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Let $$S(x,y,z)=\frac14\sqrt{(x+y+z) (-x+y+z) (x-y+z) (x+y-z) }\tag1$$ (note that it's Heron's formula for the area of a triangle with sides of lengths $x,y,z$).

I'm trying to evaluate the following integral in a closed form: $$\mathcal U=\int_0^1\int_0^x\int_{x - y}^{x + y}\sqrt{S(x,y,z) } dz dy dx.\tag2$$ I wasn't able to evaluate it symbolically (either manually or using Mathematica), but using numerical integration and heuristic methods, I found a plausible form: $$\mathcal U\stackrel{\color{gray}?}=\frac\pi{12\sqrt{2 }}+\frac{\Gamma\left(\frac14\right)^4\sqrt{10 }}{1440 \pi^2}\left(\left(\small\frac12-\frac1{\sqrt{5 \phi }}\right)\textbf K(\alpha)^2-\textbf K(\alpha) \textbf E(\alpha)\right)\\ \quad\quad\;\approx0.117599420842157114228246644831065494814051852697…,\tag3$$ where $\textbf K(\cdot),\textbf E(\cdot)$ are the complete elliptic integrals of the first and the second kind, and $\alpha=\frac1{\phi \sqrt2}+\frac1{\sqrt{2 \phi}}$.

Can we prove that $(3)$ is indeed the true value of the integral?


Update: You might ask, what heuristic methods could possibly give us the conjectured value? Here is a brief explanation.

First, calculate a sequence of values of $\int\!\int\!\int S^{2n}dz dy dx,n\in\mathbb N;$ their evaluation is straightforward and they are all rational numbers. Then, use FindSequenceFunction to find a possible recurrence relation for that sequence. We need at least $30$ elements of the sequence to get a result: $$\small{(5+2 n)^2 (7+4 n) (9+4 n) (11+4 n) (27+20 n)\cdot a(n+2)=\\4 (2+n) (7+4 n) \left(6750+16947 n+15764 n^2+6416 n^3+960 n^4\right)\cdot a(n+1)+\\128 (1+n)^2 (2+n) (3+2 n) (3+4 n) (47+20 n)\cdot a(n)}\tag4$$ Of course, there is no guarantee that this relation holds for all larger $n$, but we cross our fingers 🤞 and conjecture that it does. Next, use FunctionExpand to find a possible explicit expression for the general term determined by this recurrence relation; it's complicated and involves generalized hypergeometric functions; we also need to manually adjust a periodic factor to ensure the result remains real even for non-integer $n$. After some simplifications we arrive at this: $$\small\tfrac{4^n (20 n+27) \Gamma (2 n+2)\sqrt\pi}{(n+1) (16 n+12) \Gamma \left(2 n+\frac{7}{2}\right)}\,{_5F_4}\left(1,n+\tfrac{3}{2},n+\tfrac{3}{2},n+\tfrac{3}{2},n+\tfrac{47}{20};n+\tfrac{27}{20},n+\tfrac{7}{4},n+2,n+\tfrac{9}{4};-4\right)\tag5$$ Then we cross our fingers again 🤞 and conjecture that the same formula holds for all non-integer $n$ as well. Next, substitute $n=1/4$ and use FunctionExpand again to expand the generalized hypergeometric functions in terms of elliptic integrals. The result is quite large and unwieldy ($19$ terms), but it appears to match the integral numerically, which is a sign we might be on the right track. We cross our fingers one more time 🤞 and conjecture that only a few terms of the expression are actually linearly independent over $\mathbb Z,$ and the result can be significantly simplified. Using FindIntegerNullVector and high enough numeric precision, we find a possible basis, and a linear combination of its elements that numerically matches the integral. After some radical denestings (using ResourceFunction["RadicalDenest"]) and other simplifications, this gives us the conjectured value $(3).$

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    $\begingroup$ Why do you care about $\mathcal{U}$? $\endgroup$
    – FShrike
    Jul 5, 2023 at 10:51
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    $\begingroup$ @FShrike Don't you find it an inherently beautiful and immediately appealing question that you wish you have thought about a long time ago? I do. 😊 $\endgroup$ Jul 6, 2023 at 1:59
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    $\begingroup$ This is a beautiful problem and the path to solution more than interesting $\endgroup$ Jul 6, 2023 at 3:48
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    $\begingroup$ @VladimirReshetnikov Well, I was initially wondering why you chose the bounds you chose, but now I realise these are basically the maximal bounds you can have to keep the integrand real valued $\endgroup$
    – FShrike
    Jul 6, 2023 at 10:11
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    $\begingroup$ Your $\alpha$ is essentially a singular value with $\alpha=k(e^{-\pi/\sqrt {5}})$. The calculation of $K, E$ for this value of $k$ should match the answer given by David H. $\endgroup$
    – Paramanand Singh
    Jul 9, 2023 at 6:49

