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Consider $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{0}^{\infty}\biggr(\frac{1}{\sqrt{(1-x)^2+y^2+z^2}}-\frac{1}{\sqrt{x^2+y^2+z^2}}\biggr)^2dxdydz$$

A numerical study suggest the integral converges to $2\pi$. See: Link to numerical studyHowever I am unable to show this. When y,z are zero, we have a singular point at x=1. There the integrand behaves like $$\biggr(\frac{1}{(1-x)}-1\biggr)^2$$ right? And even in the Principal sense, the integral of this doesn't converge.

I was wondering if someone could help me show the value of the integral analytically.

Update: I just noticed that we have two singular points: Have another at x=0 when y,z=0 so maybe these two are canceling in the principal sense leading to a finite value for the integral.

Update 2: Made a mistake on integral boundaries: x goes from 0 to infinity and have made the corrections above as per comment below.

Thanks.

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  • $\begingroup$ Just wondering about the integral boundaries: $[0,\infty]$ refers to $z$ or to $x$? $\endgroup$ – denklo May 31 at 11:52
  • $\begingroup$ Sorry made a mistake on order of integration: x goes from 0 to infinity as you point out. I have made the correction above and noted the mistake. $\endgroup$ – Dominic May 31 at 12:04
  • $\begingroup$ The behavior of singularities is more complicated in multiple dimensions. Sometimes singularities like the one you point out can be handled by the residue theorem, since they appear as poles in the complex extension of the integrand. The term $\sqrt{x^2 + y^2 + z^2}$ screams spherical coordinates to me. Have you tried that? $\endgroup$ – Mortified Through Math May 31 at 12:30
  • $\begingroup$ Have you attempted to switch to spherical coordinates ? $\endgroup$ – Jean Marie May 31 at 12:33
  • $\begingroup$ (ctd...) or to oblate or prolate spheroidals coordinates... en.wikipedia.org/wiki/Oblate_spheroidal_coordinates, en.wikipedia.org/wiki/Prolate_spheroidal_coordinates $\endgroup$ – Jean Marie May 31 at 12:35
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It's not a problem that the integral does not converge on one line $y=z=0$, because a contribution from a single line is negligible for an integral over a volume.

Let us make a change of coordinates: $$ x=r\cos\theta, \qquad y = r\sin\theta\cos\phi, \qquad z=r\sin\theta\sin\phi$$ $$ 0\le r\le\infty, \qquad 0\le\theta\le\pi, \qquad 0\le\phi\le \pi$$ we have $$ x^2+y^2+z^2 = r^2 $$ $$ (1-x)^2+y^2+z^2 = 1 - 2r\cos\theta + r^2$$ $$ dxdy dz = r^2\sin\theta\, dr d\theta d\phi$$ so \begin{align} I &= \int_0^\infty dr \int_0^{\pi}d\theta \int_0^{\pi}d\phi\, r^2\sin\theta\left(\frac{1}{\sqrt{1 - 2r\cos\theta + r^2}} - \frac{1}{r}\right)^2 =^{\eta=\cos\theta} \\ &= \pi \int_0^\infty dr \int_{-1}^{1}d\eta \, \left(\frac{r^2}{1 - 2r\eta + r^2} - \frac{2r}{\sqrt{1 - 2r\eta + r^2}} +1\right) = \\ &= \pi \int_0^\infty dr \left.\left(-\frac{r}{2} \ln(1-2r\eta + r^2) +2\sqrt{1 - 2r\eta + r^2} +\eta\right)\right|_{\eta=-1}^{\eta=1} = \\ &= \pi \int_0^\infty dr \left(-r \ln |1-r| + r \ln(1 + r) +2|1-r| - 2(1 + r)+12\right) = \\ &= \pi \int_0^1 dr \left(-r \ln (1-r) + r\ln(1 + r) - 4r + 2\right) + \\ &\quad + \pi \int_1^\infty dr \left(-r \ln (r-1) + r\ln(1 + r) - 2\right) = \\ &= \pi \Big(3r-2r^2 - \frac12(1-r^2)\ln\frac{1+r}{1-r}\Big)\Big|_{r=0}^{r=1} + \pi \Big(-2r - \frac12(1-r^2)\ln\frac{r+1}{r-1}\Big)\Big|_{r=1}^{r=\infty} = \\ &=\pi + \pi = 2\pi \end{align}

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  • $\begingroup$ Beautiful. Thank you. $\endgroup$ – Dominic Jun 1 at 7:58

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