Question 45 in Chapter 19 in Michael Spivak's book "Calculus" involving an improper integral This is Problem 45 in Chapter 19 in Michael Spivak's book "Calculus".


*(a) Suppose that $\frac {f(x)} x$ is integrable on every interval [a, b] for $0$ < a < b, and that $\lim_{x\to0}f(x)=A$ and $\lim_{x\to\infty}f(x)=B$. Prove that for all $\alpha$, $\beta$ > $0$ we have


$\int_0^\infty \frac {f(\alpha x) - f(\beta x)}{x}dx = (A-B)log(\frac \beta \alpha)$.
(b) Now suppose instead that $\int_0^\infty\frac{f(x)}xdx$ converges for all $a>0$ and that $\lim_{x\to0}f(x)=A$. Prove that
$\int_0^\infty \frac {f(\alpha x) - f(\beta x)}{x}dx=Alog(\frac \beta \alpha)$.
 A: HINT:
Since $\frac{f(x)}{x}$ is an arbitrary integrable function, it can be approximated in the $\ell^1$ norm by a compactly supported smooth function $\frac{g(x)}{x}$.  So, for all $\epsilon>0$, 
$$\int_a^b \left|\frac{f(x)}{x}-\frac{g(x)}{x}\right|\,dx<\epsilon$$
Then use
$$\int_{x_1}^{x_2} \int_\alpha^\beta g'(xy)\,dy\,dx=\int_{x_1}^{x_2}\int_\alpha^\beta \frac1x\frac{\partial g(xy)}{\partial y}\,dy\,dx= \int_\alpha^\beta \int_{x_1}^{x_2} \frac1y\frac{\partial g(xy)}{\partial x}\,dx\,dy$$
A: Another approach: Observe that 
$$
\int_a^b {f(\alpha x)-f(\beta x)\over x}\,dx =\int_{\alpha a}^{\beta a}{f(t)\over t}\,dt - \int_{\alpha b}^{\beta b}{f(t)\over t}\,dt.
$$
The first integral on the right side of the above display can be written as
$$
\int_{\alpha a}^{\beta a}{f(t)-A\over t}\,dt+A\log(\beta/\alpha),
$$
and the integral in this display converges to $0$ as $a\to 0$. Likewise $\int_{\alpha b}^{\beta b}{f(t)\over t}\,dt=B\log(\beta/\alpha)+o(1)$ as $b\to\infty$. Taken together these show that
$$
\lim_{a\to 0,b\to\infty}\int_a^b{f(\alpha x)-f(\beta x)\over x}\,dx=(A-B)\log(\beta/\alpha).
$$
This is (a). Similar reasoning works for part (b).
