I would like to show that $$I:=\int_{\mathbb{R}}\int_{\mathbb{R}}\int_{\mathbb{R}}\frac{sin^{6}(x_{1})}{x_{1}^{6}}(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{2}dx_{1}dx_{2}dx_{2}<\infty$$ However when I try to evaluate it numerically, with python, it seems to be false...
How would you show that I is finite or infinite ?
Thank you very much
Perhaps one thing you might consider is the following. Observe that the function being integrated is always non-negative. Thus, if we choose a smaller domain to integrate over and show that the integral over that smaller domain is infinite we have that the integral above must diverge as well. Now let us restrict ourselves to the domain $\frac{\pi}{4}\leq x_1\leq \frac{\pi}{2}$, and $0\leq x_2\leq 1$, and $1<x_3<\infty$. Now if we consider the function $f(x_1,x_2,x_3)=\frac{\sin^6(x_1)}{x_1^6}(x_1^2+x_2^2+x_3^2)^2$, we have that we can lower bound this function as $$ 0<\frac{1}{8}\cdot\frac{64}{\pi^6}\left(\frac{\pi^2}{16}+1\right)^2\leq f(x_1,x_2,x_3) $$ the lower bound is achieved by seeing the smallest value of $\sin(x_1)=\frac{\sqrt{2}}{2}$, so we have that a lower bound on $\sin^6(x_1)=\frac{1}{8}$, then we plug in $x_1=\frac{\pi}{2}$ for $\frac{1}{x_1^6}$ to lower bound this gives the $\frac{64}{\pi^6}$, then for the $x_1^2+x_2^2+x_3^2$ term we just get the lower bound by plugging in the smallest values of $x_1=\frac{\pi}{4}$, $x_2=0$, and $x_3=1$. Then we can conclude that the integral $I$ diverges as we can bound it below by something going to infinity. Since we have that $$ I\geq \int_{\pi/4}^{\pi/2}\int_0^1\int_1^\infty f(x_1,x_2,x_3)dx_1dx_2dx_3$$
$$ \geq \int_{\pi/4}^{\pi/2}\int_0^1\int_1^\infty\frac{1}{8}\cdot\frac{64}{\pi^6}\left(\frac{\pi^2}{16}+1\right)^2dx_1dx_2dx_3=\infty $$ where we see that the integral goes to infinity since we are integrating a constant function over a domain of infinite measure.