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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!

Vim
  • 13,640

1 Answers1

1

Given $t > s > 0$,

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

Since

\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

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

Therefore

\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 $t \to +\infty$.

kobe
  • 41,901
  • 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? – Vim Mar 27 '15 at 16:57
  • @Vim Uniform continuity permits the interchange of the limit and the integrals. – Mark Viola Mar 27 '15 at 17:03
  • The second term I understand how it tends to zero. But I'm uncertain of the first term. – Vim Mar 27 '15 at 17:03
  • @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 $s\to 0^+$. – kobe Mar 27 '15 at 17:07
  • @kobe. Brilliant explanation and your answer ! Thank you again for the time you spent! – Vim Mar 27 '15 at 17:11
  • @Vim no problem. Just making a small note here but you can change $\max{|a|,|b|}$ to just $b$ -- after all, $b > a > 0$. :-) – kobe Mar 27 '15 at 23:02