Evaluating the value of $\int_0^{\infty} \frac {f(x)-f(2x)}{x} dx$

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 doesnt converge.

Obviously my line of reasoning here is flawed somewhere but I cant see where.

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

• This is just a special case of Frullani's integral. – projectilemotion Dec 6 '18 at 16:56
• If $f(x) = L \in \mathbb{R}$ then $f(x) - f(2x) = 0.$ – Digitalis Dec 6 '18 at 16:57
• "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 $N$. – Robert Israel Dec 6 '18 at 18:16
• I've tried changing the endpoints and tried to see if it satisfies some sort of criterion like Dirichlet, yet with no success. – dllegend Dec 6 '18 at 18:47
• 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. – 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 $$a=2$$.

• I haven't been told that $f$ is differentiable though. – 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 $$0 < a < 1$$.

• 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? – dllegend Dec 8 '18 at 10:46