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$$\int\limits_0^\infty f(x)\,dx=\lim\limits_{x\to\infty}\int\limits_0^x f(z)\,dz = \lim\limits_{x\to\infty}\frac{x\int\limits_0^x f(z)\,dz}{x}= \lim\limits_{x\to\infty}\frac{\left(x\int\limits_0^x f(z)\,dz\right)'}{x'}= \lim\limits_{x\to\infty}\left(\int\limits_0^x f(z)\,dz+xf(x)\right)= \lim\limits_{x\to\infty}xf(x)=0$$ But this is true only if $\lim\limits_{x\to\infty}x\int\limits_0^x f(z)\,dz=\infty$.

So how can we prove or refute that $\int\limits_0^{\infty} f(x)\,dx=0\Rightarrow\lim\limits_{x\to\infty}xf(x)=0$?

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  • $\begingroup$ Depending on your constraints on $f$, this could be false. What kind of function are you requiring $f$ to be? $\endgroup$ – Cameron Williams Dec 8 '13 at 17:22
  • $\begingroup$ This is false. In fact, $\lim_{x\to\infty} f(x)=0$ is false. If you know that the limit exists, then it is indeed zero. But the existence of the limit is not automatic. $\endgroup$ – Andrés E. Caicedo Dec 8 '13 at 17:22
  • $\begingroup$ No, the issue is not on the numerator. In fact, we do not need the numerator to tend to infinity to apply l'Hôpital's rule here. The issue is that $\lim_{x\to\infty}xf(x)$ may fail to exist, and as pointed out, in fact $\lim_{x\to\infty}f(x)$ may fail to exist. $\endgroup$ – Andrés E. Caicedo Dec 8 '13 at 17:32
  • $\begingroup$ @Cameron Williams $f(x)$ is simply a continuous function. $\endgroup$ – Constructor Dec 8 '13 at 17:32
  • $\begingroup$ @Andres Caicedo I know that $\int\limits_0^{\infty} f(x)\,dx=0$. I want to prove or refute that $\lim\limits_{x\to\infty}xf(x)=0$. $\endgroup$ – Constructor Dec 8 '13 at 17:34
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$$ \int_0^\infty \sin(x^2) \, dx = \sqrt{\frac{\pi}{8}} $$

So if you just subtract off anything else with integral $\sqrt{\pi/8}$ that vanishes at infinity, you get a counterexample.

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  • $\begingroup$ Nice counterexample. Thank you. $\endgroup$ – Constructor Dec 8 '13 at 18:02
  • $\begingroup$ I must be dense, but in what universe is the integral of the lhs even defined? $\endgroup$ – Igor Rivin Dec 8 '13 at 20:16
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    $\begingroup$ @Igor: It's an ordinary, improper Riemann integral. What is surprising about it? That the integrand doesn't converge to zero? The relevant point to note is that it's the area under the graph not matters, not the height. The graph if $\sin(x^2)$ consists of regions alternating between above and below the $x$-axis: adding up their areas gives an alternating sum whose terms converge to zero. $\endgroup$ – Hurkyl Dec 8 '13 at 22:11

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