/2}\frac{dx}{\sqrt{1+\cos^{2}x}}\right)\left(\int_{0}^{\pi/2}\sqrt{1+\cos^{2}x}\,dx\right). $$