Evaluate $\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{\sqrt{3-x^2-y^2-z^2} }\text{d}x \text{d}y\text{d}z$

Question

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.

Answer 1

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

Answer 2

Another way is to start with the easy first iteration: $$ I=\int_{0}^{1} \int_{0}^{1} \int_{0}^{1} \frac{1}{\sqrt{3-x^2-y^2-z^2} }\text{d}x \text{d}y\text{d}z=\int_{0}^{1} \int_{0}^{1}\operatorname{arccot}\sqrt{2-x^2-y^2} $$ and then switching to polar coordinates: $$I=2\int_0^{\frac\pi 4}\int_0^{\sec\theta }r\operatorname{arccot}\sqrt{2-r^2}\,drd\theta=\int_0^{\frac\pi 4}\int_{2-\sec^2\theta}^2\operatorname{arccot}\sqrt{u}\,dud\theta $$ Luckily $\operatorname{arccot}\sqrt u$ has a nice anti-derivative and we get $$I=\frac{\sqrt2}4 \pi+\frac34\pi\operatorname{arccot}\sqrt2-I_1+I_2-3I_3$$ where $$I_1=\int_0^{\frac\pi4}\sqrt{2-\sec^2\theta}\,d\theta$$ $$I_2=\int_0^{\frac\pi4}\sec^2\theta\operatorname{arccot}\sqrt{2-\sec^2\theta}\,d\theta$$ and $$I_3=\int_0^{\frac\pi4}\operatorname{arccot}\sqrt{2-\sec^2\theta}\,d\theta.$$ I think, each integral $I_k$ above is an Ahmed-type integral and can be evaluated by similar techniques. You may see this answer. If I did not do any mistakes, the final (exact) result is $$I=\frac{3}8\pi^2+\frac32\pi-\frac{3\sqrt2}4 \pi-\frac94\pi\operatorname{arccot}\sqrt2$$

Answer 3

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

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

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

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

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

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

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

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

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

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

$\rho=2u$

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

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

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

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

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

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

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

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

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

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

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

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

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$v=2\gamma$

$dv=2\ d\gamma$

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

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

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

Answer 4

Alternatively, convert to spherical coordinates and exploit the symmetry within the cube.

$$\begin{align*} I &= 6 \int_0^\tfrac\pi4 \int_0^{\arctan(\sec\theta)} \int_0^{\sec\phi} \frac{\rho^2 \sin\phi}{\sqrt{3-\rho^2}} \, d\rho \, d\phi \, d\theta \\ &= 18 \int_0^\tfrac\pi4 \int_0^{\arctan(\sec\theta)} \int_0^{\arcsin\left(\tfrac{\sec\phi}{\sqrt3}\right)} \sin^2u \sin\phi \, du \,d\phi \, d\theta & \rho=\sqrt3\sin u \\ &= \int_0^\tfrac\pi4 \int_0^{\arctan(\sec\theta)} \left[9 \arcsin\left(\frac{\sec\phi}{\sqrt3}\right)\sin\phi- 3\tan\phi\sqrt{2-\tan^2\phi}\right] \, d\phi\,d\theta \\ &= \int_0^\tfrac\pi4 \int_0^{\sec\theta} \left[\frac{9v\arcsin\sqrt{\frac{1+v^2}3}}{\left(1+v^2\right)^{3/2}}-\frac{3v\sqrt{2-v^2}}{1+v^2}\right] \, dv \, d\theta & v=\tan\phi \\ &= 3\sqrt3 \int_0^\tfrac\pi4 \int_{\sqrt{\tfrac3{1+\sec^2\theta}}}^{\sqrt3} \left(\arcsin w - w\sqrt{1-w^2}\right) \, \frac{dw}{w^2} \, d\theta & w=\sqrt{\frac{1+\phi^2}3} \\ &= \frac{9\pi}4 \left(\operatorname{arccsc}\sqrt3-\frac{\sqrt2}3\right) + 3 \underbrace{\int_0^\tfrac\pi4 \sqrt{1-\tan^2\theta} \, d\theta}_J \\&\qquad - 9\underbrace{\int_0^\tfrac\pi4 \frac{\arcsin\sqrt{\frac{1+\sec^2\theta}3}}{\sqrt{1+\sec^2\theta}} \, d\theta}_K + 3\sqrt3\int_0^\tfrac\pi4\int_\tfrac1{\sqrt3}^{\sqrt{\tfrac{1+\sec^2\theta}3}}\frac{2w}{\sqrt{1-w^2}}\,dw\,d\theta & \text{by parts} \\ &= \frac{9\pi}4\left(\operatorname{arccsc}\sqrt3+\frac{\sqrt2}3\right) - 3J - 9K \end{align*}$$

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$$\begin{align*} J&= \int_0^\tfrac\pi4 \frac{2\sqrt{\tan\theta}}{1+\tan\theta} \,d\theta & \theta\to\frac\pi4-\theta \\ &= \int_0^1 \frac{4t^2}{\left(1+t^2\right)\left(1+t^4\right)} \, dt & t^2=\tan\theta \\ &= 2\int_0^1\left(\frac{\frac1{t^2}+1}{\frac1{t^2}+t^2} - \frac1{1+t^2}\right) \, dt &\text{partial fractions} \\ &= 2\int_0^\infty\frac{d\tau}{2+\tau^2}-2\int_0^1\frac{dt}{1+t^2} & \tau=\frac1t-t \\ &= \frac{\pi\left(\sqrt2-1\right)}2 \end{align*}$$

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$$\begin{align*} K&= \sqrt3 \int_{\sqrt{\tfrac23}}^1 \frac{\arcsin s}{\left(3s^2-1\right)\sqrt{3s^2-2}} \, ds & \sqrt3\,s=\sqrt{1+\sec^2\theta} \\ &= \sqrt3 \int_{\sqrt2}^\infty \frac{\arctan \sigma}{\left(2\sigma^2-1\right)\sqrt{\sigma^2-2}} \,d\sigma & s=\frac\sigma{\sqrt{1+\sigma^2}}\\ &= \sqrt3\int_0^\infty \frac{\arctan\sqrt{\sigma^2+2}}{\left(2\sigma^2+3\right)\sqrt{\sigma^2+2}}\,d\sigma & \sigma\to\sqrt{\sigma^2+2}\\ &= \frac\pi2 \operatorname{arccot}\sqrt2 - \frac{\pi^2}{24} \end{align*}$$

The latter two integrals are nearly duplicated in this question; the method in Sangchul Lee's answer makes short work of it.