I have to calculate the integral

$$I=\iiint_{[-1,1]^3} \frac{dx dy dz}{r^2} $$

where $r^2 = x^2 + y^2 + z^2 $.

My strategy is to split the integral domain into two parts: the unit sphere and the supplement domain. From the unit sphere, I get analytically $4\pi $; for the supplement, I used Monte Carlo. The problem is that the Monte Carlo part is not very efficient, while I want to get the integral to 8 digits.

Could anyone suggest a better scheme?

  • 1
    $\begingroup$ Have you tried spherical coordinates? $\endgroup$
    – zkutch
    Commented Aug 24, 2022 at 2:13
  • 1
    $\begingroup$ Python has a function nquad which does such stuff pretty accurately. $\endgroup$
    – Kurt G.
    Commented Aug 24, 2022 at 5:42
  • $\begingroup$ If you use Matlab, you could use integral3. $\endgroup$
    – Dan Doe
    Commented Aug 24, 2022 at 13:02
  • $\begingroup$ Good strategy. This removes the singularity at the origin from the region of integration, so numerical methods become more efficient. I agree with the other comments that any built-in integrator should give you a very accurate result. I am actually surprised that Monte Carlo is inefficient. (In the end you are just integrating a smooth, non-singular function over a bounded domain. ) $\endgroup$ Commented Aug 24, 2022 at 15:01
  • $\begingroup$ See the set-up here for integrating over $[0,1]^3$ in spherical coordinates. $\endgroup$
    – user170231
    Commented Aug 25, 2022 at 16:10

2 Answers 2


Per symmetry, the integral over $[-1,1]^3$ is $8$ times that over $[0,1]^3$; the latter is $6$ times the integral over the tetrahedron $0<z<y<x<1$. Thus $$\iiint_{[-1,1]^3} \frac{dx dy dz}{r^2}= 48J$$ where \begin{align} J&=\int_0^1 \int_0^x \int_0^y \frac{1}{x^2+ y^2+z^2}\ \overset{z=yt}{dz}\ dy \ dx\\ &= \int_0^1 \int_0^1\int_0^x \frac{y}{x^2+ y^2(1+t^2)}\ \overset{y=x \sinh v}{dy\ \ dx } \ dt\\ &= \int_0^{\sinh^{-1}1}\int_0^1 \frac{\tanh v }{1+ \tanh^2 v \ t^2} dt\ dv = \int_0^{\sinh^{-1}1} \tan^{-1}(\tanh v) \overset{u=e^{-2v}}{dv}\\ &=\ \frac12 \int_{(\sqrt2-1)^2}^1 \frac1u{\tan^{-1}\frac{1-u}{1+u}}\ {du} =\frac12 \int_{(\sqrt2-1)^2}^1 \frac{\frac\pi4-\tan^{-1}u}{u}{du}\\ &=-\frac\pi8\ln(\sqrt2-1)^2-\frac12\int_0^1 \frac{\tan^{-1}u}udu + \frac12\int_0^{(\sqrt2-1)^2} \frac{\tan^{-1}u}udu \\ &=- \frac\pi4\ln(\sqrt2-1)-\frac12G+\frac12 \Im\text{Li}_2(i(\sqrt2-1)^2) \end{align}


$$\iiint_{[-1,1]^3}\frac{d\mu}{x^2+y^2+z^2} = \iint_{[-1,1]^2}\frac{2}{\sqrt{x^2+y^2}}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\,dx\,dy $$ equals $$ 8\iint_{[0,1]^2}\frac{1}{\sqrt{x^2+y^2}}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\,dx\,dy $$ or $$ 16\iint_{0\leq y\leq x\leq 1}\frac{1}{\sqrt{x^2+y^2}}\arctan\left(\frac{1}{\sqrt{x^2+y^2}}\right)\,dx\,dy \stackrel{y\mapsto tx}{=} 16\iint_{[0,1]^2}\frac{1}{\sqrt{1+ t^2}}\arctan\left(\frac{1}{x\sqrt{1+ t^2}}\right)\,dx\,dt $$ or $$ 16\int_{0}^{1}\left(\frac{\pi}{2\sqrt{1+t^2}}-\frac{\arctan\sqrt{1+t^2}}{\sqrt{1+t^2}}+\frac{\log(2+t^2)}{2(1+t^2)}\right)\,dt $$ or $$ 8\pi\log(1+\sqrt{2})+16\int_{0}^{1}\left(-\frac{\arctan\sqrt{1+t^2}}{\sqrt{1+t^2}}+\frac{\log(2+t^2)}{2(1+t^2)}\right)\,dt. $$ The last integral can be written in terms of polylogarithms, or just numerically evaluated through Newton-Cotes or Gaussian quadrature. The outcome, of course, is larger than $4\pi$ and it equals $\color{blue}{15.348248444887\ldots}$


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