The integration of a rapidly decreasing function is bounded? We say function $f$ is rapidly decreasing iff $\forall\,\,l,k>0\,\,|x|^{l}|f^{k}(x)| < \infty$ on $\mathbb{R}$. Now I need to prove the integration $$\int_{\mathbb{R}}|x|^{l}|f(x)|\,dx<\infty$$
It makes intuitive sense for me, since $f$ vanishes rather quickly, so I tried to build up a connection between rapidly decreasing function and compactly supported function. I hope I can prove that the above integration is only integrating over a finite interval, but after search, it turned out that not all rapidly decreasing function is compactly supported (am I right?). I just got stuck here, could you please show me other approaches? 

Hint:  fix $l>0$.
Suppose $$\sup_{x\in \mathbb{R}}(|x|^l|f(x)|)=M_1 $$ and
$$\sup_{x\in \mathbb{R}}(|x|^{l+2}|f(x)|)=M_2. $$
$$\int_{\mathbb{R}} |x^l||f(x)|\,dx = \int_{-1}^1 |x^l||f(x)|\,dx+\int_{\mathbb{R}\setminus [-1,1]} |x^l||f(x)|\,dx$$ $$\leq 2M_1 + \int_{\mathbb{R}\setminus [-1,1]} |x^l||f(x)|\,dx.$$
So you only need to bound the last term, the non-compact part.  For that, write
$$\int_{\mathbb{R}\setminus [-1,1]} |x^l||f(x)|\,dx = \int_{\mathbb{R}\setminus [-1,1]} |x^{l+2}||f(x)|\frac{1}{|x|^2}\,dx. $$
Go on from here.
