Possible closed form of $\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)(1+z^2)\sqrt{3-x^2-y^2-z^2} }\text{d}x\text{d}y\text{d}z$

Question

Reviewing the link, I consider to evaluate $$ \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)(1+z^2)\sqrt{3-x^2-y^2-z^2} }\text{d}x\text{d}y\text{d}z. $$ And I quickly discover $$ \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{(1+x^2)(1+y^2)(1+z^2)\sqrt{3-x^2-y^2-z^2} }\text{d}x\text{d}y\text{d}z\\ =\frac{9\sqrt{2}\pi }{4}\int_{\sqrt{\frac53}}^{\sqrt{2} } \frac{\arctan(x)}{\left ( 2x^2-3 \right )\sqrt{3x^2-5} }\text{d}x -\frac{\pi^3}{\sqrt{6} }+\frac{3\pi^2}{2\sqrt{6} }\arctan(2\sqrt{6} ). $$ I don't know whether this is helpful or not. But I instinctively know it has a sufficient simple result, which only appears $\pi,\arctan$ and some quadratic irrationals.

Hopefully you are glad to reach for my hand.

Answer 1

Let us denote by $J$ the integral to be computed. We have $$ \begin{aligned} J&=\int_{\sqrt{5/3}}^{\sqrt2} \frac{\arctan x}{( 2x^2-3 )\sqrt{3x^2-5}} \; dx \\ &= \int_{\sqrt{1/2}}^{\sqrt{3/5}} x\cdot\frac{\arctan (1/x)}{( 2-3x^2 )\sqrt{3 - 5x^2}} \; dx = \int_{\sqrt{1/2}}^{\sqrt{3/5}} x\cdot \frac{\frac\pi 2-\arctan x}{( 2-3x^2 )\sqrt{3 - 5x^2}} \; dx \\ &= \frac\pi2\cdot\frac 12\cdot \int_{1/2}^{3/5} \frac{du}{( 2-3u )\sqrt{3 - 5u}} \; du - \frac 12 \int_{1/2}^{3/5} \frac{\arctan \sqrt u}{( 2-3u )\sqrt{3 - 5u}} \; du \\ &=\frac \pi4\left( \frac{\sqrt2}3-\frac{2\sqrt3}9\arctan\sqrt{\frac32} \right) - \frac 1{6\sqrt 5} \underbrace{ \color{blue}{ \int_{1/2}^{3/5} \frac{\arctan \sqrt u}{\left( \frac 23-u \right)\sqrt{\frac 35 - u}} \; du}} _{=:\color{blue}{K}} \ . \end{aligned} $$ So let us compute $K$. Recall from a related post the excellent presentation of pisco the way to compute such integrals. He also cites the book of Lewin,

Polylogarihm and Associated Functions, Leonard Lewin, page 115-117.

We have with the conventions and definitions in loc. cit. the relation $$ \color{blue}{\int_0^x\frac{\arctan\sqrt t}{(a-t)\sqrt{b-t}}\; dt} = \frac 1{\sqrt{a-b}} S\left(\ \arctan \sqrt{\frac{b-x}{a-b}}\ ,\ \arctan \sqrt{\frac{b+1}{a-b}}\ ,\ \arctan \sqrt{\frac1a}\ \right) \ , $$ where $S(\alpha, \beta,\gamma)$ is defined and computed as follows. First of all associate the following "modulus" $k=k(\alpha,\beta,\gamma)$. $$ \begin{aligned} d_1 &= \sqrt{\cos^2\alpha \cos^2\gamma-\cos^2\beta}\ ,\\ d_2 &= \sin\alpha \sin\gamma\ ,\\ k &=\frac{d_1-d_2}{d_1+d_2}\ . \end{aligned} $$ Then set for this value of $k=k(\alpha,\beta,\gamma)$: $$ \begin{aligned} S(\alpha,\beta,\gamma) &=f(\alpha,k)-f(\beta,k)+f(\gamma,k)-f(0,k)-\alpha^2+\beta^2-\gamma^2 \ ,\\ f(\alpha,k) &=\sum_{n\ge 1}\frac1{n^2}k^n\cos(2n\alpha) =\Re\sum_{n\ge 1}\frac1{n^2}(ke^{2\alpha i})^n =\Re\operatorname{Li}_2\Big(ke^{2\alpha i}\Big) \ . \end{aligned} $$ Then $S$ has the following properties, see again loc. cit.: $$ \begin{aligned} S(0,\beta,\gamma) &= \pi(\beta-\gamma)\ ,\\ S(\alpha,\pi-2\alpha,\alpha) &= 6\cdot S\left(\alpha,\frac\pi 3,\frac \pi6\right) \ . \end{aligned} $$ In our case we have

