$\int_{0}^{\infty} f(x)dx$ converges and $f(x),g(x)$ continuous and bounded in $[0,\infty)$, must $\int_{0}^{\infty} f(x)g(x)dx$ converge? I am trying to disprove the claim in the title of this question:

If $\int_{0}^{\infty} f(x)dx$ converges and $f(x),g(x)$ are continuous, bounded functions in $[0,\infty)$, $\int_{0}^{\infty} f(x)g(x)dx$ converges.

But I can't find any counterexamples.
Help would be appreciated!
 A: Let $f(x)=\frac{\sin x}x$, $g(x)=\sin x$. Then $f,g$ satisfy your hypothesis. But 
$$
\int_0^\infty f(x)g(x)dx=\int_0^\infty \frac{\sin^2(x)}x\,dx=\infty.
$$
To see this, consider the points $\{(2k+1)\pi/2:\ k\in\mathbb{N}\}$; there exists $\delta>0$ such that $\sin^2 t\geq 1/2$ for all $t\in [(2k+1)\pi/2-\delta, (2k+1)\pi/2+\delta]$. Then
$$
\int_0^\infty\frac{\sin^2(x)}x\,dx\geq\sum_{k=0}^\infty\int_{(2k+1)\pi/2-\delta}^{(2k+1)\pi/2+\delta}\frac1{4x}\,dx\geq\sum_{k=0}^\infty\frac{2\delta}{4[(2k+1)\pi/2+\delta]}=\infty.
$$
A: Take a function $f$ whose graph consists of spikes centered at the positive integers that do not overlap, together with portions of the $x$-axis, with the following properties: 


*

*The area bounded by the $n$th spike and the $x$-axis is less than $1\over n$.

*The area of the "squared spike" is greater than ${1\over2n}$. 

*Spikes centered at odd positive integers are above the $x$-axis

*Spikes centered at even positive integers are  below the $x$-axis.


Then $\int_0^\infty f(x)\, dx$ converges (it can be computed as a convergent alternating series).  Now consider $g=f$.
I believe $f(x)=g(x)=\sin(x^2)$ furnishes an example (it has properties similar to the above).
