Is $\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{1}{1 - \cos x\cos y\cos z}\, dx\, dy\, dz$ finite?

Question

Is $$\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{1}{1 - \cos x\cos y\cos z}\, dx\, dy\, dz$$ finite? Mathematica is throwing me an error, but I suspect the integral converges, since the Taylor expansion of $\cos x\cos y\cos z$ should have a leading term of $1$ followed by a $x^{2} + y^{2} + z^{2}$ and $\frac{1}{x^{2} + y^{2} + z^{2}}$ is integrable in 3 dimensions if I switch to polar.

Answer 1

While it is a bit surprising, the integral should converge. Around each of the corners, $(0, 0, 0)$, $(\pi, \pi, 0)$, $\dots$, etc, we have (very loose bounds) $$ \begin{aligned} \cos x\cos y\cos z &\le |\cos x||\cos y||\cos z| \\ &\le \left( \frac{|\cos x| + |\cos y| + |\cos z|}{3} \right)^3 \\ &\le \left( 1 - \frac{ x^2 + y^2 + z^2 } {9} \right)^3 \quad \mathrm{for}\; |x|, |y|, |z| \in (0, \pi/2) \\ &\le 1 - \frac{ x^2 + y^2 + z^2 } { 12 }. \end{aligned} $$ So $$ \begin{aligned} \frac{1}{1-\cos x\cos y\cos z} &\le \frac{12}{ x^2 + y^2 + z^2 } = \frac{12}{r^2}. \end{aligned} $$

$$ \begin{aligned} \int_0^{\pi/2}\int_0^{\pi/2}\int_0^{\pi/2}\frac{1}{1-\cos x\cos y\cos z} \,dx \, dy \, dz &\le \frac{\pi}{2} \int_0^{\pi/2} \frac{12}{r^2} r^2 \, dr = 3\pi^2. \end{aligned} $$ Together there are 64 such integrals, and the total integral is finite (actually 32 of them are not singular, but this is just an estimate).

For Mathematica, try this

NIntegrate[1/(1 - Cos[x] Cos[y] Cos[z]), {x, -Pi, Pi}, {y, -Pi, Pi}, {z, -Pi, Pi},
    Method -> "AdaptiveMonteCarlo", PrecisionGoal -> 3]

Answer 2

Hopefully you can work out the places where I have left out explicit explanation. But to start, the Weierstrass substitution on the inner integral helps: $$ \begin{align} &\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\int_{-\pi}^{\pi}\frac{1}{1-\cos(x)\cos(y)\cos(z)}\,dx\,dy\,dz\\ &=8\int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{\pi}\frac{1}{1-\cos(x)\cos(y)\cos(z)}\,dx\,dy\,dz\qquad t=\tan(x/2)\\ &=8\int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{\infty}\frac{1}{1-\frac{1-t^2}{1+t^2}\cos(y)\cos(z)}\,\frac{2}{1+t^2}dt\,dy\,dz\\ &=16\int_{0}^{\pi}\int_{0}^{\pi}\int_{0}^{\infty}\frac{1}{t^2(1+\cos(y)\cos(z))+1-\cos(y)\cos(z)}\,dt\,dy\,dz\\ &=16\int_{0}^{\pi}\int_{0}^{\pi}\left[\frac{1}{\sqrt{1-\cos^2(y)\cos^2(z)}}\arctan\left(\sqrt{\frac{1+\cos(y)\cos(z)}{1-\cos(y)\cos(z)}}\,t\right)\right]_{0}^{\infty}\,dy\,dz\\ &=16\int_{0}^{\pi}\int_{0}^{\pi}\frac{1}{\sqrt{1-\cos^2(y)\cos^2(z)}}\frac{\pi}{2}\,dy\,dz\\ &=8\pi\int_{0}^{\pi}\int_{0}^{\pi}\frac{1}{\sqrt{1-\cos^2(y)\cos^2(z)}}\,dy\,dz\qquad u=\cos(y)\cos(z)\\ &=8\pi\int_{0}^{\pi}\int_{\cos(z)}^{-\cos(z)}\frac{1}{\sqrt{1-u^2}}\,\frac{du}{-\sin(y)\cos(z)}\,dz\\ &=8\pi\int_{0}^{\pi}\int_{\cos(z)}^{-\cos(z)}\frac{1}{\sqrt{1-u^2}}\,\frac{du}{-\sqrt{1-\left(\frac{u}{\cos(z)}\right)^2}\cos(z)}\,dz\\ &=8\pi\int_{0}^{\pi}\int_{-\cos(z)}^{\cos(z)}\frac{1}{\sqrt{1-u^2}\sqrt{\cos^2(z)-u^2}}\,du\,dz\\ &=32\pi\int_{0}^{\pi/2}\int_{0}^{\cos(z)}\frac{1}{\sqrt{1-u^2}\sqrt{\cos^2(z)-u^2}}\,du\,dz\\ \end{align}$$

Now the inner integral has behavior like $\frac{A}{\sqrt{\cos(z)-u}}$ near $u=\cos(z)$. So the inner integral will be finite. From here, getting an asymptotic with respect to $u$ on the integrand should leave you with one last integral in $z$ that similarly has behavior like $\frac{1}{\sqrt{\pi/2-z}}$ near $z=\pi/2$.