Lebesgue Integral

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

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

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