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. 

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 $[1,2]$.