2 Answers 2

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Define the function $S:\{(x,y,z)\in\mathbb{R}^{3}\mid0\le y\le x\land x-y\le z\le x+y\}\rightarrow\mathbb{R}$ bye the expression

$$S{\left(x,y,z\right)}:=\frac14\sqrt{(x+y+z)(-x+y+z)(x-y+z)(x+y-z)},$$

and let $\mathcal{U}$ denote the value of the triple integral

$$\mathcal{U}=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{x-y}^{x+y}\mathrm{d}z\,\sqrt{S{\left(x,y,z\right)}}\approx0.117599.$$


Observe that $S$ has the following scaling property:

$$\begin{align} S{\left(x,xt,xu\right)} &=\frac14\sqrt{(x+xt+xu)(-x+xt+xu)(x-xt+xu)(x+xt-xu)}\\ &=\frac14\sqrt{x^{4}(1+t+u)(-1+t+u)(1-t+u)(1+t-u)}\\ &=x^{2}\cdot\frac14\sqrt{(1+t+u)(-1+t+u)(1-t+u)(1+t-u)}\\ &=x^{2}S{\left(1,t,u\right)}.\\ \end{align}$$

We can use this property to trivialize one of the integrations of $\mathcal{U}$ and reduce it to a double integral:

$$\begin{align} \mathcal{U} &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\int_{x-y}^{x+y}\mathrm{d}z\,\sqrt{S{\left(x,y,z\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\,x\int_{1-t}^{1+t}\mathrm{d}u\,x\sqrt{S{\left(x,xt,xu\right)}};~~~\small{\left[y=xt, z=xu\right]}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\int_{1-t}^{1+t}\mathrm{d}u\,x^{2}\sqrt{x^{2}S{\left(1,t,u\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}t\int_{1-t}^{1+t}\mathrm{d}u\,x^{3}\sqrt{S{\left(1,t,u\right)}}\\ &=\int_{0}^{1}\mathrm{d}t\int_{1-t}^{1+t}\mathrm{d}u\int_{0}^{1}\mathrm{d}x\,x^{3}\sqrt{S{\left(1,t,u\right)}}\\ &=\int_{0}^{1}\mathrm{d}t\int_{1-t}^{1+t}\mathrm{d}u\,\frac14\sqrt{S{\left(1,t,u\right)}}.\\ \end{align}$$

With a couple more simple substitutions, we can rewrite this double integral in a slightly less cumbersome form:

$$\begin{align} \mathcal{U} &=\frac14\int_{0}^{1}\mathrm{d}t\int_{1-t}^{1+t}\mathrm{d}u\,\sqrt{S{\left(1,t,u\right)}}\\ &=\frac14\int_{0}^{1}\mathrm{d}t\int_{1-t}^{1+t}\mathrm{d}u\,\sqrt{\frac14\sqrt{(1+t+u)(-1+t+u)(1-t+u)(1+t-u)}}\\ &=\frac18\int_{0}^{1}\mathrm{d}t\int_{1-t}^{1+t}\mathrm{d}u\,\sqrt[4]{(1+t+u)(-1+t+u)(1-t+u)(1+t-u)}\\ &=\frac18\int_{0}^{1}\mathrm{d}t\int_{1-t}^{1+t}\mathrm{d}u\,\sqrt[4]{\left[(1+t)^{2}-u^{2}\right]\left[u^{2}-(1-t)^{2}\right]}\\ &=\frac18\int_{0}^{1}\mathrm{d}t\int_{\frac{1-t}{1+t}}^{1}\mathrm{d}x\,(1+t)\sqrt[4]{\left[(1+t)^{2}-(1+t)^{2}x^{2}\right]\left[(1+t)^{2}x^{2}-(1-t)^{2}\right]};~~~\small{\left[u=(1+t)x\right]}\\ &=\frac18\int_{0}^{1}\mathrm{d}t\int_{\frac{1-t}{1+t}}^{1}\mathrm{d}x\,(1+t)^{2}\sqrt[4]{\left(1-x^{2}\right)\left[x^{2}-\left(\frac{1-t}{1+t}\right)^{2}\right]}\\ &=\frac18\int_{0}^{1}\mathrm{d}y\,\frac{2}{(1+y)^{2}}\int_{y}^{1}\mathrm{d}x\,\frac{4}{(1+y)^{2}}\sqrt[4]{\left(1-x^{2}\right)\left(x^{2}-y^{2}\right)};~~~\small{\left[t=\frac{1-y}{1+y}\right]}\\ &=\int_{0}^{1}\mathrm{d}y\int_{y}^{1}\mathrm{d}x\,\frac{\sqrt[4]{\left(1-x^{2}\right)\left(x^{2}-y^{2}\right)}}{\left(1+y\right)^{4}}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\frac{\sqrt[4]{\left(1-x^{2}\right)\left(x^{2}-y^{2}\right)}}{\left(1+y\right)^{4}}.\\ \end{align}$$


