# a question on improper integrals

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.

• 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? Sep 30, 2022 at 14:23
• @user23571113 Yes it does include 0 and the existence is in the sense of improper Riemann integrals Sep 30, 2022 at 20:31

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.

• 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 $$F(x)=\int\limits_0^x f(t)\,dt$$ Oct 3, 2022 at 19:27
• Oh! It really does help, thank you! Oct 4, 2022 at 6:37
• If you like you can include that in your nice solution. The formula is proved by applying integration by parts to $\int tf(t)\,dt$ Oct 4, 2022 at 9:48
• Not $\limsup g(x)$ but $\lim g(x)$ Oct 5, 2022 at 12:39