Prove or disprove: if $\int_1^\infty f(x)\,dx$ converges, then $\lim\limits_{x\to\infty}f(x) =0$.

Question

I received the following question:

7. If $\int_1^\infty f(x)\,dx$ converges, then $\lim\limits_{x\to\infty}f(x) =0$.

Image.

I know that it is false, but i can't come up with a counter example, Thanks in advance

Answer 1

Here is an example from the book Counterexamples in Analysis.

For each integer $n > 1$ let $g(n) := 1$, on the closed intervals $\left[n - \frac{1}{n^2},n\right]$ and $\left[n , n + \frac{1}{n^2}\right]$ define $g$ linear and equal to $0$ at the nonintegral endpoints. For $x \geq 1$ define $g(x) := 0$ where $g(x)$ is not defined. Then the function $$f(x) := g(x) + \frac{1}{x^2} $$

is positive and continuous for $x \geq 1$, $\displaystyle\lim_{x \to \infty} f(x) \neq 0$ and the integral $$\int_1^{+\infty} f(x) dx$$

converges.

Answer 2

Ummm just take $f(x)=0 ,$ if $x\notin\mathbb{N}$ and $f(x)=1 ,$ if $x\in\mathbb{N}$

So the integral is $0$ (countable discontinuties)

However $f(x)$ does not converge to $0$ as $x$ increase to $\infty$.