By the general binomial theorem, a binomial with negative integer exponent has the following power series expansion:

$$\left(1-z\right)^{-p}=\sum_{n=0}^{\infty}\binom{p+n-1}{n}z^{n};~~~\small{p\in\mathbb{N}\land|z|<1}.$$

Using the lemma above to expand the $(1+y)^{-4}$ factor of the integrand of $\mathcal{U}$ as a binomial series, we then use technique of switching the order of summation and integration to find

$$\begin{align} \mathcal{U} &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\left(1+y\right)^{-4}\sqrt[4]{\left(1-x^{2}\right)\left(x^{2}-y^{2}\right)}\\ &=\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,\sum_{n=0}^{\infty}\binom{n+3}{n}(-y)^{n}\sqrt[4]{\left(1-x^{2}\right)\left(x^{2}-y^{2}\right)}\\ &=\sum_{n=0}^{\infty}(-1)^{n}\binom{n+3}{n}\int_{0}^{1}\mathrm{d}x\int_{0}^{x}\mathrm{d}y\,y^{n}\sqrt[4]{\left(1-x^{2}\right)\left(x^{2}-y^{2}\right)}\\ &=\sum_{n=0}^{\infty}(-1)^{n}\binom{n+3}{n}\int_{0}^{1}\mathrm{d}x\int_{0}^{1}\mathrm{d}v\,x^{n+1}v^{n}\sqrt[4]{\left(1-x^{2}\right)x^{2}\left(1-v^{2}\right)};~~~\small{\left[y=xv\right]}\\ &=\sum_{n=0}^{\infty}(-1)^{n}\binom{n+3}{n}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\,2^{-2}t^{n/2}u^{(n-1)/2}\sqrt[4]{t\left(1-t\right)\left(1-u\right)};~~~\small{\left[x^{2}=t,\,v^{2}=u\right]}\\ &=\frac14\sum_{n=0}^{\infty}(-1)^{n}\binom{n+3}{n}\int_{0}^{1}\mathrm{d}t\int_{0}^{1}\mathrm{d}u\,t^{n/2+1/4}\left(1-t\right)^{1/4}u^{(n-1)/2}\left(1-u\right)^{1/4}\\ &=\frac14\sum_{n=0}^{\infty}(-1)^{n}\binom{n+3}{n}\int_{0}^{1}\mathrm{d}t\,t^{n/2+1/4}\left(1-t\right)^{1/4}\int_{0}^{1}\mathrm{d}u\,u^{(n-1)/2}\left(1-u\right)^{1/4}\\ &=\frac14\sum_{n=0}^{\infty}(-1)^{n}\binom{n+3}{n}\operatorname{B}{\left(\frac{n}{2}+\frac54,\frac54\right)}\operatorname{B}{\left(\frac{n+1}{2},\frac54\right)}\\ &=\sum_{n=0}^{\infty}\frac{(-1)^{n}}{4}\cdot\frac{(n+3)!}{n!\,3!}\cdot\frac{\Gamma{\left(\frac{n}{2}+\frac54\right)}\,\Gamma{\left(\frac54\right)}}{\Gamma{\left(\frac{n}{2}+\frac52\right)}}\cdot\frac{\Gamma{\left(\frac{n+1}{2}\right)}\,\Gamma{\left(\frac54\right)}}{\Gamma{\left(\frac{n}{2}+\frac74\right)}}\\ &=\frac16\left[\Gamma{\left(\frac54\right)}\right]^{2}\sum_{n=0}^{\infty}(n+2)\,(-1)^{n}\,\frac{\Gamma{\left(\frac{n}{2}+\frac54\right)}}{\Gamma{\left(\frac{n}{2}+\frac74\right)}}\\ &=\frac{\left[\Gamma{\left(\frac54\right)}\right]^{2}}{8\,\Gamma{\left(\frac52\right)}}\sum_{n=0}^{\infty}(n+2)\,(-1)^{n}\,\frac{\Gamma{\left(\frac{n}{2}+\frac54\right)}\,\Gamma{\left(\frac12\right)}}{\Gamma{\left(\frac{n}{2}+\frac74\right)}}\\ &=\frac18\operatorname{B}{\left(\frac54,\frac54\right)}\sum_{n=0}^{\infty}(n+2)\,(-1)^{n}\operatorname{B}{\left(\frac{n}{2}+\frac54,\frac12\right)},\\ \end{align}$$

