I have the following spherical density distribution:

$\rho(x, z) = \frac{1}{\sqrt{x^2 + z^2}\left(1+\sqrt{x^2+z^2}\right)^2}$

which I have broken into a "line of sight" dimension $z$ and a "transverse" dimension $x$. I wish to integrate this profile along the line of sight to obtain the projected 2d-profile $\Sigma(x)$.

The published result is

enter image description here

(ignore the normalization prefactors)

I cannot figure out how this is obtained. For the indefinite integral, Wolfram gives

enter image description here

Trying to take the upper limit for the $x>1$ case via Mathematica gives

enter image description here

which does not simplify to the published result, as far as I can tell. Presumably there is a manual technique that Im unfamiliar with... anyone see a way to properly do this?

I should say, my end goal is to compute this integral for the general upper bound $Z$, but I wanted to first make sure I could at least reproduce the case for $Z = \infty$.


I observe in Mathematica (11.3)

Integrate[1/(Sqrt[x^2 + z^2] (1 + Sqrt[x^2 + z^2])^2), {z, 0, \[Infinity]}, Assumptions -> {x > 1}]

yields $$ \frac{1}{x^2-1}-\frac{\sec ^{-1}(x)}{\left(x^2-1\right)^{3/2}} \text{,} $$

Integrate[1/(Sqrt[x^2 + z^2] (1 + Sqrt[x^2 + z^2])^2) /. x -> 1, {z, 0, \[Infinity]}]

yields $$ \frac{1}{3} \text{,} $$ and

Integrate[1/(Sqrt[x^2 + z^2] (1 + Sqrt[x^2 + z^2])^2), {z, 0, \[Infinity]}, Assumptions -> {0 < x < 1}]

yields $$ -\frac{2 \sqrt{1-x^2}+\log \left(\frac{1-\sqrt{1-x^2}}{\sqrt{1-x^2}+1}\right)}{2 \left(1-x^2\right)^{3/2}} \text{.} $$

I don't necessarily recognize the identities used to complete the identification with the published result, but I do see most of the parts of the published result, so the identity of the remainders may be possible.

Further, to your intention to compute $\int_0^{\text{zz}} \dots \,\mathrm{d}z$,

Assuming[{x > 1, zz \[Element] Reals, zz > 0}, FullSimplify[ Integrate[1/(Sqrt[x^2 + z^2] (1 + Sqrt[x^2 + z^2])^2), {z, 0, zz}] ] ]

yields (for $x > 1$) $$ \frac{\text{zz} \left(\sqrt{x^2+\text{zz}^2}-1\right)}{\left(x^2-1\right) \left(x^2+\text{zz}^2-1\right)}+\frac{\tan ^{-1}\left(\frac{\text{zz}}{\sqrt{\left(x^2-1\right) \left(x^2+\text{zz}^2\right)}}\right)-\tan ^{-1}\left(\frac{\text{zz}}{\sqrt{x^2-1}}\right)}{\left(x^2-1\right)^{3/2}} $$

and I imagine the other two pieces are similar.

  • $\begingroup$ This is fantastic, thanks so much. Ill hunt around for those identities. Your use of assumptions was superior to mine I suppose, I couldn't get Mathematica to do the definite integral earlier. How long did they take to run on your machine? The $0<x<1$ case runs for some time before ultimately returning the Integrate call back to me. And the general case $0<x<zz$ returns a much more complicated expression involving a Hypergeometric2F1. I do agree with your first two, though. $\endgroup$ Jul 11 '20 at 0:35
  • $\begingroup$ On second thought, decided I didn't care about the identities, but I checked it numerically, and yep they're spot on the real answer, and the general form falls below the projection at $\infty$ in the way that I expect $\endgroup$ Jul 11 '20 at 0:52
  • $\begingroup$ Just curious, what are your machine specs? I'd like to be able to get the general expression myself, but as I mentioned my mathematica gives something ugly for the identical input. Wondering if it is timing out as is effectively skipping some simplification it did in your case, and/or it is because of more than just a version discrepancy $\endgroup$ Jul 13 '20 at 12:44
  • $\begingroup$ @Anonymous : This computation was performed on M'ma, running on Linux (x64) 4.15.0, running on an AMD Phenom II X6 1055T (2.8 GHz, 6 core, about 10 years old) and using not more than 218 MB RAM. $\endgroup$ Jul 13 '20 at 16:41

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