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I am trying to evaluate the following integral: $$I =\int_{0}^1 \int_{0}^1 \int_{0}^1 \frac{1}{x^2+y^2+z^2}dxdydz.$$

Here is my trial. By using spherical coordinate, let $x = r\sin(\theta)\cos(\phi), y = r\sin(\theta)\sin(\phi), z = r\cos(\phi).$ Then one can write $I$ as $$ I = \int_{\Omega} \frac{1}{r^2} r^2\sin(\theta)drd\theta d\phi = \int_{\Omega} \sin(\theta)drd\theta d\phi$$ where $\Omega$ is something I can't figure out... Here $\Omega$ is $$ \Omega = \{(r,\theta,\phi) : 0\le x,y,z \le 1\}.$$ Since $0\le z=r\cos(\phi) \le1$, we have $$ 0 \le \arccos(1/r) \le \phi \le \pi/2. $$ Since $0\le y=r\sin(\theta)\cos(\phi) \le 1$, we have $$ 0 \le \theta \le \arcsin(1/(r\cos(\phi)).$$ But I don't see how such classification help in evaluating $I$.

Any suggestions/comments/answers will be very appreciated. Thanks in advance.

induction601
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  • It would help to notice that the $r^2$ terms cancel in the spherical integral. –  Apr 10 '18 at 20:42
  • @majormaki Thanks for your comment. I was aware of this cancellation but it didn't help to me go further. – induction601 Apr 10 '18 at 20:50
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    Well then, integrate $\sin(\theta)$ over $\Omega$: find the bounds. The integrand does not depend on the angular variables so the answer will be a multiple of $-\cos(\theta)$. –  Apr 10 '18 at 20:54
  • I would say you can divide the cube into 3 equal regions and by the symmetry of the function, the integral is the same over all three. Cut the box along the lines $x = y, y = z, x = 0, z = 1$ Even then, the integral seems to get messy. – Doug M Apr 10 '18 at 20:58
  • The integral is equal to $$\int_{0}^{\infty}\frac{\pi^{3/2}}{4t^2}\operatorname{erf}^3 (t),dt, $$ though I am not sure whether it will ever help. – Sangchul Lee Apr 10 '18 at 21:00
  • @SangchulLee Thanks for your comment. This helps a bit as a starting. Could you provide me a reference for this? – induction601 Apr 10 '18 at 21:02
  • @SangchulLee: the integral boils down to a fashionable dilogarithmic integral. – Jack D'Aurizio Apr 10 '18 at 21:13
  • This is a consequence of an easy transformation $$ \frac{1}{x^2+y^2+z^2} = \int_{0}^{\infty} 2t e^{-(x^2+y^2+z^2)t^2} , dt, $$ but Jack D'Aurizio's answer tells that an earnest computation seems to work best. – Sangchul Lee Apr 10 '18 at 21:45

2 Answers2

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I might have a better idea than parametrizing a cube in spherical coordinates (ugh), which is to further exploit symmetry: $$ \mathcal{J}=\iiint_{(-1,1)^3}\frac{d\mu}{x^2+y^2+z^2}=8\iiint_{(0,1)^3}\frac{d\mu}{x^2+y^2+z^2}=48\iiint_{0\leq z\leq y\leq x\leq 1}\frac{d\mu}{x^2+y^2+z^2} $$ Now the substitution $z=yc, dz=y\,dc$ turns the RHS into $$ 48\int_{0}^{1}\int_{0}^{x}\int_{0}^{1}\frac{y}{x^2+y^2+c^2 y^2}\,dc \,dy\,dx $$ and the substitution $y=xb, dy=x\,db$ turns the last integral into $$ 48\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}\frac{x^2 b}{x^2+b^2 x^2+c^2 b^2 x^2}\,dc \,db\,dx = 48\int_{0}^{1}\int_{0}^{1}\frac{ b}{1+b^2+c^2 b^2}\,dc \,db$$ or $$ 24\int_{0}^{1}\frac{\log(2+c^2)}{1+c^2}\,dc=24\int_{0}^{\pi/4}\log(2+\tan^2\theta)d\theta $$ which is an integral in a single-variable, requiring the dilogarithms machinery to be computed in closed form. Numerically, $\mathcal{J}\approx\frac{353}{23}$. In explicit terms

