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Let $f$ be continuous in [0, 1]. Assume $$0 < a < b$$

Prove that the following limit exists and determine what it is:

$$\lim_{\delta\rightarrow 0}\int_{\delta a}^{\delta b} \frac{f(x)}{x} dx$$

I got stuck on this question and I can't seem to figure it out. It seems to me that there's no reason for this limit to always exist. For instance, if I take

$$f(x) = 1$$

The integral of the harmonic function diverges near 0, so by Cauchy's critrion, the limit shouldn't exist.

So what am I missing? Does anyone have an idea?

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  • $\begingroup$ I suppose you meant $\;0<a<b\le1\;$ ... $\endgroup$ – DonAntonio May 17 '18 at 20:28
  • $\begingroup$ What does exactly $\delta a$ or $\delta b$ denotes ? $\endgroup$ – onurcanbkts May 18 '18 at 3:30
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You can predict that the limit is $f(0) \ln(b/a)$ by observing that for small enough $\delta$, by continuity of $f$,

$$\int_{\delta a}^{\delta b} \frac{f(x)}{x}dx \approx \int_{\delta a}^{\delta b} \frac{f(0)}{x}dx$$

To prove it, note that

$$\left| \int_{\delta a}^{\delta b} \frac{f(x)}{x} dx - f(0)\ln\frac{b}{a} \right| = \left| \int_{\delta a}^{\delta b} \frac{f(x)}{x} dx - f(0) \int_{\delta a}^{\delta b} \frac{dx}{x} \right| = \left| \int_{\delta a}^{\delta b} \frac{f(x) - f(0)}{x} dx \right| \le \int_{\delta a}^{\delta b} \left|\frac{f(x) - f(0)}{x}\right| dx$$

Let $\epsilon > 0$. Since $f$ is continuous, there is $\alpha > 0$ such that $0 < x < \alpha \implies |f(x) - f(0)| < \epsilon'$, where $$\epsilon' = \frac1{\ln \frac{b}{a}} \epsilon$$

Let $$\delta_0 = \frac{\alpha}{b}$$

Let $\delta < \delta_0$. For $x < \delta b$, we have $x < \alpha$, so $|f(x) - f(0)| < \epsilon'$. Hence, following the above series of inequalities,

$$\left| \int_{\delta a}^{\delta b} \frac{f(x)}{x} dx - f(0)\ln\frac{b}{a} \right| < \epsilon' \int_{\delta a}^{\delta b} \frac{dx}{x} = \epsilon$$

So we are done.

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Hint:

By the integral MVT, there exists $\xi \in (\delta a, \delta b)$ such that $\xi \to 0$ as $\delta \to 0$ and

$$\int_{\delta a}^{\delta b} \frac{f(x)}{x} \, dx = f(\xi)\log \frac{\delta b}{\delta a} $$

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Do a change of variables by setting $y = \frac{x}{\delta}$. Then $y \in [a,b]$ and $dx = \delta dy$. Hence

$$ \int_{\delta a}^{\delta b } \frac{f(x) }{x} dx = \int_{a}^b \frac{f(\delta y)}{y} dy \to f(0) \int_a^b \frac{dy}{y} = f(0) \ln \frac{b}{a}, $$ where you need to use the continuity of $f$ to prove the passage to the limit.

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