3

Let f be a positive continious function such that $\displaystyle \int_0^\infty f(x)dx=\infty$. Prove that $$\int_1^\infty\frac{f(x)}{\int_0^x f(t)dt}dx=\infty$$

2 Answers2

5

Let $F(x):=\int_0^xf(t)dt$. Then $F$ is positive, $C^1$, and $\displaystyle\lim_{A\rightarrow +\infty}F(A)=\int_0^{+\infty}f(t)dt=+\infty$. Hence $$ \int_1^A\frac{f(x)}{\int_0^xf(t)dt}dx=\int_1^A\frac{F'(x)}{F(x)}dx=\log F(x)\Big|_1^A=\log F(A)-\log F(1)\longrightarrow +\infty $$ when $A$ tends to $+\infty$.

Julien
  • 44,791
2

For any $x$, since $\displaystyle \int_0^{\infty} f(t) dt$ diverges, there exists $y>x$ such that $$\int_0^y f(t) dt \geq 2 \int_0^x f(t) dt$$ Hence, we have $$\overbrace{\int_x^y \dfrac{f(s)}{\displaystyle \int_0^s f(t) dt} ds \geq \int_x^y \dfrac{f(s) ds}{\displaystyle \int_0^y f(t) dt}}^{\because\text{Integrating a positive fn in denominator over a larger set}} = \dfrac{\displaystyle \int_x^y f(s) ds}{\displaystyle \int_0^y f(t) dt} = 1 - \dfrac{\displaystyle \int_0^x f(s) ds}{\displaystyle \int_0^y f(t) dt} \geq \dfrac12$$ Now conclude from this.

PS: This is adapted from Did's answer to this question.

Community
  • 1
  • why $\int_x^y \frac {f(s)ds}{\int_0^y f(t)dt}=\frac{\int_x^y f(s)ds}{\int_0^y f(t)dt}$? –  May 12 '13 at 03:11
  • 1
    @theOrthanormalBeginner Because the denominator is a contant independent of $s$. –  May 12 '13 at 03:11
  • thanks. pardon for illiteracy but if the integral is bigger than $\frac {1}{2}$ for every $y>x \in [1,\infty)$ how does it solve the problem? –  May 12 '13 at 03:20
  • 1
    @theOrthanormalBeginner If the integral were to converge, then given any small $\epsilon > 0$, there exists a $x$ such that $\int_x^{\infty} f(s) dx < \epsilon$. However, we have now shown that for any $x$, we have $\int_x^{\infty} f(s) ds \geq \int_x^y f(s) ds \geq \dfrac12$, making it impossible to make it arbitrarily small. –  May 12 '13 at 03:23