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I'm having trouble with an integral and I would like some pointers on how to "take" it:

$$ \int \limits_{-\infty}^{\infty}\frac{3\gamma a^{2}d^{3}\mathbf r}{4 \pi \left( r^{2} + \frac{\gamma^{2}}{c^{2}}(\mathbf r \cdot \mathbf u)^{2} + a^{2}\right)^{\frac{5}{2}}} $$ Here $\mathbf u$, $a$ and $\gamma$ are constants, and the integrand converges to a Dirac delta $\delta(\mathbf r)$ as $a\rightarrow 0$. The integral must be equal to 1.

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up vote 4 down vote accepted

First choose a direction for u, along the z-axis. Then the integral is

$$ I = \int {1\over (x^2 + y^2 + A z^2 + B)^{5/2} } dx dy dz $$

Rescale z by $\sqrt{A}$ to get rid of A and restore rotational invariance.

$$ I = {1\over \sqrt{A}} \int {1\over (x^2 + y^2 + z^2 + B)^{2.5}} dx dy dz $$

Now you do find the B dependence immediately from rescaling x y and z by $\sqrt{B}$ (or from dimensional analysis-- B has units length squared):

$$ I = {1\over \sqrt{A} B} \int {1\over (r^2 + 1)^{2.5}} d^3r $$

The only thing undetermined is the transcendental factor, which is just a number. You evaluate it by doing it radially and doing a string of substitutions:

$$ \sqrt{A}B I = 4\pi \int_0^{\infty} {r^2\over (r^2 +1)^{2.5} } dr $$

first $u=r^2 + 1$ gives

$$ \sqrt{A}B I = 2\pi \int_1^\infty {\sqrt{u-1}\over (u)^{2.5}} du $$

Then $v = {1\over u}$ makes it,

$$ \sqrt{A}B I = 2\pi \int_0^1 \sqrt{1-v} dv = {4\pi\over 3} $$


$$ I = {4\pi \over 3 \sqrt{A}B} $$

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Thank you! The integral is equal to one: $$ u = r^{2} + 1 \Rightarrow r = \sqrt{u - 1}. $$ – user8817 Jul 20 '12 at 12:36
Yes, that is probably cleaner. – Emilio Pisanty Jul 20 '12 at 15:34
@EmilioPisanty: It is not just cleaner, your way doesn't work--- the spherical integration will probably not produce a function with elementary antiderivative without the rescaling of z step. – Ron Maimon Jul 20 '12 at 17:28
Mathematica has no problems doing the integration. I can follow the steps for the $\theta$ integration and though I haven't tried doing the $r$ integral the antiderivative is complicated but elementary. – Emilio Pisanty Jul 22 '12 at 10:41
@EmilioPisanty: I am surprised--- the angular integration without rescaling of Z leads to complicated square roots in the denominator, and I don't see why the antiderivative should exist. Did you try getting an antiderivative for the r-integral using mathematica, or just doing the integral over both theta and r? Mathematica will do a definite integral using hypergeometric functions in intermediate steps, and then reduce with hypergeometric special values and identities, so it isn't clear that the antiderivative is elementary from the fact that Mathematica gets the right answer. – Ron Maimon Jul 22 '12 at 19:36

Set the $z$ axis along the direction of $\mathbf{u}$ and use spherical coordinates, which reduces your integral to something like $$\int_0^\infty dr\int_0^\pi d\theta\int_0^{2\pi}d\phi\frac{r^2 \sin(\theta)}{\left(a^2 +r^2(1+\frac{\gamma^2}{c^2}\cos^2(\theta))\right)^{5/2}}.$$ Do the $\phi$ integral first and then the $\theta$ integral, transforming to $u=\cos(\theta)$. Be careful to use absolute values for the roots when necessary. After that the $r$ integral should be tough but doable.

EDIT to take some discussion off comments. I gave

Integrate[Sin[[Theta]]/(a^2+r^2 (1+[Gamma]^2 Cos[[Theta]]^2))^(5/2),[Theta]]/.[Theta]->[Pi]

to Mathematica to get $$\frac{3 \left(a^2+r^2\right)+2 r^2 \gamma ^2}{3 \left(a^2+r^2\right)^2 \left(a^2+r^2+r^2 \gamma ^2\right)^{3/2}},$$ and then doing the radial integration by

Integrate[r^2 (3 (a^2+r^2)+2 r^2 [Gamma]^2)/(3 (a^2+r^2)^2 (a^2+r^2+r^2 [Gamma]^2)^(3/2)),r]

gives $$\frac{r^3}{3 a^2 \left(a^2+r^2\right) \sqrt{a^2+r^2 \left(1+\gamma ^2\right)}}$$ for the antiderivative. I agree that the roots make one suspect nonelementary antiderivatives but it is only one root so that elliptic integrals are out. Once one has the antiderivative, of course, it is routine to check that it does differentiate to what it should.

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Thank you. But after some replacements I obtained $$ \frac{c}{\gamma u r (a^{2} + r^{2})^{2}}\int \frac{dt}{(1 + t^{2})^{\frac{5}{2}}}, $$ where $$ t = \frac{tru\gamma}{c\sqrt{a^{2} + r^{2}}}. $$ But after integration I got zero. Can you tell me where I made the mistake? – user8817 Jul 19 '12 at 13:52
For one, you're probably integrating $t$ from $-\infty$ where you should be integrating from $r=0$. Secondly, you can't factor out the $r$-dependent term from the integral! – Emilio Pisanty Jul 19 '12 at 16:10
Excuse me. I meaned $$ t = \frac{rv \gamma u}{c\sqrt{a^{2} + r^{2}}}, $$ where $$ v = cos (\theta ). $$ – user8817 Jul 19 '12 at 16:49
is this the $r$ integral or the $\theta$ integral you're trying to do? – Emilio Pisanty Jul 19 '12 at 17:00
If you're doing the $\theta$ integral then you're on the right track, but you need to be careful with the limits of integration. Setting $t=\tan(\eta)$ should reduce the integral to doable trigonometric integrals. But watch the limits of integration and watch the integrand's parity! – Emilio Pisanty Jul 19 '12 at 17:06

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