A question related to a convergence of an integral Let $f$ be $f\in C([0,\infty ])$, such that
$\lim_{x \to \infty} f(x) = L $.
Calculate
$$ \int _{0}^\infty \frac{f(x)-f(2x)}{x}dx $$
Help?
 A: The answer from @Sami Ben Romdhane is incomplete since it contains 
$$ \int_1^2\frac{f(x)}{x}dx. $$
Here is the complete answer.
Let $0<A<2A<B$ and define
\begin{eqnarray*}
 g(A,B)&=&\int_A^B\frac{f(x)-f(2x)}{x}dx=\int_A^B\frac{f(x)}{x}dx-\int_A^B\frac{f(2x)}{x}dx\\
&=&\int_A^B\frac{f(x)}{x}dx-\int_{2A}^{2B}\frac{f(x)}{x}dx\\
&=&\int_A^{2A}\frac{f(x)}{x}dx-\int_{B}^{2B}\frac{f(x)}{x}dx.
\end{eqnarray*}
Then by the Integral Mean Value Theorem in $[A, 2A]$ and $[B, 2B]$, we have
$$ \int_A^{2A}\frac{f(x)}{x}dx=f(c_1)\int_A^{2A}\frac{1}{x}dx=f(c_1)\ln 2, \int_B^{2B}\frac{f(x)}{x}dx=f(c_2)\int_B^{2B}\frac{1}{x}dx=f(c_2)\ln 2 $$
where $A<c_1<2A, B<c_2<2B$. Thus
$$ g(A,B)=(f(c_1)-f(c_2))\ln2. $$
Therefore
\begin{eqnarray*}
\int_0^\infty\frac{f(x)-f(2x)}{x}dx&=&\lim_{A\to 0^+,B\to\infty}g(A,B)\\
&=&\lim_{A\to 0^+,B\to\infty}(f(c_1)-f(c_2))\ln2\\
&=&(f(0)-L)\ln 2.
\end{eqnarray*}
A: By an obvious change variable we have
$$\int_1^A\frac{f(x)-f(2x)}{x}dx=\int_1^A\frac{f(x)}{x}dx-\int_2^{2A}\frac{f(x)}{x}dx=\int_1^2\frac{f(x)}{x}dx-\int_A^{2A}\frac{f(x)}{x}dx$$
Now since $\displaystyle\lim_{x\to\infty}f(x)=L$ we prove easily that
$$\lim_{A\to\infty}\int_A^{2A}\frac{f(x)}{x}dx=L\log2\tag{*}$$
so
$$\int_1^\infty\frac{f(x)-f(2x)}{x}dx=\int_1^2\frac{f(x)}{x}dx-L\log2$$
Edit Here I explain the equality $(*)$:
We have $\displaystyle\lim_{x\to\infty}f(x)=L$ so for $\epsilon>0$ there's $A>0$ and if $x\ge A$ we have $|f(x)-L|\le\epsilon$, hence
$$\left|\int_A^{2A}\frac{f(x)}{x}dx-L\log2\right|=\left|\int_A^{2A}\frac{f(x)-L}{x}dx\right|\le\int_A^{2A}\frac{|f(x)-L|}{x}dx\le\epsilon\log2$$
so we have the desired equality.
A: Condition $\lim_{x\to\infty}f(x)=L$ is not enough. To see that the problem is underdetermined, take $f(x)=L+\frac{C}{1+x}$ with any prescribed constant $C$.
For such $f$, the improper integral 
$$\int_0^{\infty}\frac{f(x)-f(2x)}{x}dx=C\cdot\ln{2}$$
with an arbitrary given $C$ while $f(\infty)=L$. Combining this with the answer from @xpaul, we get a correct formulation of the OP problem: Given the values $f(0)$ and $f(\infty)$ of some function $f\in C[0,\infty]$, calculate the improper integral
$$\int_0^{\infty}\frac{f(x)-f(2x)}{x}dx$$
when it does converge.
