I want to evaluate the value of $\int_0^{\infty} \frac {f(x)-f(2x)}{x} dx$ where $f\in C([0,\infty])$ and $lim_{x\to \infty} f(x)=L$

I narrowed down the problem to showing $\int_0^\infty \frac{f(x)}{x} dx$ convgeres.

If the above integral indeed converge we have :

$$0=\int_0^{\infty} \frac {f(x)}{x} dx-\int_0^{\infty} \frac {f(t)}{t} dt=\int_0^{\infty} \frac {f(x)}{x} dx-\int_0^{\infty} \frac {f(2x)}{x} dx=\int_0^{\infty} \frac {f(x)-f(2x)}{x} dx$$

After using the substitution $t=2x$.

However, I cant see why $\int_0^{\infty} \frac {f(x)}{x} dx$ must converge, We can take $f(x)=1$ satisfying all of the conditions above and the integral obviously doesn`t converge.

Obviously my line of reasoning here is flawed somewhere but I can`t see where.

I'll be glad if someone can help me solve this problem.

  • 5
    $\begingroup$ This is just a special case of Frullani's integral. $\endgroup$ – projectilemotion Dec 6 '18 at 16:56
  • $\begingroup$ If $f(x) = L \in \mathbb{R}$ then $f(x) - f(2x) = 0.$ $\endgroup$ – Digitalis Dec 6 '18 at 16:57
  • $\begingroup$ "If the above integral indeed converge..." But since it doesn't converge, you can't conclude anything. However, you might get somewhere by changing the endpoints from $0$ and $\infty$ to $\epsilon$ and $N$. $\endgroup$ – Robert Israel Dec 6 '18 at 18:16
  • $\begingroup$ I've tried changing the endpoints and tried to see if it satisfies some sort of criterion like Dirichlet, yet with no success. $\endgroup$ – dllegend Dec 6 '18 at 18:47
  • $\begingroup$ I've also noticed that with the limit, we can conclude that the function is uniformly continuous however, I fail to see how that helps me. $\endgroup$ – dllegend Dec 6 '18 at 18:48

Define $g(a):=\int_0^\infty\frac{f(x)-f(ax)}{x}dx$ so $g(1)=0$ and $g'(a)=-\int_0^\infty f'(ax)dx=\frac{f(0)-L}{a}$. If the numerator is finite, $g(a)=(f(0)-L)\ln|a|$. Now just substitute $a=2$.

  • $\begingroup$ I haven't been told that $f$ is differentiable though. $\endgroup$ – dllegend Dec 6 '18 at 18:44

Let $g_a(y):=\int_0^y\frac{f(x)-f(ax)}{x}dx $. We want to show that $\lim_{x \to \infty} f(x) = L$ implies that $\lim_{y \to \infty} g_a(y)$ exists.

For any $c > 0$ there is an $x(c)$ such that $x > x(c) \implies |f(x)-L| < c$.

Then, assuming that $a > 1$ and $y > x(c)$,

$\begin{array}\\ g_a(y) &=\int_0^y\frac{f(x)-f(ax)}{x}dx\\ &=\int_0^y\frac{f(x)-L-(f(ax)-L)}{x}dx\\ &=\int_0^y\frac{f(x)-L}{x}dx-\int_0^y\frac{f(ax)-L}{x}dx\\ &=\int_0^y\frac{f(x)-L}{x}dx-\int_0^{ay}\frac{f(x)-L}{x}dx\\ &=-\int_y^{ay}\frac{f(x)-L}{x}dx\\ \end{array} $

If $z > y$ then

$\begin{array}\\ g_a(z)-g_a(y) &=-\int_z^{az}\frac{f(x)-L}{x}dx+\int_y^{ay}\frac{f(x)-L}{x}dx\\ &=-\int_z^{az}\frac{f(x)}{x}dx+\int_z^{az}\frac{L}{x}dx+\int_y^{ay}\frac{f(x)}{x}dx-\int_y^{ay}\frac{L}{x}dx\\ &=-\int_z^{az}\frac{f(x)}{x}dx+(\ln(az)-\ln(z))+\int_y^{ay}\frac{f(x)}{x}dx-(\ln(ay)-\ln(y))\\ &=-\int_z^{az}\frac{f(x)}{x}dx+\ln(a)+\int_y^{ay}\frac{f(x)}{x}dx-\ln(a)\\ &=-\int_z^{az}\frac{f(x)}{x}dx+\int_y^{ay}\frac{f(x)}{x}dx\\ &=-\int_y^{ay}\frac{f(xz/y)}{x}dx+\int_y^{ay}\frac{f(x)}{x}dx\\ &=\int_y^{ay}\frac{f(x)-f(xz/y)}{x}dx\\ \text{so}\\ |g_a(z)-g_a(y)| &=|\int_y^{ay}\frac{f(x)-f(xz/y)}{x}dx|\\ &\le|\int_y^{ay}\frac{|f(x)-f(xz/y)|}{x}dx\\ &\le|\int_y^{ay}\frac{|2c|}{x}dx \qquad\text{for large enough } y\\ &=2c\int_y^{ay}\frac{1}{x}dx\\ &=2c(\ln(ay)-\ln(a))\\ &=2c\ln(a)\\ \end{array} $

and this can be made arbitrarily small by making $c$ small enough.

Note: I'm probably just rediscovering a standard proof but I though that it would be fun to work this through without looking anything up.

Note 2: This can be modified to show that $\int_0^{\infty} g(x)(f(x)-f(ax))dx$ exists where $g(x)$ is any function such that $\int_{y}^{ay}g(x)dx$ is bounded as $y \to \infty$. In this case, where $g(x) = \frac1{x}$, $\int_{y}^{ay}g(x)dx =\ln(a) $.

Note 3: The proof can be easily modified to work if $0 < a < 1$.

  • $\begingroup$ Could you clarify your reasoning for bounding the functions in the integral "for some large enough $y$"? Are you implying a use of a Cauchy Criterion? $\endgroup$ – dllegend Dec 8 '18 at 10:46

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