Ytukyg

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

$$intlimits_{phi=0}^{phi=2pi}intlimits_{theta=0}^{theta=pi}sin^3theta cos^2{r(sin theta sin alphacos phi+cos thetacos alpha)+c} , dtheta , d phi$$ for $0 leq alphaleq pi$, $cin mathbb{R}$, 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

Let's break this down into something simpler. Using the addition formula for cosine, we have that
$$cos(rsinthetasinalphacosphi+rcosthetacosalpha+c)\=cos(rsinthetasinalphacosphi)cos(rcosthetacosalpha+c)\-sin(rsinthetasinalphacosphi)sin(rcosthetacosalpha+c)$$
When we integrate this from $phi=0$ to $2pi$, 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
$$sin^3(theta)cos(rsinthetasinalphacosphi)cos(rcosthetacosalpha+c)$$
from $theta=0$ to $pi$ and $phi=0$ to $2pi$. Now we may make use of some symmetry to say that this integral is equal to half the value of the integral of
$$sin^3(theta)cos(rsinthetasinalphacosphi)cos(rcosthetacosalpha+c)\+sin^3(pi-theta)cos(rsin(pi-theta)sinalphacosphi)cos(rcos(pi-theta)cosalpha+c)$$
Using the fact that $sin(pi-theta)=sin(theta)$ and that $cos(pi-theta)=-cos(theta)$, and the fact that $cos(a+b)+cos(a-b)=2cos(a)cos(b)$, the integral you seek is equal to the integral of
$$sin^3(theta)cos(rsinthetasinalphacosphi)cos(rcosthetacosalpha)cos(c)$$
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 $2pi$ causes the terms with odd $m$ to vanish, leaving the series
$$2pi cos(c)sum_{m,n=0}^infty frac{(-1)^{n}r^{4m+2n}sin^{2m}(alpha)cos^n(alpha)sin^{2m+3}(theta)cos^n(theta)(2m-1)!!}{2^m m!(4m)!(2n)!}$$
Integrating again from $theta=0$ to $pi$ causes terms with odd $n$ to vanish, leaving
$$4pi cos(c)sum_{m,n=0}^infty frac{r^{4m+2n}sin^{2m}(alpha)cos^{2n}(alpha)(m+1)!(2m-1)!! (2n-1)!!}{m!(4m)!(2n)!(2m+2n+1)!!}$$
...which I cannot further simplify.