# 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 $\mathbb{R}$.
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 $(1,\infty)$.