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 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.
 A: 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
$0 < a < 1$.
A: 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 $a=2$.
