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The problem is:

If $f(x)\in C[0,+\infty)$, $\displaystyle\lim_{x\to+\infty}f(x)=k\in\mathbb R$, and $b>a>0$, prove:
$$\int_{0}^{+\infty}\frac{f(ax)-f(bx)}{x}dx=[f(0)-k]\ln(\frac ba)$$

My attempt:
This integral has two singularities, one at $x=0$ and the other at $x=+\infty$, so we'd better split it into two parts, each with only one singularity. Let $A>0$, and let $I$ denote the integral, then
$$\begin{align} I&=\Big(\int_{0}^{A}+\int_{A}^{+\infty}\Big)\Big(\frac{f(ax)-f(bx)}{x}dx\Big) \\&=\Big(\int_{0}^{A}+\int_{A}^{+\infty}\Big)\frac{f(ax)}{x}dx-\Big(\int_{0}^{A}+\int_{A}^{+\infty}\Big)\frac{f(bx)}{x}dx \\&=\Big(\int_{0}^{aA}+\int_{aA}^{+\infty}\Big)\frac{f(x)}{x}dx-\Big(\int_{0}^{bA}+\int_{bA}^{+\infty}\Big)\frac{f(x)}{x}dx \\&=\Big(\int_{0}^{aA}+\int_{bA}^{0}+\int_{aA}^{+\infty}+\int_{+\infty}^{bA}\Big)\frac{f(x)}{x}dx \end{align}$$ If I keep going and cancel these integral limits out (and I know it's probably not doable because infinity is involved here and may play tricks on us), I simply get $I=0$, of course that's not the desired result.
What's more, I hope to get the limit $k$ involved, but there seems no obvious way to do so.
Can you help me? Any help or hint (if not too obscure) will be appreciated. Best regards!


marked as duplicate by Julián Aguirre, Community Mar 27 '15 at 16:47

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


Given $t > s > 0$,

$$\int_s^t \frac{f(ax) - f(bx)}{x}\, dx = \int_s^t \frac{f(ax)}{x}\, dx - \int_s^t \frac{f(bx)}{x}\, dx = \int_{as}^{at} \frac{f(x)}{x}\, dx - \int_{bs}^{bt} \frac{f(x)}{x}\, dx.$$


\begin{align}\int_{as}^{at} \frac{f(x)}{x}\, dx - \int_{bs}^{bt} \frac{f(x)}{x}\, dx &= \int_{as}^{bs} \frac{f(x)}{x}\, dx + \int_{bs}^{at} \frac{f(x)}{x}\, dx - \int_{bs}^{bt} \frac{f(x)}{x}\, dx \\ & = \int_{as}^{bs} \frac{f(x)}{x}\, dx - \int_{at}^{bt} \frac{f(x)}{x}\, dx\\ & = \int_a^b \frac{f(sx)}{x}\, dx - \int_a^b \frac{f(tx)}{x}\, dx\\ & = \int_a^b \frac{f(sx) - f(tx)}{x}\, dx, \end{align}

we have

$$\int_s^t \frac{f(ax) - f(bx)}{x}\, dx = \int_a^b \frac{f(sx) - f(tx)}{x}\, dx.$$


\begin{align}&\int_s^t \frac{f(ax) - f(bx)}{x}\, dx - [f(0) - k]\ln \frac{b}{a}\\ & =\int_a^b \frac{f(sx) - f(tx)}{x}\, dx - \int_a^b \frac{f(0) - k}{x}\, dx\\ & = \int_a^b \frac{f(sx) - f(0)}{x}\, dx - \int_a^b \frac{f(tx) - k}{x}\, dx \tag{*}. \end{align}

Now since $f\in C[0, +\infty)$ and $\lim_{x\to +\infty} f(x)$ exists, $f$ is uniformly continuous on $[0, +\infty)$. Using uniform continuity of $f$, show that (*) tends to $0$ as $s \to 0^+$ and $t \to +\infty$.

  • $\begingroup$ Thank you! I have read to your last step. But I'm not sure what property of uniform continuity should I use here? Can you explain a bit further? $\endgroup$ – Vim Mar 27 '15 at 16:57
  • $\begingroup$ @Vim Uniform continuity permits the interchange of the limit and the integrals. $\endgroup$ – Mark Viola Mar 27 '15 at 17:03
  • $\begingroup$ The second term I understand how it tends to zero. But I'm uncertain of the first term. $\endgroup$ – Vim Mar 27 '15 at 17:03
  • $\begingroup$ @Vim: Given $\varepsilon > 0$, choose a corresponding $\delta$ in the definition of uniform continuity of $f$. If $0 < s < \delta/\max\{|a|,|b|\}$, then $|sx| < \delta$ for all $x \in [a,b]$, whence $|f(sx) - f(0)| < \varepsilon$ for all $x \in [a,b]$. Then $$\left|\int_a^b \frac{f(sx) - f(0)}{x}\, dx\right| < \int_a^b \frac{\varepsilon}{x}\, dx = \varepsilon\ln\frac{b}{a}.$$ Since $\varepsilon$ was arbitrary, the first term tends to $0$ as $s\to 0^+$. $\endgroup$ – kobe Mar 27 '15 at 17:07
  • $\begingroup$ @kobe. Brilliant explanation and your answer ! Thank you again for the time you spent! $\endgroup$ – Vim Mar 27 '15 at 17:11

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