• $a=\frac 23$, $b=\frac35$, $x_1=\frac 12$, $x_2=\frac 35=b$,
• so $a-b=\frac1{15}$, $b-x_1=\frac 1{10}$, $b-x_2=0$,
• and associate $\alpha=\alpha_1=\arctan\sqrt{\frac{b-x_1}{a-b}}=\arctan\sqrt{\frac{1/10}{1/15}}=\arctan\sqrt{\frac 32}$, later we also need $\cos^\alpha=\frac 1{1+\tan^2\alpha}=\frac 1{5/2}=\frac 25$, and $\sin^\alpha=\frac 35$,
• $\alpha_2=\arctan\sqrt{\frac{b-x_2}{a-b}}=\arctan 0=0$,
• $\beta=\arctan\sqrt{\frac{b+1}{a-b}}=\arctan\sqrt{\frac{8/5}{1/15}}=\arctan\sqrt{24}=\pi-2\alpha$,
• $\gamma=\arctan\sqrt{\frac1a}=\arctan\sqrt{\frac32}$.

and have to compute $$ \begin{aligned} \color{blue}{K}&= \frac 1{\sqrt{a-b}} \left[\ S\left(\ \arctan \sqrt{\frac{b-x}{a-b}}\ ,\ \arctan \sqrt{\frac{b+1}{a-b}}\ ,\ \arctan \sqrt{\frac1a}\ \right)\ \right]_{x=x_1}^{x=x_2} \\ &= \sqrt{15}S\left(\ 0\ ,\ \arctan \sqrt {24}\ ,\ \sqrt{\frac32}\ \right) \\ &\qquad -\sqrt{15}S\left(\ \arctan\sqrt{\frac32}\ ,\arctan \sqrt {24}\ ,\ \arctan\sqrt{\frac32}\ \right) \\ &=\sqrt{15}(S_1-S_2)\ ,\\[3mm] S_1 &:=S(\alpha_1,\beta,\gamma)=S(0,\beta,\gamma)=\pi(\beta-\gamma)=\pi\left(\arctan \sqrt{24} - \arctan\sqrt{\frac32}\right) \\ &=\pi\arctan\frac{3\sqrt 6}{14}\ , \\ S_2 &=S(\alpha_2,\beta,\gamma)=S(\alpha,2\pi-\alpha,\alpha)= 6\cdot S\left(\alpha,\frac\pi 3,\frac \pi6\right)\ , \\ &\qquad \text{ and only $S_2$ has to be computed to complete.} \\ &\qquad \text{ Above, there are two way to do this. Same modulus $k$.} \\ k\left(\alpha,\beta,\gamma\right) &%=\frac{d_1-d_2}{d_1+d_2} =\frac {\sqrt{\frac25\cdot\frac 25-\frac 1{25}}-\frac 35} {\sqrt{\frac25\cdot\frac 25-\frac 1{25}}+\frac 35} = \frac{\sqrt 3-3}{\sqrt 3+3} =-\frac 12(\sqrt 3-1)^2=\sqrt 3-2\ , \\ k\left(\alpha,\frac\pi 3,\frac \pi6\right) &=\frac {\sqrt{\frac25\cdot\frac 34-\frac 14}-\frac 12\cdot\sqrt{\frac 35}} {\sqrt{\frac25\cdot\frac 34-\frac 14}+\frac 12\cdot\sqrt{\frac 35}} = \frac{\sqrt{\frac 15}-\sqrt{\frac 35}}{\sqrt{\frac 15}+\sqrt{\frac 35}} =-\frac 12(\sqrt 3-1)^2=\sqrt 3-2\ , \end{aligned} $$

We have now an explicit expression in terms of dilogarithmic values and "easier" data for the integral $K=\sqrt{15}(S_1-S_2)$, thus also for $J$. Here, $S_1$ is also simple, but $S_2$ involves the (real part of the) dilogarithm $\operatorname{Li}_2$ computed in points like $$ (\sqrt 3-2)\cdot\left(\sqrt{\frac 25}-i\sqrt{\frac 35}\right)^2\ ,\\ (\sqrt 3-2)\cdot\frac 12(-1+\sqrt 3)\ ,\qquad (\sqrt 3-2)\cdot\frac 12(1+i\sqrt 3)\ ,\qquad (\sqrt 3-2)\cdot 1\ . $$ And there is no K-theoretic trick to get something like a multiple of $\pi$.