where the beta function and the gamma function are defined in the usual way by their respective integral representations,

$$\operatorname{B}{\left(x,y\right)}=\int_{0}^{1}\mathrm{d}t\,t^{x-1}\left(1-t\right)^{y-1};~~~\small{x>0\land y>0},$$

$$\Gamma{\left(z\right)}:=\int_{0}^{\infty}\mathrm{d}t\,t^{z-1}\exp{\left(-t\right)};~~~\small{z>0},$$

and are related through the identity

$$\operatorname{B}{\left(x,y\right)}=\frac{\Gamma{\left(x\right)}\,\Gamma{\left(y\right)}}{\Gamma{\left(x+y\right)}};~~~\small{x>0\land y>0}.$$


Define $f$ on $(-1,1)$ by the power series

$$f{\left(z\right)}=\sum_{n=0}^{\infty}z^{n}\operatorname{B}{\left(\frac{n}{2}+\frac54,\frac12\right)};~~~\small{|z|<1}.$$

Differentiating term-by-term, we have

$$f^{\prime}{\left(z\right)}=\sum_{n=0}^{\infty}nz^{n-1}\operatorname{B}{\left(\frac{n}{2}+\frac54,\frac12\right)},$$

and it follows that

$$2f{\left(z\right)}+zf^{\prime}{\left(z\right)}=\sum_{n=0}^{\infty}\left(n+2\right)z^{n}\operatorname{B}{\left(\frac{n}{2}+\frac54,\frac12\right)}.$$

We can convert the series representation for $f$ into an integral by expressing the beta function through its integral representation and summing under the integral sign:

$$\begin{align} f{\left(z\right)} &=\sum_{n=0}^{\infty}z^{n}\operatorname{B}{\left(\frac{n}{2}+\frac54,\frac12\right)}\\ &=\sum_{n=0}^{\infty}z^{n}\int_{0}^{1}\mathrm{d}t\,t^{n/2+1/4}\left(1-t\right)^{-1/2}\\ &=\sum_{n=0}^{\infty}z^{n}\int_{0}^{1}\mathrm{d}t\,\frac{t^{n/2+1/4}}{\sqrt{1-t}}\\ &=\sum_{n=0}^{\infty}z^{n}\int_{0}^{1}\mathrm{d}u\,\frac{2u^{n+1}\sqrt{u}}{\sqrt{1-u^{2}}};~~~\small{\left[t=u^{2}\right]}\\ &=\int_{0}^{1}\mathrm{d}u\,\sum_{n=0}^{\infty}\frac{2z^{n}u^{n+1}\sqrt{u}}{\sqrt{1-u^{2}}}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{2u\sqrt{u}}{\left(1-zu\right)\sqrt{1-u^{2}}}.\\ \end{align}$$

