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How to evaluate $$ I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{\sqrt{3-x^2-y^2-z^2} }\text{d}x \text{d}y\text{d}z? $$ Some simple calculation shows that $$ I=\frac{\sqrt{2} -1}{4}\pi+\frac{\pi^2}{4} -\frac{3\pi}{4}\arctan\left ( \sqrt{2} \right )+\frac12\int_{0}^{1}\arctan\left ( \frac{1}{\sqrt{x} } \right ) \left ( \arcsin\left ( \frac{1}{\sqrt{2-x} } \right ) -\arctan\left ( \sqrt{1-x} \right ) \right ) \text{d}x $$ which is out of my capability.

Setness Ramesory
  • 5,172

2 Answers2

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Utilize the identities

$$\arcsin\frac1{\sqrt{2-x}} = \arctan\frac1{\sqrt{1-x}} = \frac\pi2 - \arctan \sqrt{1-x} \\ \arctan\frac{1-x}{1+x} = \frac\pi4 - \arctan x$$

and substitute $x=1-y^2$ to help simplify the remaining integral,

$$\begin{align*} J &= \int_0^1 \arctan \frac1{\sqrt x} \left(\arcsin\frac1{\sqrt{2-x}} - \arctan \sqrt{1-x}\right) \, dx \\ &= \int_0^1 \arctan \frac1{\sqrt x} \left(\frac\pi2 - 2 \arctan \sqrt{1-x}\right) \, dx \\ &= \int_0^1 2 \arctan \frac1{\sqrt x} \arctan \frac{1-\sqrt{1-x}}{1+\sqrt{1-x}} \, dx \\ &= \int_0^1 4y \arctan \frac1{\sqrt{1-y^2}} \arctan\frac{1-y}{1+y} \, dy \\ &= \frac12 \int_0^1 y \left(\pi - 2\arctan\sqrt{1-y^2}\right) \left(\pi - 4\arctan y\right) \, dy \\ &= \frac{\pi^2}4 - 3\pi \int_0^1 y \arctan y \, dy + 4 \int_0^1 y \arctan y \arctan\sqrt{1-y^2} \, dy \tag{$*$} \\ &= - \frac{\pi^2}2 + \frac{3\pi}2 + 4 \underbrace{\int_0^1 y \arctan y \arctan\sqrt{1-y^2} \, dy}_{=:K} \end{align*}$$

where in $(*)$ we use

$$\int_0^1 y \arctan \sqrt{1-y^2} \, dy \stackrel{z=\sqrt{1-y^2}}= \int_0^1 z \arctan z \, dz$$

Integration by parts yields an Ahmed-like integral:

$$\begin{align*} K &= \int_0^1 \frac{\left(y^2+1\right)\arctan y-y}2 \cdot \frac y{\left(2-y^2\right)\sqrt{1-y^2}} \, dy \\ &= \frac12 \int_0^1 \left(\frac{3y\arctan y}{2-y^2} - 3\arctan y - \frac{y^2}{2-y^2}\right) \, \frac{dy}{\sqrt{1-y^2}} \\[2ex] \implies I &= \frac32 \left(1-\frac1{\sqrt2}\right)\pi - \frac{3\pi}4 \arctan\sqrt2 + 3 \int_0^1 \frac{y \arctan y}{\left(2-y^2\right) \sqrt{1-y^2}} \, dy \end{align*}$$

via Euler substitution plus the closed form of $\displaystyle\int_0^1\frac{\arctan x}{\sqrt{1-x^2}}\,dx$.

This last integral can be evaluated as follows.

$$\begin{align*} & \int_0^1 \frac{y \arctan y}{\left(2-y^2\right) \sqrt{1-y^2}} \, dy \\ &= \int_0^1 \int_0^1 \frac{y^2}{\left(2-y^2\right) \left(1+y^2z^2\right) \sqrt{1-y^2}} \, dz \, dy \\ &= \frac\pi{\sqrt2} \int_0^1 \frac{dz}{1+2z^2} - \frac\pi2 \int_0^1 \frac{dz}{(1+2z^2)\sqrt{1+z^2}} \\ &= \frac\pi2\arctan\sqrt2 - \left(\frac{\pi^2}8 - \frac\pi2 \arctan\left(3-2\sqrt2\right)\right) \\ &= \frac\pi2 \arctan\frac{9+4\sqrt2}7 - \frac{\pi^2}8 \end{align*}$$

