Question the convergence of $$\iiint_{x^2+y^2+z^2\geq1}\frac{e^{\sin(x+y+z)}}{(x^2+y^2+z^2)^p}$$ in dependence of $p$.
In class we did $$\iiint_{x^2+y^2+z^2\geq1}\frac{e^{-x^2-y^2-z^2}}{\sqrt{x^2+y^2+z^2}}$$ Just used spherical coordinates(also defining the sets $D_n=\{(x,y,z)|1<x^2+y^2+z^2<n^2\}$) and it was straight forward to solve the triple integral being $$\lim_{n\to \infty}\int_{0}^{2\pi}d\varphi\int_{0}^{\pi}\sin \theta d\theta\int_{1}^{n^2}e^{-r^2}rdr.$$ But in my example the $\sin(x+y+z)$ makes it confusing to solve using spherical coordinates.Am I not seeing something trivial in solving this integral?
The explicit value of the integral may be difficult to find, but its convergence depends only on the convergence of: $$\iiint_{x^2+y^2+z^2\geq 1}\frac{d\mu}{(x^2+y^2+z^2)^p}=\frac{4\pi}{2p-3}\qquad \left(p>\frac{3}{2}\right)$$ since $\exp\sin(x+y+z)$ is always bounded between the positive constants $\frac{1}{e}$ and $e$.
By assuming $p>\frac{3}{2}$ and exploiting an isometry, we may check that the original integral equals:
$$ I_p=\iiint_{x^2+y^2+z^2\geq 1}\frac{\exp\sin(z\sqrt{3})\,d\mu}{(x^2+y^2+z^2)^p}=2\pi\int_{1}^{+\infty}\int_{0}^{\pi}\rho^{2-2p}\sin\phi\exp\sin(\rho\sqrt{3}\cos\phi)\,d\phi\,d\rho.$$
That can be computed as a fast-convergent series by expanding $\sin\phi\exp\sin(\rho\sqrt{3}\cos\phi)$ as a Fourier series (with coefficients depending on values of Bessel functions), but I think that wasn't required. Anyway, the main term should be: $$ \frac{4\pi}{2p-3}\sum_{n\geq 1}\frac{1}{4^n n!^2} = \frac{4\pi\cdot I_0(1)}{2p-3}.$$