Then,

$$\begin{align} f^{\prime}{\left(z\right)} &=\frac{d}{dz}\int_{0}^{1}\mathrm{d}u\,\frac{2u\sqrt{u}}{\left(1-zu\right)\sqrt{1-u^{2}}}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{\partial}{\partial z}\frac{2u\sqrt{u}}{\left(1-zu\right)\sqrt{1-u^{2}}}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{2u^{2}\sqrt{u}}{\left(1-zu\right)^{2}\sqrt{1-u^{2}}},\\ \end{align}$$

and so

$$\begin{align} \sum_{n=0}^{\infty}\left(n+2\right)z^{n}\operatorname{B}{\left(\frac{n}{2}+\frac54,\frac12\right)} &=2f{\left(z\right)}+zf^{\prime}{\left(z\right)}\\ &=2\int_{0}^{1}\mathrm{d}u\,\frac{2u\sqrt{u}}{\left(1-zu\right)\sqrt{1-u^{2}}}+z\int_{0}^{1}\mathrm{d}u\,\frac{2u^{2}\sqrt{u}}{\left(1-zu\right)^{2}\sqrt{1-u^{2}}}\\ &=\int_{0}^{1}\mathrm{d}u\,\frac{2\left(2-zu\right)u\sqrt{u}}{\left(1-zu\right)^{2}\sqrt{1-u^{2}}}.\\ \end{align}$$

In the limit as $z\to-1^{+}$, this becomes

$$\sum_{n=0}^{\infty}\left(n+2\right)(-1)^{n}\operatorname{B}{\left(\frac{n}{2}+\frac54,\frac12\right)}=2\int_{0}^{1}\mathrm{d}u\,\frac{\left(2+u\right)u\sqrt{u}}{\left(1+u\right)^{2}\sqrt{1-u^{2}}}.$$


For any nonnegative integer $n$, observe how the following class of integrals transform:

$$\begin{align} \int_{0}^{1}\mathrm{d}x\,\frac{1}{\left(1+x\right)^{n}\sqrt{x\left(1-x^{2}\right)}} &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\left(1+x\right)^{n}\sqrt{x\left(1-x\right)\left(1+x\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{1}{\left(1+x\right)^{n+1}\sqrt{x\left(\frac{1-x}{1+x}\right)}}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{2}{\left(1+y\right)^{2}}\cdot\frac{1}{\left(\frac{2}{1+y}\right)^{n+1}\sqrt{y\left(\frac{1-y}{1+y}\right)}};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\ &=\int_{0}^{1}\mathrm{d}y\,\frac{\left(1+y\right)^{n}}{2^{n}\sqrt{y\left(1-y^{2}\right)}}.\\ \end{align}$$

Also for any nonnegative integer $n$, we find

$$\begin{align} \int_{0}^{1}\mathrm{d}x\,\frac{x^{n}}{\sqrt{x\left(1-x^{2}\right)}} &=\int_{0}^{1}\mathrm{d}x\,\frac{x^{n-1/2}}{\sqrt{1-x^{2}}}\\ &=\int_{0}^{1}\mathrm{d}t\,\frac{t^{n/2-1/4}}{2\sqrt{t}\sqrt{1-t}};~~~\small{\left[x=\sqrt{t}\right]}\\ &=\frac12\int_{0}^{1}\mathrm{d}t\,t^{n/2-3/4}\left(1-t\right)^{-1/2}\\ &=\frac12\operatorname{B}{\left(\frac{n}{2}+\frac14,\frac12\right)}.\\ \end{align}$$

We then obtain the following result:

$$\begin{align} \int_{0}^{1}\mathrm{d}x\,\frac{\left(2+x\right)x\sqrt{x}}{\left(1+x\right)^{2}\sqrt{1-x^{2}}} &=\int_{0}^{1}\mathrm{d}x\,\frac{\left(2+x\right)x^{2}}{\left(1+x\right)^{2}\sqrt{x\left(1-x^{2}\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\,\left[x-\frac{1}{1+x}+\frac{1}{\left(1+x\right)^{2}}\right]\frac{1}{\sqrt{x\left(1-x^{2}\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\sqrt{x\left(1-x^{2}\right)}}-\int_{0}^{1}\mathrm{d}x\,\frac{1}{\left(1+x\right)\sqrt{x\left(1-x^{2}\right)}}\\ &~~~~~+\int_{0}^{1}\mathrm{d}x\,\frac{1}{\left(1+x\right)^{2}\sqrt{x\left(1-x^{2}\right)}}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\sqrt{x\left(1-x^{2}\right)}}-\int_{0}^{1}\mathrm{d}y\,\frac{\left(1+y\right)}{2\sqrt{y\left(1-y^{2}\right)}}\\ &~~~~~+\int_{0}^{1}\mathrm{d}y\,\frac{\left(1+y\right)^{2}}{2^{2}\sqrt{y\left(1-y^{2}\right)}};~~~\small{\left[x=\frac{1-y}{1+y}\right]}\\ &=\int_{0}^{1}\mathrm{d}x\,\frac{x}{\sqrt{x\left(1-x^{2}\right)}}-\frac14\int_{0}^{1}\mathrm{d}y\,\frac{1}{\sqrt{y\left(1-y^{2}\right)}}\\ &~~~~~+\frac14\int_{0}^{1}\mathrm{d}y\,\frac{y^{2}}{\sqrt{y\left(1-y^{2}\right)}}\\ &=\frac12\operatorname{B}{\left(\frac34,\frac12\right)}-\frac18\operatorname{B}{\left(\frac14,\frac12\right)}+\frac18\operatorname{B}{\left(\frac54,\frac12\right)}.\\ \end{align}$$

