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## Homework Statement

If [tex] \int_a^{\infty} \frac{f(x)}{x} dx [/tex] converges for any postiive value of a, and if f(x) tends to a limit L as x approaches 0, show that [tex] \int_0^{\infty} \frac{f(\alpha x) - f(\beta x)}{x} dx [/tex] 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 [tex] g(x) = \frac{\alpha f(x)}{x} [/tex]. We know that the integral from a to infinity of f(x)/x converges, and so we set this equal to M and so [tex] \int_a^{\infty} g(x) dx = \alpha M [/tex]. Now, [tex]\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[/tex] Since as epsilon tends to infinity so does epsilon/beta, we have [tex] \int_{\frac{a}{\alpha}}^{\infty} \frac{f(\alpha x)}{x} dx = \alpha M [/tex]. Now we need only show that [tex] \int_0^{\frac{a}{\alpha}} \frac{f(\alpha x)}{x} dx [/tex] 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 [tex] \int_0^{\infty} \frac{f(\alpha x)}{x} dx = \alpha M + N_1[/tex] (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 [tex] \beta M + N_2 [/tex] (betaM + N_2).

I don't know how to show that [tex]\int_0^{\infty} \frac{f(\alpha x) - f(\beta x)}{x} dx = L ln(\frac{\beta}{\alpha}) [/tex] 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.

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