# Big $O$ of integral and estimate

Suppose $$f: (0,+\infty) \to \mathbb{R}$$ is s.t. $$$$\tag{1} f(s) = \frac{1}{4\pi s} + \mathcal{O} \biggl (\frac{1}{\sqrt{s}} \biggr ) \quad \quad\text{ as } s \to 0^+$$$$ and $$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 $$0. Then, for each $$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. But I cannot control the LHS of this last inequality.

All suggestions are welcome!

• NO. Because $\int_{2t}^{\infty}f(x)dx$ might not exist in $\Bbb R$. For example if $f(x)=1/\sqrt x$ then $\int_{2t}^{\infty}f(x)dx=\infty.$ – DanielWainfleet Mar 21 at 5:53
• 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. – Bremen000 Mar 21 at 5:58