Assume $\int_0^\infty f(x)dx$ converges. Prove there exists $\xi\in[1,\infty)$ such that $\int_1^\infty \frac{f(x)}x dx=\int_1^\xi f(x)dx$.

Question

Assume $\int_0^\infty f(x)dx$ converges. Prove there exists $\xi\in[1,\infty)$ such that $$\int_1^\infty \frac{f(x)}x dx=\int_1^\xi f(x)dx.$$

I used various mean value theorems but couldn't get the desired result. Another attempt beyond these is:

Let $\displaystyle F(x)=\int_1^x f(t)dt$, then \begin{align*} \int_1^\infty \frac{f(x)}x dx=\int_1^\infty \frac1x ~dF(x)&=\left.\frac {F(x)}x\right|_1^\infty-\int_1^\infty F(x) ~d\frac1x\\ &=0-0+\int_1^\infty \frac {F(x)}{x^2}dx\\ &=\int_1^\infty \frac {F(x)}{x^2}dx. \end{align*} How to analyze next? Any help would be appreciated!

Answer 1

With your excellent notations denote by $M=\sup F(x)$ and $m=\inf F(x)$ which are finite since $F$ is continuous and $L=\lim _{x\to \infty} F(x)$ exists and is finite. Then since $\int_1^{\infty}\frac{1}{x^2}dx=1$ we have $$m\leq I=\int_1^{\infty}\frac{F(x)}{x^2}dx\leq M.$$ If $I=m$ or $M$ then $F(x)\equiv m$ or $M$. If not there exists $x_0$ and $y_0$ such that $$m<F(x_0)<I<F(y_0)<M$$ and one applies the intermediate value theorem to the continuous function $F$ restricted to the interval $[x_0,y_0]$ or $[y_0,x_0]$ to claim the existence of $\xi$ such that $F(\xi)=I.$

Answer 2

With the substitution $x = 1/t$ the problem is equivalent to showing that there is a $\eta \in (0, 1)$ such that $$ \int_0^1 t \cdot \frac{f(1/t)}{t^2} \, dt = \int_\eta^1 \frac{f(1/t)}{t^2} \,dt \, , $$ and that is exactly the statement of the second mean value theorem for definite integrals for the increasing function $G(t) = t$ and the integrable function $\varphi(t) = f(1/t)/t^2$.