# Boundedness of integrand in the infinite integral

Definition: Let $f$ be defined on $[a,+\infty)$, and $f$ is a Riemann integrable function on $[a,x]$ for any $x\geq a$, if $\lim\limits_{x\to+\infty}\int_a^xf(t)dt$ exists , denoted by $\int_a^{+\infty}f(x)dx$.  Suppose it equals to $J\in\mathbb{R}$,  then we call that $f$ has an infinite integral $\int_a^{+\infty}f(x)dx=J$.

Suppose a real-value function $y=f(x)$ is continuous on $[a,+\infty)$ and $\lim\limits_{x\to+\infty}f(x)$ does not exist and $\lim\limits_{x\to+\infty}f(x)\neq\infty$. If the infinite integral $\int_a^{+\infty}|f(x)|dx$ converges, i.e. $\int_a^{+\infty}f(x)dx$ absolutely converges, then $f$ is bounded on $[a,+\infty)$?   If the condition modifies that $\lim\limits_{x\to+\infty}|f(x)|\neq+\infty$, and the rest of the conditions remain the same, then the conclusion is also right or wrong?   Thanks a lot.

Let $a=1$ and $f:[1, \infty) \to \mathbb R$ defined by
$f(x)=x$ if $x \in \mathbb N$ and $f(x)=0$ if $x \in [1, \infty) \setminus \mathbb N$.
Then $\int_1^{+\infty}f(x)dx=\int_1^{+\infty}|f(x)|dx=0$, $\lim\limits_{x\to+\infty}f(x)$ does not exist, $\lim\limits_{x\to+\infty}f(x)\neq\infty$, but $f$ is not bounded on $[1, \infty)$.
• And if we prefer a continuous (differentiable, smooth) integrand, one can simply put in sufficiently narrow peaks around each integer $x$. – Greg Martin Feb 15 '17 at 8:13