# Integration over solid angle

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.

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Just to clarify; to what variable are you integrating? To $\theta$ or $\gamma$? – Stijn Mar 6 '11 at 20:44
Oops, sorry. Integrating with respect to theta. – user7877 Mar 6 '11 at 20:57
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"? – cardinal Mar 6 '11 at 21:31
It's to calculate the radiation emitted by an "ultra-relativistic particle". The question says to keep the "leading power of gamma only". – user7877 Mar 6 '11 at 22:18
@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. – Rudy the Reindeer Mar 6 '11 at 22:23

As it is not clear whether to have the larger or small $\gamma$ expansion of the integral $$I(\gamma)= \int^{\pi}_0\frac{\gamma^{10} \theta^2 \sin\theta}{(\gamma^2 \theta^2 + 1)^5} d\theta,$$ I will provide both.
For large $\gamma$, we observe that the integral is dominated for $\theta \approx \gamma^{-1}$. That is why can hope that we can expand the $\sin$ function. We will show later that this is really admissible when we see that successive terms are smaller when $\gamma\to\infty$ (the expansion will be an asymptotic expansion). The substitution $x= \gamma\theta$ yields $$I(\gamma) = \gamma^{7} \int_0^{\pi \gamma} \frac{x^2 \sin(x/\gamma)}{(1+x^2)^5} dx \sim \gamma^{6} \int_0^\infty \frac{x^3}{(1+x^2)^5} dx + O(\gamma^5).$$ Here, we have used that every term in the expansion of $\sin$ gives an additional $\gamma^{-1}$ and that the integral is dominated for $x\approx 1$ such that we could put the integration boundary to $\infty$. The remaining integral is a number which can be evaluated to $1/24$. In total we have $$I(\gamma) \sim \frac{\gamma^6}{24}.$$
For small $\gamma$, we can obtain a converging series by expanding the integrand in a Taylor series. The first term in the expansion reads $$I(\gamma) = \gamma^{10} \int_0^\pi \theta^2 \sin \theta d\theta + O(\gamma^{11}) = (\pi^2 -4) \gamma^{10} + O(\gamma^{11}).$$
Knowing the two asymptotics, we (more or less) can draw a plot of the function. It starts off as $(\pi^2 -4) \gamma^{10}$ and then turns over into $\gamma^6/24$. The crossover region is for $\gamma \approx 1$. In this spirit, I expect $$I(\gamma) \approx \frac{\gamma^{10}}{(\pi^2-4)^{-1} + 24 \gamma^4}$$ to be a reasonable good approximation to the initial integral.