# $\int_{0}^{\infty} f(x)dx$ 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!

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Why you think it is wrong? –  Fabian Feb 25 '12 at 23:57
Mostly because it appeared in a set of two questions, the second of which is true ;). So the statement is correct, then? I'll try to prove it instead. –  josh Feb 26 '12 at 0:02
there might be a reason why you cannot find a counterexample. Did you try to prove the statement? –  Fabian Feb 26 '12 at 0:03
Nothing comes to mind in regards to a proof. The intuition behind it being wrong is also that the equivalent claim for series is wrong. –  josh Feb 26 '12 at 0:05
@JerryGagelman: I can't think of a bounded function that does this. Presumably if $f(x)$ is absolutely convergent then $f(x)g(x)$ is also absolutely convergent, so I think it might not exist? –  josh Feb 26 '12 at 0:18

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 $$\int_0^\infty\frac{\sin^2(x)}x\,dx\geq\sum_{k=0}^\infty\int_{(2k+1)\pi/2-\delta}^{(2k+1)\pi/2+\delta}\frac1{4x}\,dx\geq\sum_{k=0}^\infty\frac{2\delta}{4[(2k+1)\pi/2+\delta]}=\infty.$$

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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).

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Does $sin(x^2)$ converge in $[0,\infty)$? –  josh Feb 26 '12 at 0:20
@DavidMitra: $\int_0^\infty \sin^2(x^2)\,dx$ does not converge, just substituting $u=x^2$ leads to an integral similar to the one in my answer (with $\sqrt{u}$ in the denominator). –  Martin Argerami Feb 26 '12 at 0:33
@josh Yes. It's graph consists of the "alternating above/below the $x$-axis" spikes as above . But the area of the spikes heads towards zero as $x$ heads towards infinity (the amplitude of a spike is always $1$; but the width of the spikes become small, because for large x it doesn't take much to get $x^2$ to range over an interval of length $2\pi$). The integral $\int_0^\infty \sin(x^2)\,dx$ can be computed as a convergent alternating series. According to WA, $\int_0^\infty \sin^2(x^2)\,dx$ seems to diverge. And so it does. See Martin's comment above. –  David Mitra Feb 26 '12 at 0:37
@MartinArgerami It seems that the conclusion is right in a finite interval, but it is not right in a infinite interval. Can you give me an intuitive recognition? –  89085731 Feb 26 '12 at 1:38
@DavidMitra It seems that the conclusion is right in a finite interval, but it is not right in a infinite interval. Can you give me an intuitive recognition? –  89085731 Feb 26 '12 at 2:52