Let $f\in \mathcal{C}^1(\mathbb{R}^+,\mathbb{R})$ such that $\int_{0}^{\infty}f$ converges. I read that there exists necessarly $a>0$ such that $\int_{0}^{a}tf(t)dt=a^2f(a)$ It seems natural to study the function $x\mapsto \frac{1}{x}\int_0^x tf(t)dt$ but things still jot clear to me. Any help on this would be greatly appreciated. thanks in advance.

  • $\begingroup$ Two questions: $\mathbb{R}^+$ does not include $0$ for you, does it? And does the integral of $f$ converge in the sense of Lebesgue or improper Riemann integrals? $\endgroup$ Sep 30, 2022 at 14:23
  • $\begingroup$ @user23571113 Yes it does include 0 and the existence is in the sense of improper Riemann integrals $\endgroup$
    – user536450
    Sep 30, 2022 at 20:31

1 Answer 1


If $f$ is positive then the exercise can be proved in the following way.

Denote your function $x\mapsto \frac{1}{x}\int_0^x tf(t)\,dt$ by $g$.

First of all, notice that $\displaystyle\lim_{x\to 0^+}g(x) = tf(t)\big|_{t = 0} = 0$.

Secondly, let us show that $\displaystyle\lim_{x\to \infty}g(x) = 0$. Fix $\varepsilon > 0$ and let $N$ be such that $\int_N^{\infty}f(t) < \varepsilon$. Then we have $$ \limsup_{x\to\infty}g(x) = \limsup_{x\to\infty}\frac{\int_0^x tf(t)\,dt}{x} = \limsup_{x\to\infty}\frac{\int_0^N tf(t)\,dt + \int_N^x tf(t)\,dt}{x} \le \limsup_{x\to\infty}\frac{\int_0^N tf(t)\,dt + x\int_N^{\infty} tf(t)\,dt}{x}\le \varepsilon. $$

Therefore, By Rolle's theorem there exists $a > 0$ such that $g'(a) = 0$. It remains to notice that $$ g'(a) = \left(\frac{\int_0^a tf(t)\,dt}{a}\right)' = \frac{a^2f(a) - \int_0^a tf(t)\,dt}{a^2}.$$

UPD. For a non necessarily positive $f$ one can integrate by parts to notice that $$ g(x) = \frac{1}{x}\int_0^x tf(t)\,dt = \frac{1}{x}\left[tF(t)\big|_0^x - \int_0^x F(t)\,dt\right] = \frac{1}{x}\left[xF(x) - \int_0^x F(t)\,dt\right] = {\frac{1}{x}\int_0^x\left[ F(x) - F(t)\right]\,dt,} $$ where $F$ is antiderivative of $f$.

Now $\displaystyle\lim_{x\to\infty}g(x) = 0$ can be proved using the same idea as in the positive case.

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    $\begingroup$ Nice idea to use the Rolle theorem ! (+). I think the limit $\lim_{x\to\infty} g(x)=0$ can be proved for $f$ not necessarily positive, applying the formula $$g(x)={1\over x}\int\limits_0^x[F(x)-F(t)]\,dt$$ where $$F(x)=\int\limits_0^x f(t)\,dt$$ $\endgroup$ Oct 3, 2022 at 19:27
  • $\begingroup$ Oh! It really does help, thank you! $\endgroup$ Oct 4, 2022 at 6:37
  • $\begingroup$ If you like you can include that in your nice solution. The formula is proved by applying integration by parts to $\int tf(t)\,dt$ $\endgroup$ Oct 4, 2022 at 9:48
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    $\begingroup$ Not $\limsup g(x)$ but $\lim g(x)$ $\endgroup$ Oct 5, 2022 at 12:39

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