is the improper integral over a bounded and convergent integrand always bounded?

Question

I'm looking at an improper Integral of the form

$$ \int_a^\infty f(x) dx $$

where

$$ \lim_{x \rightarrow \infty} f(x) = 0 $$ and $$ 0 \leq f(x) < K \quad \forall \: x $$ for some $ K > 0 $

The first question I have is, does the second condition imply $$ f(x) < \infty \quad \forall x $$ and the second question is does this overall imply that the improper integral is bounded $$ \int_a^\infty f(x) dx < \infty $$

EDIT: The original question has been satisfyingly answered by the counter example $\frac{1}{x}$ however I wonder if the statement may be true if $f(x)$ converges to $0$ exponentially?

Answer 1

Yes, then no. Assume we have $f(x)>0$.

$$0<f(x)<K<\infty\ \forall\ x\in\mathbb R$$

But, if we had $f(x)=\begin{cases}1&;|x|\le1\\\frac1{|x|}&;|x|>1\end{cases}$ and $K=2$,

$$\int_a^bf(x)dx\to\infty\text{ as }b\to\infty$$

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EDIT:

By comparison, if we have

$$f(x)<Ke^{-x}\ \forall\ x>N$$

then, for $N<a<b$,

$$\int_a^bf(x)dx<\int_a^bKe^{-x}dx<\infty$$

so the integral will converge.

Answer 2

I think $$f(x)={1\over x}$$ may be an answer to your question.