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Can someone explain how is the result of:

$A(x_0,y_0,z_0)=\int\limits_{-a}^{a}\int\limits_{-b}^{b}\int\limits_{-\infty}^{+\infty}\frac{1}{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}}{\rm d}z{\rm d}y{\rm d}x$

equal to (assuming $z_0=0$):

$A(x_0,y_0,0)=\int\limits_{-a}^{a}\int\limits_{-b}^{b}\ln\left[(x-x_0)^2+(y-y_0)^2\right]{\rm d}y{\rm d}x$

if we assume that all variables are reals? This is the result obtained in an article and I would like to understand how the integration with respect to $z$ from $-\infty$ to $+\infty$ can converge? Perhaps their bounds are wrong, but what would they be in order to get this answer?

user89723
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  • The inner integral is infinite. For large $\lvert z\rvert$, the integrand is larger than $1/(2\lvert z\rvert)$. I can't guess how the result is obtained. – Daniel Fischer Sep 03 '13 at 20:12
  • It is indefinite, but to get to the result presented, I assumed that perhaps the authors used different bounds (definite). Or could it be that a Dirac's delta is missing (or something similar) in the indefinite integral? – user89723 Sep 03 '13 at 21:07
  • Not indefinite, infinite, as in $+\infty$. If one uses finite bounds, I don't see how one could get the result either. It might be that if one interprets it as the Fourier transform of a tempered distribution one would get the result. – Daniel Fischer Sep 03 '13 at 21:11

1 Answers1

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The second integral should have opposite sign, or reciprocal inside the logarithm. Otherwise, for small $a,b$ the second integral has negative integrand, contrary to the first integral being positive.

Recalling that $$\int_0^M \frac{1}{\sqrt{a^2+z^2}}\,dz = \ln (\sqrt{a^2+M^2}+M) - \ln (a^2)$$ we find that as $M\to \infty$, the expression $$ \begin{split} \int\limits_{-a}^{a}\int\limits_{-b}^{b}\int\limits_{-M }^{M}\frac{1}{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}}{\rm d}z\,{\rm d}y\,{\rm d}x \\ - (2a)(2b)\ln(2M) \end{split}\tag1$$ converges to $$ \int\limits_{-a}^{a}\int\limits_{-b}^{b} \ln \frac{1}{(x-x_0)^2+(y-y_0)^2 } \,{\rm d}y\,{\rm d}x \tag2$$

In other words, (2) is what one gets from the divergent integral $$ \int\limits_{-a}^{a}\int\limits_{-b}^{b}\int\limits_{-\infty }^{\infty}\frac{1}{\sqrt{(x-x_0)^2+(y-y_0)^2+(z-z_0)^2}}{\rm d}z\,{\rm d}y\,{\rm d}x $$ by a kind of regularization, removing the term with undesired behavior. What this means physically (are we talking about 3d and 2d gravitation or electrostatic force?) is another matter.

user98130
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