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Suppose $f: (0,+\infty) \to \mathbb{R}$ is s.t. \begin{equation}\tag{1} f(s) = \frac{1}{4\pi s} + \mathcal{O} \biggl (\frac{1}{\sqrt{s}} \biggr ) \quad \quad\text{ as } s \to 0^+ \end{equation} and $f \in L^1([t_0,+\infty))$ for each $t_0>0$.

Can I conclude $$\tag{2} \int_{2t}^{+\infty} f(s) ds = \frac{ |\log(t)|}{4\pi} + \mathcal{O}(\sqrt{t}) \quad \quad \text{ as } t \to 0^+ $$ ?

I was trying in the following way: from $(1)$ I know that there exist $0 < \delta < 1/2 $ and $C>0$ s.t. $$ \frac{1}{4 \pi s} - \frac{C}{\sqrt{s}} \le f(s) \le \frac{1}{4 \pi s} + \frac{C}{\sqrt{s}} $$ for each $0<t< \delta$. Then, for each $0< t < \delta/4$, $$ \int_{2t}^{\infty} f(s)ds = \int_{2t}^{\delta/2}f(s)ds + \int_{\delta/2}^{+\infty} f(s)ds \le \int_{2t}^{\delta/2} \biggl (\frac{1}{4 \pi s} + \frac{C}{\sqrt{s}} \biggr ) ds + \|f\|_{L^1([\delta/2, + \infty))} =$$ $$= \frac{\log(\delta/2)}{4 \pi} + 2C\sqrt{\delta/2} - \frac{\log(2t)}{4\pi}-2\sqrt{2}C \sqrt{t} + \|f\|_{L^1([\delta/2, + \infty))} = $$ $$= \frac{|\log(t)|}{4\pi} -2\sqrt{2}C \sqrt{t} + \biggl ( \|f\|_{L^1([\delta/2, + \infty))} -\frac{\log(2)}{4\pi} + 2C\sqrt{\delta/2}-\frac{|\log(\delta/2)|}{4\pi} \biggr ) =$$ Now, in order to obtain one of the inequalities of $(2)$, I should prove that there exists some $0 < \delta_1 < 1/10$ and $C_1 >0 $ s.t. $$ \|f\|_{L^1([\delta/2, + \infty))} -\frac{\log(2)}{4\pi} + 2C\sqrt{\delta/2}-\frac{|\log(\delta/2)|}{4\pi} \le (-2\sqrt{2}C +C_1) \sqrt{t} $$ for every $0<t< \delta_1$. But I cannot control the LHS of this last inequality.

All suggestions are welcome!

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  • $\begingroup$ NO. Because $\int_{2t}^{\infty}f(x)dx$ might not exist in $\Bbb R$. For example if $f(x)=1/\sqrt x$ then $\int_{2t}^{\infty}f(x)dx=\infty.$ $\endgroup$ – DanielWainfleet Mar 21 at 5:53
  • $\begingroup$ Thank you for your comment. The function you wrote is not in $L^1([t_0, + \infty))$ for every $t_0 > 0$. In any case, I think thai a fuction like $f(s) = \frac{1}{4 \pi s}$ if $0<s<1$ and $f(s) = e^{-s}$ if $s \ge 1$ could be a counterexample. $\endgroup$ – Bremen000 Mar 21 at 5:58

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