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Let $f:(0,1] \to \mathbb R$ be a bounded integrable function. How to (dis)prove the existence of this limit? $$\lim_{x \to 0+} \frac{1}{|\ln(x)|} \int_x^1 \frac{f(t)}{t} \mathrm dt $$ If the limit doesn't exist, what kind of assumption(s) should be given to guarantee the limit? Also, is there a name or kind for this limit/integral? Thank you.

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Let $J(x) = \displaystyle \int_x^1 \frac{f(t)}{t}\ dt$. Note that if $|f(x)| \le B$ for all $x$, $\left|J(x)\right| \le - B \ln(x)$, so the lim sup and lim inf are finite. However, consider $\displaystyle f(x) = (-1)^{\lfloor \ln(-\ln(x)) \rfloor}$, i.e. for $\exp(-e^n) < x \le \exp(-e^{n-1})$, $f(x) = 1$ if $n$ is odd and $-1$ if $n$ is even. Then $$(-1)^{n-1} J(\exp(-e^n)) > \int_{\exp(-e^n)}^{\exp(-e^{n-1})} \dfrac{dt}{t} - \int_{\exp(-e^{n-1})}^1 \dfrac{dt}{t} = (1 - 2/e) e^n $$ so $\lim_{x \to 0+} J(x)/|\ln(x)|$ does not exist.

EDIT: It may be useful to make a change of variables: if $g(s) = f(e^{-s})$, $$\lim_{x \to 0+} \frac{J(x)}{|\ln(x)|} = \lim_{y \to \infty} \frac{1}{y} \int_0^y g(s)\ ds$$

(each limit existing if and only if the other does).

Note that if $\int_{n}^{n+1} |g(s)|\ ds$ is bounded, the limit exists and is $L$ if and only if the series $\sum_{k=1}^\infty a_k$ is Cesàro summable to $L$, where $a_1 = \int_0^1 g(s)\ ds$ and $a_{k} = \int_{k-1}^k g(s)\ ds - \int_{k-2}^{k-1} g(s)\ ds$ for $k \ge 2$.

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  • $\begingroup$ Nice. Boundedness is not sufficient for the limit to exist. (+1) $\endgroup$
    – robjohn
    Commented Aug 1, 2012 at 23:03
  • $\begingroup$ Thanks Robert, your answer is very helpful. $\endgroup$
    – Deco
    Commented Aug 2, 2012 at 9:24
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If $\displaystyle\lim_{x\to0^+}f(x)=L$, then $\displaystyle\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^1\frac{f(t)}{t}\,\mathrm{d}t=L$.

Choose an $\epsilon>0$ and find $\delta>0$ so that if $0<x<\delta$, then $|f(x)-L|<\epsilon$.

Then $$ \begin{align} &\left|L-\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^1\frac{f(t)}{t}\,\mathrm{d}t\right|\\ &=\left|\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^1\frac{f(t)-L}{t}\,\mathrm{d}t\right|\\ &=\left|\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^\delta\frac{f(t)-L}{t}\,\mathrm{d}t +\lim_{x\to0^+}\frac1{|\log(x)|}\int_\delta^1\frac{f(t)-L}{t}\,\mathrm{d}t\right|\\ &\le\left|\lim_{x\to0^+}\frac1{|\log(x)|}\int_x^\delta\frac{f(t)-L}{t}\,\mathrm{d}t\right| +\left|\lim_{x\to0^+}\frac1{|\log(x)|}\int_\delta^1\frac{f(t)-L}{t}\,\mathrm{d}t\right|\\ &\le\epsilon\lim_{x\to0^+}\left|\frac{\log(x)-\log(\delta)}{\log(x)}\right| +\lim_{x\to0^+}\frac1{|\log(x)|}\left|\int_\delta^1\frac{f(t)-L}{t}\,\mathrm{d}t\right|\\ &=\epsilon+0 \end{align} $$ Therefore, $$ \lim_{x\to0^+}\frac1{|\log(x)|}\int_x^1\frac{f(t)}{t}\,\mathrm{d}t=L $$

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    $\begingroup$ Perhaps it is worth pointing out that this condition is not necessary for the limit to exist: $\lim_{x\to0+}\frac{1}{|\log(x)|}\int_x^1\frac{\sin(1/t)}{t}\,dt=0$, even though $\lim_{x\to0+}\sin(1/x)$ does not exist. $\endgroup$ Commented Aug 1, 2012 at 22:34
  • $\begingroup$ @PinkElephants: Indeed, the condition I present is sufficient, but not necessary. $\endgroup$
    – robjohn
    Commented Aug 1, 2012 at 22:57

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