Let $\displaystyle f(x)= \dfrac {1}{\sqrt{|x|}}$ when $\displaystyle |x|>1 $ and $\displaystyle f(x)=1$ when $\displaystyle |x| \le 1$.Show that f is Lebesgue integrable on $\displaystyle \mathbb{R}$.

My work:
On $\displaystyle |x| \le 1$,since $\displaystyle f(x)=1$,so $\displaystyle \int_{[-1,1]} f dm < \infty$
but when $\displaystyle |x|>1$,I don't think it is integrable as when $\displaystyle x>1$ , $\displaystyle \int_{(1,\infty)} f dm > \int_{[2,n]} f dm$
But $\displaystyle \int_{[2,n]} f dm = 2\sqrt{n} -2\sqrt{2}$,then as $\displaystyle n\rightarrow \infty , \int_{[2,n]} f dm \rightarrow \infty$ and this implies that $\displaystyle f$ in not integrable on $\displaystyle (1,\infty)$.

Am I right?If not,can anyone show me the right way?