# $\int\limits_{0}^{2\pi}[\int\limits_{0}^{\pi}\sin^3\theta \cos^2\{r(\sin \theta \sin \alpha\cos \phi+\cos \theta\cos \alpha)+c\}d\theta] d \phi$

Could you tell me how to compute the following definite integration?

$$\int\limits_{\phi=0}^{\phi=2\pi}\int\limits_{\theta=0}^{\theta=\pi}\sin^3\theta \cos^2\{r(\sin \theta \sin \alpha\cos \phi+\cos \theta\cos \alpha)+c\} \, d\theta \, d \phi$$ for $$0 \leq \alpha\leq \pi$$, $$c\in \mathbb{R}$$, and $$r>0$$.

I tried using Mathematica, but it didn't show me the answer.

This is the two non-relativistic radiating charges' energy loss power calculated from the surface integration of Poynting vector's radial component on the infinite sized sphere. From the energy conservation, as I know the electrons' lost energy from direct dissipation calculation, I know the answer when $$c=0$$. But, I can't compute the above integration

• You can edit your title using MathJax also. Please edit that. – jayant98 Dec 19 '18 at 21:17
• @user16308 What makes you think that this has a nice answer? Where did you come up with this problem? – Franklin Pezzuti Dyer Dec 19 '18 at 21:18

Let's break this down into something simpler. Using the addition formula for cosine, we have that $$\cos(r\sin\theta\sin\alpha\cos\phi+r\cos\theta\cos\alpha+c)\\=\cos(r\sin\theta\sin\alpha\cos\phi)\cos(r\cos\theta\cos\alpha+c)\\-\sin(r\sin\theta\sin\alpha\cos\phi)\sin(r\cos\theta\cos\alpha+c)$$ When we integrate this from $$\phi=0$$ to $$2\pi$$, we can see that the second term vanishes because $$\sin$$ is an odd function and $$\cos(\phi)=-\cos(\pi-\phi)$$. Thus, you may consider the simpler integral of $$\sin^3(\theta)\cos(r\sin\theta\sin\alpha\cos\phi)\cos(r\cos\theta\cos\alpha+c)$$ from $$\theta=0$$ to $$\pi$$ and $$\phi=0$$ to $$2\pi$$. Now we may make use of some symmetry to say that this integral is equal to half the value of the integral of $$\sin^3(\theta)\cos(r\sin\theta\sin\alpha\cos\phi)\cos(r\cos\theta\cos\alpha+c)\\+\sin^3(\pi-\theta)\cos(r\sin(\pi-\theta)\sin\alpha\cos\phi)\cos(r\cos(\pi-\theta)\cos\alpha+c)$$ Using the fact that $$\sin(\pi-\theta)=\sin(\theta)$$ and that $$\cos(\pi-\theta)=-\cos(\theta)$$, and the fact that $$\cos(a+b)+\cos(a-b)=2\cos(a)\cos(b)$$, the integral you seek is equal to the integral of $$\sin^3(\theta)\cos(r\sin\theta\sin\alpha\cos\phi)\cos(r\cos\theta\cos\alpha)\cos(c)$$ Which is a little simpler than the integral that you presented; it at least allows you to remove the constant $$c$$ by factoring out $$\cos(c)$$.
A nice closed form of this integral seems unlikely. From the simpler form that I obtained, you might try using this series representation of the integrand: $$\cos(c)\sum_{m,n=0}^\infty \frac{(-1)^{m+n}r^{2m+2n}\sin^m(\alpha)\cos^n(\alpha)\sin^{m+3}(\theta)\cos^n(\theta)\cos^m(\phi)}{(2m)!(2n)!}$$ Integrating this from $$\phi=0$$ to $$2\pi$$ causes the terms with odd $$m$$ to vanish, leaving the series $$2\pi \cos(c)\sum_{m,n=0}^\infty \frac{(-1)^{n}r^{4m+2n}\sin^{2m}(\alpha)\cos^n(\alpha)\sin^{2m+3}(\theta)\cos^n(\theta)(2m-1)!!}{2^m m!(4m)!(2n)!}$$ Integrating again from $$\theta=0$$ to $$\pi$$ causes terms with odd $$n$$ to vanish, leaving $$4\pi \cos(c)\sum_{m,n=0}^\infty \frac{r^{4m+2n}\sin^{2m}(\alpha)\cos^{2n}(\alpha)(m+1)!(2m-1)!! (2n-1)!!}{m!(4m)!(2n)!(2m+2n+1)!!}$$ ...which I cannot further simplify.