Hence

$$\begin{align*} I &= \frac{3\pi}2 \left(1-\frac1{\sqrt2} - \frac{\pi}4 +\arctan\frac{9+4\sqrt2}7 - \frac12 \arctan\sqrt2\right) \\ &= \frac{3\pi}2 \left(1-\frac1{\sqrt2} - \frac{\pi}4 + \frac12 \arctan \frac5{\sqrt2}\right) \\ &= \boxed{\frac{3\pi}2 \left(1-\frac1{\sqrt2} - \frac12 \arctan \frac{\sqrt2}5\right)} \end{align*}$$

with the identity $\arctan x\pm\arctan y=\arctan\dfrac{x\mp y}{1\pm xy}$.

user170231
  • 19,334
1

One easier way to evaluate this kind of integral is to convert to spherical coordinates.

$I=\int_0^1\int_0^1\int_0^1\frac{1}{\sqrt{3-x^2-y^2-z^2}}dxdydz$

$I=\int_0^1\int_0^1\int_0^1\frac{1}{\sqrt{3-\left(x^2+y^2+z^2\right)}}dxdydz$

$\begin{array}{c} 0\le x\le1\\ 0\le y\le1\\ 0\le z\le1 \end{array} \to \begin{array}{c} 0\le\rho\le\sqrt{3}\\ 0\le\theta\le\frac{\pi}{2}\\ 0\le\phi\le\frac{\pi}{2} \end{array}$

$I=\int_0^{\sqrt{3}}\int_0^\frac{\pi}{2}\int_0^\frac{\pi}{2}\frac{1}{\sqrt{3-\rho^2}}\times\rho^2\sin\phi\ d\phi d\theta d\rho$

$I=\int_0^{\sqrt{3}}\int_0^\frac{\pi}{2}\frac{\rho^2}{\sqrt{3-\rho^2}}\left[-\cos\phi\right]_0^\frac{\pi}{2} d\theta d\rho$

$I=\int_0^{\sqrt{3}}\int_0^\frac{\pi}{2}\frac{\rho^2}{\sqrt{3-\rho^2}}d\theta d\rho$

$I=\int_0^{\sqrt{3}}\frac{\rho^2}{\sqrt{3-\rho^2}}\left[\theta\right]_0^\frac{\pi}{2}d\rho$

$I=\pi\int_0^{\sqrt{3}}\frac{\rho^2}{\sqrt{3-\rho^2}}\frac{d\rho}{2}$

$u=\frac{\rho}{2}$

$\rho=2u$

$du=\frac{d\rho}{2}$

$I=\pi\int_0^\frac{\sqrt{3}}{2}\frac{\left(2u\right)^2}{\sqrt{3-4u^2}}du$

$\sin\gamma=\frac{2u}{\sqrt{3}}$

$\gamma=\sin^{-1}\left(\frac{2u}{\sqrt{3}}\right)$

$\sqrt{3}\sin\gamma=2u$

$u=\frac{\sqrt{3}}{2}\sin\gamma$

$du=\frac{\sqrt{3}}{2}\cos\gamma\ d\gamma$

$\cos\gamma=\frac{\sqrt{3-4u^2}}{\sqrt{3}}$

$\sqrt{3}\cos\gamma=\sqrt{3-4u^2}$

$I=\pi\int_0^\frac{\pi}{2}\frac{\left(\sqrt{3}\sin\gamma\right)^2}{\sqrt{3}\cos\gamma}\times\frac{\sqrt{3}}{2}\cos\gamma\ d\gamma$

$I=\frac{3\pi}{8}\int_0^\frac{\pi}{2}4\sin^2\gamma\ d\gamma$

$I=\frac{3\pi}{8}\int_0^\frac{\pi}{2}2\left(1-\cos2\gamma\right)d\gamma$

$v=2\gamma$

$dv=2\ d\gamma$

$I=\frac{3\pi}{8}\int_0^\pi\left(1-\cos v\right)dv$

$I=\frac{3\pi}{8}\left[v-\sin v\right]_0^\pi$

$I=\frac{3\pi^2}{8}\approx3.701$

Ryan Wilson
  • 160