$$ \mathcal{J}=24\left[\text{Im}\,\text{Li}_2((3-2\sqrt{2})i)-G+\frac{\pi}{4}\log(3+2\sqrt{2})\right]$$ where $G$ is Catalan's constant.

Jack D'Aurizio
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    Sometimes the good old way works best! :) (+1) – Sangchul Lee Apr 10 '18 at 21:48
  • I hope this is not a stupid question, but when one make the substitution say: x=ys can we choose freely which way to differentiate(dx=yds but also dx=sdy)? – Zacky Apr 10 '18 at 22:14
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    @Zacky: of course not, the change of variables are chosen in such a way that $0\leq z\leq y\leq x \leq 1$ is mapped back into a cube of the $(0,1)^3$ kind. For instance since $y$ ranges from $0$ to $x$ I chose to set $y=xb, dy=x,db$, in such a way that $b$ ranges over $(0,1)$. – Jack D'Aurizio Apr 10 '18 at 22:21
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    What a great answer! Thanks a lot! – induction601 Apr 11 '18 at 02:29
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Here is another way: Let $\mathbf{F}$ be defined by

$$ \mathbf{F} = \left( \frac{x}{x^2+y^2+z^2}, \frac{y}{x^2+y^2+z^2}, \frac{z}{x^2+y^2+z^2} \right). $$

Then it is easy to check that $\operatorname{div}\mathbf{F} = \frac{1}{x^2+y^2+z^2}$. So if $I$ denotes the integral in question, then by the divergence theorem1) we get

$$ 8I = \int_{[-1, 1]^3} \operatorname{div}\mathbf{F} \, dV = \int_{\partial [-1, 1]^3} \mathbf{F} \cdot \mathbf{n} \, dA. $$

Again by symmetry,

$$ 8I = 6 \int_{\{1\}\times[-1,1]^2} \mathbf{F} \cdot \mathbf{e}_1 \, dydz = 24 \int_{[0,1]^2} \frac{dydz}{1+y^2+z^2}.$$

Applying the polar coordinate change, we have

$$ \int_{[0,1]^2} \frac{dydz}{1+y^2+z^2} = \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sec\theta} \frac{2r}{1+r^2} \, drd\theta = \int_{0}^{\frac{\pi}{4}} \log(1 + \sec^2\theta) \, d\theta,$$

which leads to the same integral as in Jack D'Aurizio's answer

$$ I = 3\int_{0}^{\frac{\pi}{4}} \log(2 + \tan^2\theta) \, d\theta. $$

1) It can be easily shown, by excising a small ball around $0$ and letting the radius zero, that the singularity of $\mathbf{F}$ at $0$ poses no issue.

Sangchul Lee
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    Another good approach. Thanks for your answer. One quick follow up question. Do you think this approach has a potential generalization? That is, $$I_n = \int_{[0,1]^n} \frac{1}{\sum_{i=1}^n x_i^2}d\mu_n.$$ I tried to mimic this approach but I stuck on computing $$\int_{[0,1]^{n-1}} \frac{d\mu_{n-1}}{1+\sum_{i=1}^{n-1}x_i^2}.$$ Using Jack D'Aurizio approaches, this ends up with computing $$ \frac{1}{n-2}\int_{[0,1]^{n-1}} \frac{\prod_{i=1}^{n-2}w_i^i}{1+\sum_{i=1}^{n-1}\prod_{j=1}^i w_{n-j}^2}d\omega_{n-1}.$$ – induction601 Apr 11 '18 at 04:47