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Assume that $ \int_{1}^{+\infty} f(x) dx $ converges (as improper Riemann integral). Does there necessarily exist some $ \xi \in (1, +\infty) $ such that $$ \int_{1}^{+\infty} \frac{f(x)}{x} dx = \int_{1}^{\xi} f(x) dx ? $$

My Attempt :

If $f(x)$ is a positive function, then we can apply the second mean value theorem for integrals on the interval $[1, n]$. This guarantees the existence of $\xi_n \in (1, n)$ such that $$ \int_{1}^{n} \frac{f(x)}{x} dx = \int_{1}^{\xi_n} f(x) dx. $$ Furthermore, the sequence $\{\xi_n\}$ must be bounded. Otherwise, if it were unbounded, there would exist a subsequence $\xi_{n_k} \to +\infty$ such that $$ \int_{1}^{\infty} f(x) dx = \int_{1}^{\infty} \frac{f(x)}{x} dx, $$ which leads to a contradiction since $f(x) > 0$ implies $\frac{f(x)}{x} < f(x)$ for $x > 1$, making the equality impossible.

Difficulty: However, the problem does not actually require $f(x)$ to be a positive function, which is where I encounter difficulties.

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  • $\begingroup$ Riemann , and $f$ is not assumed to be continuous, I don't know whether integration by parts is valid without continuity. $\endgroup$ Commented Oct 24 at 16:11

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The function $F(x) = \int_1^x f(t) \, dt$ is continuous on $[1, +\infty)$ and has a limit for $x \to +\infty$, therefore $$ m = \inf \{ F(x) \mid x \ge 1 \} \, ,\\ M = \sup \{ F(x) \mid x \ge 1 \} \, . $$ both exist as (finite) real numbers.

Integration by parts (see for example Integration by parts for Riemann integrable fucntion) gives $$ \int_1^b \frac{f(x)}{x} \,dx = \frac{F(b)}{b} + \int_1^b \frac{F(x)}{x^2} \,dx $$ for any $b > 1$ and we can take the limit $b \to +\infty$: $$ \int_1^{+\infty} \frac{f(x)}{x} \,dx = \int_1^{+\infty} \frac{F(x)}{x^2} \,dx \, . $$ We have $$ m = \int_1^{+\infty} \frac{m}{x^2} \,dx \le \int_1^{+\infty} \frac{F(x)}{x^2} \,dx \le \int_1^{+\infty} \frac{M}{x^2} \,dx = M \, . $$ Equality on either side can only hold if $F$ is constant. In that case is $f$ identically zero and the desired identity holds trivially for any $\xi > 1$.

So we can assume that $$ m < \int_1^{+\infty} \frac{f(x)}{x} \,dx < M $$ and conclude that $$ \int_1^{+\infty} \frac{f(x)}{x} \,dx = F(\xi) $$ for some $\xi > 1$.

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    $\begingroup$ Thank you! The proof that $\xi > 1$ is ingenious, and I've learned a great deal from it. $\endgroup$ Commented Oct 24 at 16:54

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