1
$\begingroup$

I need to integrate this expression:

$$\int^{\pi}_0\frac{\gamma^{10} \theta^2 \sin\theta}{(\gamma^2 \theta^2 + 1)^5} d\theta.$$

I can use the fact that gamma is very large, which I think means I should rearrange the bottom line to allow an expansion, i.e. $$\int^{\pi}_0\frac{\theta^2 \sin\theta}{(\theta^2 + \dfrac{1}{\gamma^2})^5}d\theta ,$$

but I'm still not sure how to continue from here. Any hints would be great, thanks.

  • 0
    Just to clarify; to what variable are you integrating? To $\theta$ or $\gamma$?2011-03-06
  • 0
    Oops, sorry. Integrating with respect to theta.2011-03-06
  • 2
    Well, As $\gamma \to \infty$, $\int_0^\pi \frac{\theta^2 \sin \theta}{(\theta^2 + \gamma^{-2})^5} \mathrm{d}\theta \to \infty$. So, how large is "very large"?2011-03-06
  • 0
    It's to calculate the radiation emitted by an "ultra-relativistic particle". The question says to keep the "leading power of gamma only".2011-03-06
  • 0
    @cardinal: I think as $\gamma \rightarrow \infty$, the expression inside the integral $\rightarrow 1$ and then the integral is $\pi$. But I don't think he's supposed to take the limit, he's probably right in that he needs to rearrange the expression inside the integral.2011-03-06
  • 0
    @Matt: Come again?2011-03-06
  • 0
    The evaluation for $\gamma\to\infty$ is quite involved. Are you sure that you are not supposed to obtain the leading order in $\gamma\to 0$ (which would be $\gamma^{10}$)?2011-03-06
  • 0
    For $\gamma \to \infty$, I get $\gamma^6/24$. The $\gamma^6$ basically follows from dimensional analysis. About the $1/24$ I'm not completely sure.2011-03-06
  • 0
    @cardinal: oops, don't know what I was thinking when I wrote that. Somehow I dropped the $\sin \Theta$ when I evaluated the expression in my head. Writing it down works always better for me but I'm so lazy : /2011-03-09

1 Answers 1