I need to prove (as I wrote in the title): for monotonic $f$: if the improper integral $\int_0^\infty f(x)dx$ converges, then $\lim_{x\to\infty}xf(x)=0$

hints please? tried to think of Cauchy's.....

  • $\begingroup$ small hint. /consider a rational function. What doe you know about the difference of the degrees of numerator and denominator in order for such an improper integral to be convergent? Now why is it then that even if there is a x-term in front of $f(x)$ that the limit is still zero? $\endgroup$ – imranfat Nov 11 '13 at 21:43
  • $\begingroup$ duplicate of math.stackexchange.com/questions/560894/… $\endgroup$ – Paramanand Singh Nov 12 '13 at 5:39

For large $x$ $f$ will either be nonnegative or nonpositive. So assume WLOG $f(x)\ge 0$. Then it clearly must be monotone decreasing. Note that since $\int_0^\infty f(x)\,dx<\infty$ and $f\ge 0$ we have $\lim\limits_{t\to\infty}\int_{\frac{t}{2}}^{t} f(x)\,dx=0$. Then since $f$ is monotone decreasing we have $0\le \displaystyle \frac{tf(t)}{2}\le \int_{\frac{t}{2}}^{t} f(x)\,dx\to 0$ at $t\to\infty$, giving the result.

  • $\begingroup$ can you explain the last sentence? I don't understand how did you get to it and why is it true $\endgroup$ – CnR Nov 12 '13 at 6:39
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    $\begingroup$ It's because $(b-a)\inf\limits_{x\in(a,b)} f(x)\le\int_a^b f(x)\,dx$. $\endgroup$ – JLA Nov 12 '13 at 6:47
  • $\begingroup$ very nice proof! better than the one provided for the question math.stackexchange.com/questions/560894/… of which this one is a duplicate. $\endgroup$ – Paramanand Singh Nov 12 '13 at 7:35
  • $\begingroup$ @JLA is it a known inequality or does it derive from f being monotonic? $\endgroup$ – CnR Nov 12 '13 at 22:24
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    $\begingroup$ The one I posted in the comments? It's a basic inequality, it follows from the fact that if $f\le g$, then $\int f\le\int g$. The one in the answer follows from this. $\endgroup$ – JLA Nov 12 '13 at 22:31

Let $F(x) = \int_{0}^xf(x)dx$, Note that $F'(x) = f(x)$. Because $f(x)$ is monotonic at $[0,+\infty)$, it implies $f(x)$ is ultimately either $\lt 0$ or $\gt 0$ after a specific point $x_0$ as $x\rightarrow +\infty$. Therefore, its antiderivative $F(x)$ is ultimately either decreasing or increasing for all $x \gt x_0$. Moreover since $\int_{0}^{\infty}f(x)dx$ is convergent, then $F(x)$ must be bound. To sum up, $F(x)$ is ultimately either monotonic increasing with a upper bound or monotonic decreeing with a lower bound at $[x_0,+\infty)$, where $x_0 \gt 0$.

Without losing generality, we assume F(x) is monotone increasing at $[x_0, +\infty)$, it implies $f(x) \gt 0$ at $[x_0, +\infty)$. Thus, if $lim_{x\rightarrow\infty}xf(x) = d$, Note $d \gt 0$ since $f(x) \gt 0$ and monotonic at $[0,+\infty)$. Hence by the definition of limit:

$$\text{There exist X, when } x\gt x_0 \gt X \text{ such that, }|xf(x) - d| \lt \frac{d}{2}\tag{1}$$ Note (1) implies $$xf(x) \gt \frac{d}{2} \Longrightarrow f(x) \gt \frac{N}{x}\space\space(\text{where }N=\frac{d}{2})\tag{2}$$ Also since $f(x) \gt 0$, (2) implies: $$\int_{x}^{\infty}f(x)dx \gt \int_{x}^{\infty}\frac{N}{x}dx~~(x \gt x_0 \gt X)\tag{3}$$ Note, $F(x) = \int_{0}^{x}f(x)dx + \int_{x}^{\infty}f(x)dx$. Since $\int_{x}^{\infty}\frac{N}{x}dx$ is divergent, By (3), so does $\int_{x}^{\infty}f(x)dx$, which contradict the fact F(x) is convergent. Hence, $lim_{x\rightarrow\infty}xf(x) = 0$.

  • $\begingroup$ How do you know F(x) exists? $\endgroup$ – CnR Nov 12 '13 at 6:32
  • $\begingroup$ @CnR Since $\int_{0}^{\infty}f(x)dx = \lim_{x\rightarrow\infty}\int_{0}^{x}f(x)dx$ exist, then $F(x)$ exist $\endgroup$ – SundayCat Nov 12 '13 at 6:34
  • $\begingroup$ Doesn't it only mean that f is integrable? $\endgroup$ – CnR Nov 12 '13 at 6:40
  • $\begingroup$ @CnR $\int_{0}^{\infty}f(x)dx$ exist means $f$ is integrable at interval $[0,+\infty)$. It implies that $f$ is integrable in any subinterval $[0,x]$ when $x \gt 0$. Now you can consider $x$ as variable, then you will have a function $F(x) = \int_{0}^{x}f(x)dx$ $\endgroup$ – SundayCat Nov 12 '13 at 6:47

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