Assume that $f(x)$ is continuous over $[a,+\infty)$ where $a>0$, and $\displaystyle\int_{a}^{+\infty}\dfrac{f(x)}{x}{\rm d}x$ is convergent. Can we infer that $\lim\limits_{x \to +\infty}f(x)=0$? If not, what conditions else are needed?
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1Take $f(x)=\sin(x)$. – Botond Apr 10 '19 at 12:24
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@Botond what if we add the assumption $f(x)>0$? – mengdie1982 Apr 10 '19 at 12:27
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1That is not enough, either. Just take triangle of smaller width at each integer – Clayton Apr 10 '19 at 12:31
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It is true if you add the assumption that f(x)/x is weakly decreasing. To prove this consider the antiderivative evaluated at 2x minus that evaluated at x. – shalop Apr 10 '19 at 13:10
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The limit as $x\to\infty$ may not exist at all for $f(x)$; consider $f(x)=x\sin(x^2)$. Then $$\int_a^\infty \frac{f(x)}{x}\,dx$$ is convergent (see Fresnel integrals on wikipedia), but $f(x)$ oscillates so has no limit.
At each $n\in\mathbb N$, place a triangle of height $1$ and width $1/n^2$. The base need not rest on the $x$-axis; indeed, the base could be placed $\frac{1}{n^2}$ above the $x$-axis. Now just connect the endpoints of the triangles by straight lines. This guarantees that $f(x)>0$ for all $x>0$ and that $\lim_{x\to\infty}f(x)$ does not exist.
To force the limit of $f(x)$ to be $0$, requiring that $f(x)$ is monotone suffices.
Clayton
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I already responded in the comments. That is not enough, even with strict inequality. You can take triangles of height $1$ at each integer of decreasing base length. – Clayton Apr 10 '19 at 12:34
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@mengdie1982: I've edited my response and provided a sufficient condition to guarantee the existence of the limit for $f(x)$ and that its limit will be $0$. – Clayton Apr 10 '19 at 12:50
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An example where $f$ is non-negative but $f(x)$ does not tend to is obtained as follows: let $f(x)=n^{2}(n+\frac 1 {n^{2}}-x)$ for $x \in (n,n+\frac 1 {n^{2}})$,$f(x)=n^{2}(x-n+\frac 1 {n^{2}})$ for $x \in (n-\frac 1 {n^{2}},n)$, $n=2,3,...$ and $0$ elsewhere.
Kavi Rama Murthy
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Let $f>0$ with $\lim\limits_{x \to +\infty} f(x)=b > 0$. Then we have that $\exists y$ so that $\forall x \geqslant y$ we have that $$\frac{b}{2} \leqslant f(x) \leqslant \frac{3b}{2}$$ We have that $$\int_a^{+\infty} f =\int_a^y f + \int_y^{+\infty} f$$ The first integral has a finite value, so we need to take a look at the second one only. But on that interval, we have that $$\frac{f(x)}{x} \geqslant \frac{b}{2x}$$ And the integral of the RHS is divergent, which would imply that the original integral is divergent as well, which is a contradiction. So $f$ cannot have a positive limit, which means that it's either $0$ or does not exist. Which means that we need to assume the existence of $\lim\limits_{x \to +\infty} f(x)$ as well.
Botond
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@KaviRamaMurthy Did you want to point out the error of $b \geqslant 0$? I wanted to write $b >0$, and I already fixed it. – Botond Apr 10 '19 at 12:49
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Your answer only proves that if $\lim f(x)$ exists then it cannot be $>0$. You have to prove the existence of the limit or else give a counterexample. – Kavi Rama Murthy Apr 10 '19 at 12:52
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@KaviRamaMurthy You provided an example when the integral exists, but the limit of the function does not, so I didn't see any reason to give another one. – Botond Apr 10 '19 at 12:56