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

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!

• Do we know that $f$ is positive? If the answer is yes, we can use the intermediate value theorem. Sep 3 at 7:50
• @1mdlrjcmed No, the only assumption is $\int_0^\infty f(x)dx$ converges. Sep 3 at 7:53
• @1mdlrjcmed Of course, $f$ must be integrable. Sep 3 at 8:03

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 $$m and one applies the intermediate value theorem to the continuous function $$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.$$
With the substitution $$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$$.