Limit involving integral $\frac{1}{r}\int_0^r f(s)\, ds \to0$?

Question

$$ \lim_{r\to 0}\left( r\int_0^r \frac{s^{n-1}}{s^2\log s}\, ds\right) $$

I am considering $ r $ small, r <1/2 and $n$ an integer.

For me, intuitively, I notice that the integral converges to zero. I tried to make a substitution, like $ t = | \log s | $, but I was not well sustained.

Answer 1

If $p>-1$, and $0 < r < \tfrac12$, then $$ \int_0^r s^p (\log(1/s))^\alpha \, ds \approx r^{p+1} (\log(1/r))^\alpha ,$$ where $A \approx B$ means there exists a constant $C>0$ with $C^{-1} A \le B \le C A$. The constant gets worse as $p \to 1+$.

Proof: See that for $\epsilon>0$, there exists $C>0$ such that for all $0<t<1$ and $0 < r< \frac12$ that $$ 1 \le \frac{\log(1/(rt))}{\log(1/r)} = 1 + \frac{\log(1/t)}{\log(1/r)} \le 1 + \frac{\log(1/t)}{\log(2)} \le C t^\epsilon .$$

So if you write $f(r) = r^p (\log(1/r))^\alpha$, then you realize that there exists constants $C_1, C_2>0$ and $-1 < p_1 < p < p_2 < \infty$ such that for $0<t<1$ we have $$ C_1 t^{p_2} f(r) \le f(r t) \le C_2 t^{p_1} f(r) . $$ Also $$ \int_0^r f(s) \, ds = r \int_0^1 f(rt) \, d t $$ and so $$ \frac{C_1}{1-p_2} r f(r) \le \int_0^r f(s) \, ds \le \frac{C_2}{1-p_1} r f(r) , $$ or $$ \int_0^r f(s) \, ds \approx r f(r) . $$

Answer 2

Hint : if $f$ is differentiable, $$\lim_{r \rightarrow 0} \frac{f(r)-f(0)}{r-0} = f'(0)$$

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ As an example, I'll assume $\ds{\large\color{red}{n > 2}}$: \begin{align} &\bbox[5px,#ffd]{\lim_{r \to 0}\bracks{r\int_{0}^{r}{s^{n - 1} \over s^{2}\ln\pars{s}}\,\dd s}}\,\,\,\stackrel{s\ =\ \expo{-t}}{=}\,\,\, \lim_{r \to 0}\bracks{r\int_{\infty}^{-\ln\pars{r}}{\expo{-\pars{n - 2}t} \over t}\,\dd t} \\[5mm] = & -\lim_{r \to 0}\bracks{r\int_{-\pars{n - 2}\ln\pars{r}}^{\infty}{\expo{-t} \over t}\,\dd t} = -\lim_{r \to 0}\bracks{r\,\mrm{E}_{1}\pars{-\pars{n - 2}\ln\pars{r}}} \\[5mm] = &\ -\lim_{r \to 0}\bracks{r\, {\expo{\pars{n - 2}\ln\pars{r}} \over -\pars{n - 2}\ln\pars{r}}} = {1 \over n - 2}\lim_{r \to 0}{r^{n - 1} \over \ln\pars{r}} = \bbx{\large 0} \\ & \end{align}

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$\ds{\mrm{E}_{1}}$ is the Exponential Integral Function. In evaluating the limit I used the $\ds{\mrm{E}_{1}}$ asymptotic Expansion.

Answer 4

$$\lim_{r \to 0} r\int_0^r \frac{s^{n-1}}{s^2 \log s} \text{d}s=\lim_{r \to 0} \frac{\int_0^r \frac{s^{n-1}}{s^2 \log s}\text{d}s}{\frac{1}{r}}$$ By De L'Hôpital rule $$\lim_{r \to 0} \frac{\int_0^r \frac{s^{n-1}}{s^2 \log s}\text{d}s}{\frac{1}{r}} =\lim_{r \to 0} \frac{\frac{r^{n-1}}{r^2 \log r}}{-\frac{1}{r^2}}=\lim_{r\to 0} -\frac{r^{n-1}}{\log r}=0$$ For all $n \geq 1$.