Convergence of improper integrals and logs

[JG89]

Homework Statement

If $$\int_a^{\infty} \frac{f(x)}{x} dx$$ converges for any postiive value of a, and if f(x) tends to a limit L as x approaches 0, show that $$\int_0^{\infty} \frac{f(\alpha x) - f(\beta x)}{x} dx$$ converges for alpha and beta positive and has the value L*ln(beta/alpha).

Homework Equations

The Attempt at a Solution

First we show it converges. Let $$g(x) = \frac{\alpha f(x)}{x}$$. We know that the integral from a to infinity of f(x)/x converges, and so we set this equal to M and so $$\int_a^{\infty} g(x) dx = \alpha M$$. Now, $$\int_a^{\infty} g(x) dx = \lim_{\epsilon \rightarrow \infty} \int_a^{\epsilon} g(x) dx = \lim_{\epsilon \rightarrow \infty} \int_{\frac{a}{\alpha}}^{\frac{\epsilon}{\alpha}} g(\alpha x)} dx = \lim_{\epsilon \rightarrow \infty} \int_{\frac{a}{\alpha}}^{\frac{\epsilon}{\alpha}} \frac{f(\alpha x)}{x} dx$$ Since as epsilon tends to infinity so does epsilon/beta, we have $$\int_{\frac{a}{\alpha}}^{\infty} \frac{f(\alpha x)}{x} dx = \alpha M$$. Now we need only show that $$\int_0^{\frac{a}{\alpha}} \frac{f(\alpha x)}{x} dx$$ exists. Since f(alpha x)/x converges from a/alpha to infinity for any positive a (since alpha is also positive then so is a/alpha), and has no infinite discontinuity at x = 0, then the integral of f(alpha x)/x from 0 to a/alpha exists. Let this value be equal to N_1. Then we have $$\int_0^{\infty} \frac{f(\alpha x)}{x} dx = \alpha M + N_1$$ (alphaM + N_1). Through a similar argument, we can show that the integral from 0 to infinity of f(beta x)/x exists and has the value $$\beta M + N_2$$ (betaM + N_2).

I don't know how to show that $$\int_0^{\infty} \frac{f(\alpha x) - f(\beta x)}{x} dx = L ln(\frac{\beta}{\alpha})$$ though. Any hints? Please only very small hints

Edit: What I wrote in the brackets towards the end is what I really meant to write. The latex wouldn't let me edit it.

 

[JG89]

Oh shoot, I meant to post this in the Calculus and Beyond forum. If you are a mod, can you please move it over there?

 

[JG89]

Bump...