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This is my first question here, so I hope I'm not giving too little/too much information. I need some help calculating (or even approximating) an integral which I've been wrestling with for a while.

As part of my internship, I need to calculate or even just approximate a power spectrum which boils down to the following integral:

$$I(k)=\int \frac{d^3p}{(2\pi)^3} (1 - \kappa^2)(1-\lambda^2) \frac{\left\lvert p \right\rvert^2}{(1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6)} \frac{\left\lvert p-k \right\rvert^2}{(1 + a \left\lvert p-k \right\rvert^4 + b \left\lvert p-k \right\rvert^6)}$$

where $\kappa = \hat{k}.\hat{p}$ and $\lambda = \hat{k}.\widehat{k-p}$ (Obviously $\left\lvert p-k \right\rvert^2 = p^2 + k^2 - 2 p k \cos \theta$). I need the answer in terms of $a,b$, two positive constants that I'd like to vary to describe different physical situations.

First attempt at a solution:

I moved into spherical coordinates and (I'm reasonably sure that) the integral reduces to:

$$\int \frac{d^3p}{(2\pi)^3} \frac{\left\lvert p \right\rvert^2}{1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6 } \frac{\left\lvert p \right\rvert^2\sin^4 \theta}{1 + a \left\lvert p-k \right\rvert^4 + b \left\lvert p-k \right\rvert^6 }$$

So naturally I considered performing the integration

$$\frac{1}{2\pi^2}\int_{0}^{\infty} \frac{\left\lvert p \right\rvert^6 dp }{1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6 } \int_{0}^{\pi}\frac{\sin^5 \theta \,d\theta}{1 + a \left\lvert p-k \right\rvert^4 + b \left\lvert p-k \right\rvert^6 }$$

Sadly, this doesn't really help me too much since these integrals are still pretty hard. Mathematica does manage to calculate them, but the results are really far too cumbersome to handle.

Further attempts:

As an approximation, I considered sloppily breaking the integral into $p<k$ and $p>k$, which leads me to something of the form (modulo constants):

$$\frac{1}{1 + a \left\lvert k \right\rvert^4 + b \left\lvert k \right\rvert^6 }\int_{0}^{k} \frac{\left\lvert p \right\rvert^6 dp}{1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6} +\int_{k}^{\infty} \frac{\left\lvert p \right\rvert^6 dp}{(1 + a \left\lvert p \right\rvert^4 + b \left\lvert p \right\rvert^6)^2 } $$

But I found even these "simpler" integrals too hard to calculate, and I'm not particularly fond of the approximation either.

Questions:

1) Does anyone know of a way I could solve or even approximate any of the above integrals as a function of the parameters $a$ and $b$?

2) It reminds me slightly on a convolution product, though not exactly since it's the norm of the vector $\left\lvert p-k \right\rvert$, which means I can't use the convolution theorem (can I?).

3) I know how to use the Feynman Parametrization for terms that are quadratic in the denominator. But I don't think it would work with $\left\lvert p\right\rvert^6$ (would it?).

3) The parameters $a$ and $b$ are quite small ($a$ is of the order of $10^{-2}$ and $b$ around the same), I was wondering if I could use that to my advantage, maybe perform some sort of Taylor expansion in either of them, reducing the integral to something more manageable, but I'm not sure that's allowed, and at any rate such integrals seem to diverge.

Thanks!

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    $\begingroup$ Contour integration should work fine. The tedious part is finding the poles and corresponding residues. $\endgroup$ – Mark Viola May 23 '17 at 0:46
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    $\begingroup$ This is a very good first question. I have some suggestions, although not an actual answer. The angular integral is theoretically calculable using the tangent half-angle formulae, although being a degree-5 rational function's not enormously promising. Another possibility is to use some kind of Feynman thing to write the original thing as a single squared denominator at the expense of introducing yet another integral. Is there a regime of $k$ you are particularly interested in? $\endgroup$ – Chappers May 23 '17 at 0:59
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    $\begingroup$ These integrals can be evaluated numerically in about 500 ns. That means you can create a graph of these as a function of $a$ and $b$ in a few seconds. If this is for work, I suggest getting the problem done that way and moving on to your next task. $\endgroup$ – user14717 May 23 '17 at 1:12
  • $\begingroup$ Thanks for your replies! @Chappers I technically managed to calculate the angular part, but Mathematica's solution was really far too complicated to be useful to me. I thought of the Feyman Trick, but you'd need to complete some sort of fourth or sixth power and I'm not sure that can even be done! (can it?) As for the regimes of $k$, I think the most important would be $0.01<k<1000$. Could I use that to my advantage? $\endgroup$ – Philip Cherian May 23 '17 at 8:16
  • $\begingroup$ @user14717 Do you mean calculating it for a bunch of $a,b$ and then interpolating and using the results? I'll try that for now, it's true that it'll help me advance, but I'd still like an approximate "closed form" solution to the integral, since I'd need to use it later on in other integrals... $\endgroup$ – Philip Cherian May 23 '17 at 8:18

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