# property about improper integrals [duplicate]

Let be $$f:[0,\infty[ \rightarrow \mathbb{R}$$ a decreasing and continuos, $$f(x)>0 , \forall x$$. If $$\displaystyle \int_0^{\infty}f(x)dx$$ converges then $$\displaystyle \lim_{x \rightarrow \infty} xf(x) = 0.$$

How to prove that?

Ok, I know that if $$G(x) = \displaystyle \int_0^{x}f(t)dt$$ then $$\displaystyle \lim_{x \rightarrow \infty} G(x) = L$$, How can I use that for arrive in $$xf(x)$$??

## merged by quid♦Mar 30 at 20:11

This question was merged with If monotone decreasing and $\int_0^\infty f(x)dx <\infty$ then $\lim_{x\to\infty} xf(x)=0.$ because it is an exact duplicate of that question.

• – Martin R Mar 27 at 14:28