In Chapter 19 of Spivak's Calculus, Problem 46b asks the reader the following question:
Suppose that $\displaystyle \int_a^{\infty}\frac{f(x)}{x}dx$ converges for all $a \gt 0$ and that $\displaystyle \lim_{x \to 0}f(x)=A$. Prove that \begin{align}\int_0^{\infty}\frac{f(\alpha x)-f(\beta x)}{x}dx=A\log(\frac{\beta}{\alpha})\end{align}
Using some general integral manipulations /substitutions, I came up with the following expression:
$$\int_{\alpha \varepsilon}^{\beta\varepsilon}\frac{f(u)}{u}du-\int_{\alpha N}^{\beta N}\frac{f(u)}{u}du$$
I can see that $\int_{\alpha \varepsilon}^{\beta\varepsilon}\frac{f(u)}{u}du \to A\log(\frac{\beta}{\alpha})$ as $\varepsilon \to 0^+$…but I am having difficulties understanding why $\int_{\alpha N}^{\beta N}\frac{f(u)}{u}du \to 0$ as $N \to \infty$ (at least, I assume that is what I need to show). Any suggestions would be appreciated!
Steps:
\begin{align} \int_{\varepsilon}^{N}\frac{f(\alpha x)-f(\beta x)}{x}dx&=\int_{\varepsilon}^N\frac{f(\alpha x)}{x}dx-\int_{\varepsilon}^N\frac{f(\beta x)}{x}\\&=\int_{\alpha \varepsilon}^{\alpha N}\frac{f(u)}{u}du-\int_{\beta \varepsilon}^{\beta N}\frac{f(u)}{u}du \\
\end{align}
Drawing a picture, for example, with $\alpha \varepsilon \lt \beta \varepsilon \lt \alpha N \lt \beta N$ should help visualize why this next step is true:
$$=\int_{\alpha\varepsilon}^{\beta \varepsilon}\frac{f(u)}{u}du-\int_{\alpha N}^{\beta N}\frac{f(u)}{u}du$$
Best Answer
Just to summarize Robearz comments (and close out this post):
We need to show that $\displaystyle \lim_{N \to \infty} \int_{\alpha N}^{\beta N}\frac{f(u)}{u}= 0$.
Note that $\displaystyle \int_{\alpha N}^{\beta N}\frac{f(u)}{u}=\int_a^{\beta N}\frac{f(u)}{u}-\int_a^{\alpha N}\frac{f(u)}{u}$ for an arbitrary $a \in (0,\infty)$.
Given that $\beta N \to \infty$ and $\alpha N \to \infty$ as $N \to \infty$, we have by assumption that:
$\displaystyle \lim_{N\to \infty}\int_a^{\beta N}\frac{f(u)}{u}=L_a$ and $\displaystyle \lim_{N\to \infty}\int_a^{\alpha N}\frac{f(u)}{u}=L_a$. Therefore, we have that:
\begin{align}\lim_{N\to \infty} \int_{\alpha N}^{\beta N}\frac{f(u)}{u}&=\lim_{N\to \infty}\int_a^{\beta N}\frac{f(u)}{u}-\int_a^{\alpha N}\frac{f(u)}{u}\\&=\lim_{N\to \infty}\int_a^{\beta N}\frac{f(u)}{u}-\lim_{N\to \infty}\int_a^{\alpha N}\frac{f(u)}{u}\\&=L_a-L_a=0 \end{align}