If $\int_{1}^{\infty}f\left(x\right)dx$ converges absolutely then $\int_{1}^{\infty}f^{2}\left(x\right)dx$ converges? Let $f:[0,\infty)\to\mathbb{R}$ be a continuous function. I was asked to prove/disprove the following statement:
If $\int_{1}^{\infty}f\left(x\right)dx$ converges absolutely then $\int_{1}^{\infty}f^{2}\left(x\right)dx$ converges as well. 
I figure that my only way of proving this is by direct comparison. However, for that to work I need $f(x)\leq1$  for sufficiently large $x$, but $\int_{1}^{\infty}f\left(x\right)dx$ converging (Even absolutely) does not guarantee this.
If the statement is false I'd appreciate a hint on how to construct a counter example, rather then one pulled out of thin air.

If
$f(x)$
has very narrow peaks
that grow in size
so that the sum of the area of
the $n$-th peak
converges,
the sum of
the peaks of
$f^2(x)$
can diverge.
For example,
if
$f(x)
= n^a
$
for
$n \le x \le n+n^{-b}
$,
then
$\int_n^{n+1} f(x) dx
=n^{a-b}
$
and
$\int_n^{n+1} f^2(x) dx
=n^{2a-b}
$.
To get the first sum to converge,
we need
$a-b < -1$
and to get the
second sum to diverge
we need
$2a-b > -1$.
If we choose,
for some $c > 0$,
$a-b=-1-c$
and
$2a-b=-1+c$,
then
$a = 2c$
and
$b=1+3c$
will work.
Choosing
$c=1/4$,
we get
$a = 1/2$
and
$b = 7/4$.
For this,
$n^{a-b}
=n^{-5/4}
$
and the sum of these converges,
and
$n^{2a-b}
=n^{-3/4}
$
and the sum of these diverges.
Round off the corners
if you don't like boxes.