Returning finally to the evaluation of $\mathcal{U}$, we have

$$\begin{align} \mathcal{U} &=\frac18\operatorname{B}{\left(\frac54,\frac54\right)}\sum_{n=0}^{\infty}\left(n+2\right)(-1)^{n}\operatorname{B}{\left(\frac{n}{2}+\frac54,\frac12\right)}\\ &=\frac14\operatorname{B}{\left(\frac54,\frac54\right)}\int_{0}^{1}\mathrm{d}x\,\frac{\left(2+x\right)x\sqrt{x}}{\left(1+x\right)^{2}\sqrt{1-x^{2}}}\\ &=\frac14\operatorname{B}{\left(\frac54,\frac54\right)}\left[\frac12\operatorname{B}{\left(\frac34,\frac12\right)}-\frac18\operatorname{B}{\left(\frac14,\frac12\right)}+\frac18\operatorname{B}{\left(\frac54,\frac12\right)}\right]\\ &=\frac{\pi}{12\sqrt{2}}-\frac{\left[\Gamma{\left(\frac14\right)}\right]^{4}}{576\sqrt{2}\,\pi}.\blacksquare\\ \end{align}$$


Now, it would be interesting if your conjectured value is also correct, in which case my result provides an exact value for that combination of elliptic integrals in terms of gamma functions. Perhaps that can be explored later. Cheers!

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    $\begingroup$ +1 for the effort! $\endgroup$
    – Dominique
    Jul 7, 2023 at 9:14
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    $\begingroup$ Regarding the conjectured form it can be proved that if $n$ is a positive rational number then we have $$\frac{\pi\sqrt{n}} {4}=K(k)E(k)-A_nK^2(k)$$ where $k$ is the modulus corresponding to nome $e^{-\pi/\sqrt{n}} $ (ie $K'/K=1/\sqrt{n}$) and $A_n$ is some specific algebraic number dependent on $n$. The expression in question is for $n=5$. For $n=1$ we have $\pi/4=K(k)E(k)-(1/2)K^2(k)$ with $k=1/\sqrt{2}$. $\endgroup$
    – Paramanand Singh
    Jul 9, 2023 at 22:07
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    $\begingroup$ So I think your final closed form is the simplest and most preferred. One can just use many well known values of $k$ and corresponding $A_n$ to generate forms like the one in question for various values of $n$. $\endgroup$
    – Paramanand Singh
    Jul 9, 2023 at 22:09
  • $\begingroup$ +1 But think the complete solution involves also @ParamanandSingh answer wich explains the OP's conjecture. $\endgroup$
    – User
    Oct 17, 2023 at 9:26
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In this post we discuss the genesis of $\alpha, \phi$ present in the closed form solution given in question.

The ideas discussed below were developed by Ramanujan in his famous paper Modular equations and approximations to $\pi$. He dealt with Dedekind eta function $$\eta(q) =q^{1/24}\prod_{n\geq 1}(1-q^n)\tag{1}$$ and its logarithmic derivative $$P(q) =24q\frac{d}{dq}\log\eta(q)=1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}\tag{2}$$ If $k$ is the elliptic modulus and $K, E$ be complete elliptic integrals corresponding to nome $q$ then both the functions above can be expressed in terms of these elliptic integrals and moduli. The relations below were well known before Ramanujan's time \begin{align} \eta(q^2)&=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{3}\\ P(q^2)&=\left(\frac{2K}{\pi}\right)^2\left(\frac{3E}{K}+k^2-2\right)\tag {4} \end{align} Jacobi proved in his Fundamenta Nova that if $r$ is a positive rational number then there is a unique algebraic number $k_r\in(0,1)$ such that $K(k'_r) /K(k_r) =\sqrt{r} $ and such values $k_r$ are called singular moduli. Many such values of singular moduli have been calculated in explicit form involving radicals for various positive integer values of $r$. Further Selberg and Chowla proved that $K(k_r) $ can be expressed in closed form involving $\pi$ and values of gamma function at rational points. In particular the values of elliptic integral $K(k_r) $ were known / given by Legendre for $r=1,2,3$.

