0
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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 $[1, \infty)$.

$\endgroup$
  • $\begingroup$ And if we prefer a continuous (differentiable, smooth) integrand, one can simply put in sufficiently narrow peaks around each integer $x$. $\endgroup$ – Greg Martin Feb 15 '17 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.