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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?2012-02-25
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    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.2012-02-26
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    there might be a reason why you cannot find a counterexample. Did you try to prove the statement?2012-02-26
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    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.2012-02-26
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    Another reason why it seems wrong is that the Dirchlet test seems to 'demand' a lot more, but here we don't demand much of f and g.2012-02-26
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    @mixedmath: $(-1)^n/\sqrt{n}$2012-02-26
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    @josh: then you can find a counterexample similar to the one for series? (hint: $f(x)$ has to be oscillating)2012-02-26
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    I know $f$ has to be oscillating... but I can't think of any trigonometrical trick that would be the equivalent of $(-1)^n$ in series, and keep continuity.2012-02-26
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    It seems like $sin(x)/\sqrt(x)$ ought to work, but I can't prove it. I'll think about that.2012-02-26
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    Okay, I got it. It took noticing that $sin^2(x)=1-cos(2x)$2012-02-26
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    the integral of $\sin(x)/\sqrt{x}$ converges. To prove it, split the integral up in pieces $[n \pi, (n+1)\pi]$ ...2012-02-26
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    One doesn't even need $f$ to be oscillating: suppose $f= 1$ on $[0,1]$ and $f(x) = x^{-3/2}$ for $x> 0$. Can you multiply $f$ by function that destroys the convergence of the $x^{-3/2}$ part?2012-02-26
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    @josh: if you have solved it please add an answer to your question.2012-02-26
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    @Fabian: I cannot, being a guest user. Could any one of you write an answer? Thanks.2012-02-26
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    @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?2012-02-26
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    @josh : you're right. I was (wrongly) thinking about functions on a bounded interval when I hurriedly wrote that comment. That's math.2012-03-02

2 Answers 2

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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 $[0,\infty)$?2012-02-26
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    @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).2012-02-26
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    @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.2012-02-26
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    @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?2012-02-26
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    @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?2012-02-26
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    @Gingerjin I'm not sure what you're referring to. The integral $\int_0^\infty \sin(x^2)\, dx$ converges. I imagine one can show that $\int_1^\infty \sin^2(x^2)\, dx ={1\over2}\int_1^\infty {\sin(u)\over \sqrt u}\, du $ diverges by comparing it with the integral in Martin's example.2012-02-26