The evaluation of $E$ occurring in $(4)$ is rather more troublesome. Ramanujan instead focused on the function $P(q^2)$ and proved that if $q=e^{-\pi\sqrt{r}} $ then the value of $P(q^2)$ could be expressed as $$P(q^2)=\frac{3}{\pi\sqrt{r}}+A_r\cdot\left(\frac{2K(k_r)}{\pi}\right)^2\tag{5}$$ where $A_r$ is some algebraic number dependent on $r$. Using this equation with $(4)$ one can get a closed form for $E(k_r) $. Comparing the equations $(4),(5)$ we get $$\frac{3}{\pi\sqrt{r}}=\left(\frac{2K(k_r)}{\pi}\right)^2\left(\frac{3E(k_r)}{K(k_r)}+k_r^2-2-A_r\right)$$ or $$\frac{\pi} {4\sqrt{r}}=K(k_r)E(k_r)+B_rK^2(k_r)\tag{6}$$ where $B_r$ is an algebraic number dependent on $r$.

Ramanujan described the process to get the values $A_r$ and gave some concrete results for small integer values of $r$. The formulas given by Ramanujan were later proved by Bruce C. Berndt and his collaborators but no new formulas of this kind have appeared for any new values of $r$.

The value of $\alpha$ given in question is a singular modulus corresponding to nome $e^{-\pi/\sqrt{5}}$ ie $\alpha=k(e^{-\pi/\sqrt{5}})$ and this corresponds to $r=1/5$ (which is linked with $r=5$).

We have $$k_5=\frac{\sqrt{\sqrt {5}-1}-\sqrt{3-\sqrt{5}}}{2}=\frac{1}{\sqrt{2\phi}}-\frac{1}{\phi\sqrt{2}}$$ and $$k'_5=\frac{\sqrt{\sqrt {5}-1}+\sqrt{3-\sqrt{5}}}{2}=\frac{1}{\sqrt{2\phi}}+\frac{1}{\phi\sqrt{2}}=\alpha$$ Another radical needed in our calculations is $$2k_5k'_5=\sqrt{5}-2=2\phi-3$$ The value of $K(\alpha)$ is $$K(\alpha) =\sqrt{5}K(k_5)$$ For $r=5$ Ramanujan proved that $$5P(q^{10})-P(q^2)=\frac{4KL}{\pi^2}(3+kl+k'l')\sqrt{\frac{1+kl+k'l'}{2}}$$ where moduli $k, l$ correspond to nomes $q, q^5$. Putting $q=\exp(-\pi/\sqrt{5})$ so that $k=\alpha, l=k'$ we get $$5P(e^{-2\pi\sqrt{5}})-P(e^{-2\pi/\sqrt{5}})=\frac{4KK'}{\pi^2}(3+2kk')\sqrt{\frac{1+2kk'}{2}}$$ We have $K=K(\alpha) $ and $K'=K(\alpha) /\sqrt{5}$ and $2kk'=\sqrt{5}-2$ and hence we get $$5P(e^{-2\pi\sqrt{5}})-P(e^{-2\pi/\sqrt{5}})=\frac{4K^2(\alpha)}{\pi^2}\frac{2\sqrt{\phi}}{\sqrt{5}}$$ Another identity we need here is $$rP(e^{-2\pi\sqrt{r}})+P(e^{-2\pi/\sqrt {r}})=\frac{6\sqrt{r}}{\pi}$$ Using these two equations (with $r=5$) we get $$P(q^2)=\frac{3\sqrt{5}}{\pi}-\frac{4K^2}{\pi}\cdot\frac{\phi}{\sqrt{5\phi}}$$ and comparing with $(4)$ we get $$\frac{\pi\sqrt{5}}{4}=K(\alpha)E(\alpha) - \left(\frac{1}{2}-\frac{1}{\sqrt{5\phi}}\right)K^2(\alpha)$$

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  • $\begingroup$ (+1) Nice improvement. $\endgroup$
    – User
    Oct 17, 2023 at 9:27

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