I have to stop here, computations were done without double check, there may be some, but the computational line applies at any rate.

Answer 2

Though I know it probably can't be solved elegantly, I still want to mention something about the integral relation given by me. In fact, the following generalization holds.

For non-negative integers $\alpha,\beta$ satisfying $\alpha\ge\beta$, define $$I(\alpha,\beta)=\int_{[0,1]^{\alpha-\beta}} \frac{\text{d}x_1...\text{d}x_{\alpha-\beta}}{(1+x_1^2)...(1+x_{\alpha-\beta}^2)\sqrt{\alpha+\beta-x_1^2-...-x_{\alpha-\beta}^2}}.$$ If $\beta=\alpha$, it equals $$ I(\alpha,\alpha)=\frac{1}{\sqrt{2\alpha}}. $$ Then $$\sum_{0\le\beta\le\alpha}c_\beta\pi^{\beta}I(\alpha,\beta)=0.$$ Where the coefficients $c_\beta\in\mathbb{Z}$.

• For example, let $\alpha=2$. We simply calculates $$ I(2,0)=\int_{0}^{1} \int_{0}^{1} \frac{\text{d}x\text{d}y}{(1+x^2)(1+y^2)\sqrt{2-x^2-y^2} } =\frac\pi2\arctan\left ( \frac{1}{2\sqrt{2}} \right ),\quad I(2,1)=\int_{0}^{1} \frac{\text{d}x}{(1+x^2)\sqrt{3-x^2} } =\frac12\arctan\left ({\sqrt{2}} \right ),\quad I(2,2)=\frac12. $$ And immediately we have $$ 2I(2,0)-4\pi I(2,1)+\pi^2 I(2,2)=0. $$
• For $\alpha=3$, $$ 2I(3,0)-9\pi I(3,1)+6\pi^2 I(3,2)-\pi^3 I(3,3)=0. $$ And this is just the claimed one.
• For $\alpha=4$, $$ 2I(4,0)-16\pi I(4,1)+20\pi^2 I(4,2)-8\pi^3 I(4,3)+\pi^4 I(4,4)=0. $$ And we have $$ \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{8\pi-\arctan\left(\sqrt{\frac{5-x^2-y^2-z^2}{3-x^2-y^2-z^2} } \right) }{(1+x^2)(1+y^2)(1+z^2)\sqrt{5 -x^2-y^2-z^2} }\text{d}x\text{d}y\text{d}z =5\pi^2\sqrt{2}\int_{\sqrt{\frac75}}^{\sqrt{\frac32} } \frac{\arctan(x)}{\left ( 3x^2-4 \right )\sqrt{5x^2-7} }\text{d}x -\frac{3\pi^4}{4\sqrt{2} }+\frac{\pi^3}{\sqrt{2} }\arctan(4\sqrt{3}). $$
• For $1\le\alpha\le10$, the coefficients are listed below. \begin{matrix}\hline \alpha& c_0 &c_1 & c_2& c_3& c_4& c_5 & c_6 & c_7& c_8& c_9& c_{10}\\\hline 1& 2& -1& .& . & . & . & .& .& . & .& .\\ 2 & 2& -4& 1& .& . & . & .& .& .& .& .\\ 3& 2& -9& 6& -1&. & . & . & . & . & . & .\\ 4& 2& -16& 20& -8& 1& . & .& . & . & . & .\\ 5& 2& -25& 50& -35& 10& -1& . & . &. & . & .\\ 6 & 2& -36& 105& -112& 54& -12& 1& .& . & .& .\\ 7& 2& -49& 196& -294& 210& -77 & 14 & -1 & . & .&. \\ 8& 2& -64& 336& -672& 660& -352& 104& -16& 1 & .&.\\ 9 & 2& -81& 540& -1386& 1782& -1287& 546& -135 & 18& -1 &. \\ 10 & 2& -100& 825& -2640& 4290& -4004& 2275& -800 & 170 & -20 & 1\\ \hline \end{matrix}