$\int_0^{\infty} f(x)dx$ converges and $\lim_{x\rightarrow \infty} f(x)$ exists, then $\lim_{x\rightarrow \infty} f(x)=0$?

Question

Is the following true or false:

If $\int_0^{\infty} f(x)dx$ converges and $\lim_{x\rightarrow \infty} f(x)$ exists, then $\lim_{x\rightarrow \infty} f(x)=0$?

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This should be doable without series.

Answer 1

For all $n\in \mathbb{N}$ larger than 1,

Let $$ f(x)= \begin{cases} 0 \ for \ x\leq n \\ n^4x-n^5 \ for \ n<x\leq n+1/n^3 \\ -n^4+2n+n^5 \ for \ n+1/n^3<x\leq n+2/n^3 \end{cases}$$ By some calculation $\int_n^{n+1}f(x)dx=1/n^2$. Then $\int_0^\infty f(x) = \sum_{n=2}^\infty 1/n^2$

Answer 2

Assume that $$\lim_{x\to\infty}f(x)=c>0$$ Then there exists $R$ such that $f(x)>c/2$ for all $x>R$.

Let $$\int_0^Rf(x)dx=M$$

Then for all $S>M$, we have $$\int_0^{R+2(S-M)/c} f(x)dx=M+\int_R^{R+2(S-M)/c} f(x)dx>M+\frac{c}{2}\frac{2(S-M)}{c}=S$$

Hence $\int_0^{\infty}f(x)dx$ diverges.

The proof is similar for $c<0$.

Hence we conclude that $c=0$.