Evaluate the improper integral $\int_{0}^{\infty}{f(x)-f(2x)\over x}dx$, where $\lim_{x \to \infty} f(x) = L$ 
Find $$\int_{0}^{\infty}{f(x)-f(2x)\over x}\, \mathrm{d}x$$ if $f\in C([0,\infty])$ and $\lim\limits_{x\to \infty}{f(x)=L}$.

I tried denoting $\displaystyle \int{f(x)\over x}dx=F(x)$, but I don't know what to do with $F(\infty)-F(0)$. I also tried somehow to use l'Hopital's rule but don't seem to come by a plausible justification for it, nor do I arrive at a clear expression. I would really appreciate your help in this. 
 A: If one assumes that $f$ is sufficiently well-behaved (and in particular, differentiable), one can solve this by differentiating under the integral sign (this is sometimes called Feynman's trick).
Define
$$I(a) := \int_0^{\infty} \frac{f(x) - f(a x)}{x} \,dx.$$
Differentiating with respect to $a$ (and justifying passing the derivative sign through the improper integral) gives (for $a > 0)$) that
\begin{align}
I'(a)
&= \frac{d}{da} \int_0^{\infty} \frac{f(x) - f(a x)}{x} \,dx \\
&= \int_0^{\infty} \frac{d}{da} \left(\frac{f(x) - f(a x)}{x}\right) dx \\
&= -\int_0^{\infty} f'(ax) \,dx \\
&= -\left.\frac{1}{a} f(ax)\right\vert_0^{\infty} \\
&= \frac{1}{a}(f(0) - L) .
\end{align}
By the F.T.C., integrating w.r.t. $a$ gives
$$I(2) - I(1) = \int_1^2 \frac{1}{a}(f(0) = L)\, da = (f(0) - L) \log a \vert_1^2 = (f(0) - L) \log 2.$$
On the other hand, $I(2)$ is the integral we want to evaluate and
$$I(1) = \int_0^{\infty} \frac{f(x) - f(x)}{x} dx = 0.$$
Putting this all together gives, as desired,
$$\color{#bf0000}{\int_0^{\infty} \frac{f(x) - f(2 x)}{x} \,dx = (f(0) - L) \log 2}.$$

Here's a different way to present essentially the same argument: The integrand is a difference of the values of a particular expression at two points, which is precisely the form of one side of the F.T.C. Reverse-engineering gives
$$\frac{f(x) - f(2 x)}{x} = -\int_1^2 f'(y x) \,dy.$$
Justifying a reversal of the order of integration again gives
\begin{align}
\color{#bf0000}{\int_0^{\infty} \frac{f(x) - f(2 x)}{x} dx}
&=  \int_0^{\infty} \left[-\int_1^2 f'(y x) \,dy\right] dx \\
&= -\int_1^2 \int_0^{\infty} f'(y x) \,dx \,dy \\
&= -\int_1^2 \left.\frac{1}{y} f(y x)\right\vert_0^{\infty} dy \\
&= -(L - f(0)) \int_1^2 \frac{dy}{y} \\
&\color{#bf0000}{= (f(0) - L) \log 2} .
\end{align}
More generally, the same argument gives (for $f$ satisfying the same conditions as in the question) that
$$\int_0^{\infty} \frac{f(bx) - f(ax)}{x} dx = (L - f(0)) \log \frac{b}{a}.$$
A: There are possible singularities at the origin and at infinity; by definition the integral is the limit of $\int_\delta^A$ as $\delta\to0$ and $A\to\infty$. You have
$$\begin{eqnarray*}
\int_\delta^A\frac{f(t)}{t}-\int_\delta^A\frac{f(2t)}{t}&=\int_\delta^A\frac{f(t)}{t}-\int_{2\delta}^{2A}\frac{f(t)}t
\\&=\int_{\delta}^{2\delta}\frac{f(t)}t-\int_{A}^{2A}\frac{f(t)}t
\\&=\int_1^2\frac{f(\delta t)}{t}-\int_1^2\frac{f(At)}{t}
\\&\to(f(0)-L)\log(2),
\end{eqnarray*}$$
by uniform convergence on $[1,2]$.
