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consider a continuous function $f:[0.\infty)\rightarrow \mathbb{R},$ with the property that $f\sim \frac{\sin(x)}{x}, $ as $x\rightarrow \infty$.

We know that the function $\frac{\sin(x)}{x}$ is improper Riemann integrable. I would like to know if one can deduce that $f$ must be integrable under this conditions as well over the interval $(0,\infty)$, i.e. is it true that

$$\int_0^{\infty}f(x)dx<\infty?$$

Or more generally. If $f \sim g $ as $x\rightarrow \infty$ and f,g are ciontinuous and $\int_{0}^{\infty}g(x)dx <\infty$. Can one deduce that $\int_0^{\infty}f(x)dx<\infty$

edit: $f\sim g$ iff $\lim_{x\rightarrow \infty}\frac{f(x)}{g(x)}=1.$

Best wishes

Hasti Musti
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2 Answers2

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Put $\displaystyle f(t)=\frac{\sin(t)}{t}$, and $g(t)=f(t)$ for $0\leq t\leq \pi$, $\displaystyle g(t)=\frac{\sin(t)}{t}+\frac{\sin(t))^2}{t\log t}$ for $t\geq \pi$. Note that $g$ is continous on $[0,+\infty[$, and that we have $\displaystyle g(t)=f(t)(1+\frac{\sin(t)}{\log t})$ for $t\geq \pi$, hence $f\sim g$ if $t\to +\infty$. Now $\displaystyle \int_{\pi}^{+\infty}\frac{(\sin(t)^2}{t\log t}dt=+\infty$, as for $k\geq 2$ $$\int_{k\pi}^{(k+1)\pi} \frac{(\sin(t)^2}{t\log t}dt\geq \frac{\int_0^{\pi}\sin(t)^2dt}{(k+1)\pi (\log (k+1)\pi)}\geq \frac{c}{k\log k}$$ for a constant $c>0$, and the fact that the series of general term $\displaystyle \frac{1}{k\log k}$ is divergent. Hence $\displaystyle \int_0^{+\infty}g(t)dt$ is divergent.

Kelenner
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I think that you can prove it this way: since $f(x)$ is continuous and $f\sim \frac{\sin(x)}{x}$ as $x\rightarrow \infty$, then you can suppose that $\exists k, x_0\geq 0 \in \mathbb{R}$ s.t. $f(x)\leq k\frac{\sin(x)}{x}$, $\forall x\geq x_0$. Then, since $\int_{x_0}^{\infty}\frac{\sin(x)}{x}dx<\infty$, also $\int_{x_0}^{\infty}k\frac{\sin(x)}{x}dx<\infty$ and therefore $$\int_{x_0}^{\infty}f(x) dx\leq \int_{x_0}^{\infty}k\frac{\sin(x)}{x}dx<\infty$$

cholo14
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