The converse of $\exists\alpha>1, \forall x\in \mathbb{R}: |x|^\alpha |f(x)| It is easy to show, that (for continuous functions f)
 $$\exists c>0,\exists\alpha>1, \forall x\in \mathbb{R}: |x|^\alpha |f(x)| <c \implies \int |f|dx <\infty$$
The question is, whether or not this is also a neccessary condition. I could not come up with any counterexamples ($1/(x \log x)$ is not integrable yet). But I am also not able to prove it. And it is difficult to search for even though someone has probably already asked this question.

UPDATE: It appears this is not true in general even for smooth functions $f\in C^\infty$ due to the ability to create smooth bumps of fixed hight in regular intervals with a decreasing base, making them integrable (See comments and answers). So in order to avoid that: what about monotonous functions?
I am trying to understand wether or not there is a function with a decrease "between" $1/x$ and $1/x^\alpha$. Sorry for moving the goalposts.

Proof, that it is sufficient:
$$\int |f(x)| dx \le \int_{|x|<1} |f(x)| dx + \int_{|x|>1} |x|^{-\alpha}|x|^\alpha |f(x)|dx<2\|f\|_{\infty,[-1,1]} + c\int_{|x|>1} |x|^{-\alpha}dx<\infty$$
Fix due to the helpful question whether or not $|x|^{-\alpha}$ is integrable.
 A: If it were, then $|f(x)|<c/|x|^{\alpha}$ for $|x|>1$ which entails that $f(x)\rightarrow 0$ as $x\rightarrow\infty$. So we are asking whether integrable functions will eventually converge to zero, there are plenty of counterexamples.
One counterexample would be, $f(x)=1$ for $n\leq x\leq n+1/2^{n}$ and zero otherwise.
For the updated question, take $f(x)=1/(2(\log 2)^{2})$ for $0\leq x\leq 2$ and $f(x)=1/(x(\log x)^{2})$ for $x\geq 2$, and for $x<0$, make it as an even function.
So $f\in L^{1}({\mathbb{R}})$. If there were some $C>0$ and $\alpha>1$ such that $x^{\alpha}f(x)\leq C$ for all $x\geq 2$, then $x^{\alpha-1}/(\log x)^{2}\leq C$. But we can have $\log x\leq C'x^{(\alpha -1)/4}$, then $x^{\alpha -1}/(\log x)^{2}\geq C''x^{(\alpha-1)/2}$. Taking limit as $x\rightarrow\infty$ gives you a contradiction.
A: A better example than my measure-zero one in the comment: Pick your favorite summable sequence $(a_n)$ and your favorite positive number $b$, and construct a function whose grap has a right triangle with base $a_n$ and height $b$. It's integral will be $\sum_n a_n b/2 <+\infty$, but it does not go to zero at infinity.
