TITLE: a question on improper integrals
QUESTION [1 upvotes]: 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.

REPLY [2 votes]: 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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