A question related to a convergence of an integral

Question

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?

Answer 1

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*}

Answer 2

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.

Answer 3

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.