Let $f:[1, \infty) \rightarrow \mathbb{R}$ s.t $f$ is differentiable at $[1, \infty)$ and $\lim_{x \to \infty} f(x) = 0.$
Suppose that $f'$ is Riemann-integrable and the improper integral $\int_{1}^{\infty} xf'(x) \space dx$ converges absolutely.
Prove\disprove: $\int_{1}^{\infty} f(\xi) \space d\xi$ converges absolutely.
Since I've observed & proved the discrete version of this argument, I suspect it is true at its "continues" version as well (which is much more applicable and therefore, in my opinion, more interesting).
So far I couldn't find a counter-example as well as I struggle to think of a proof strategy for this.
I'd like to have a bit of guidance here, thanks.
For this kinf of question, it is often useful to think $f$ as the integral of its derivative (since the information you have is on the derivative). The convergence toward $0$ of $f$, you have, \begin{align*} \int_1^{+\infty} |f(x)| \, dx & = \int_1^{+\infty} \left|\int_x^{+\infty} f'(t) \, dt\right| \, dx\\ & \leqslant \int_1^{+\infty} \int_1^t |f'(t)| \, dx \, dt \textrm{ by Fubini-Tonelli,}\\ & = \int_1^{+\infty} (t - 1)|f'(t)| \, dt\\ & < +\infty. \end{align*} It proves that $\displaystyle \int_1^{+\infty} f(x) \, dx$ converges (absolutely by the way), and you conclude with an integration by parts.
Edit : other method without Fubini-Tonelli. Notice that if $x \mapsto xf(x)$ converges at infinity toward some limit $l$, then $l \neq 0 \Rightarrow f(x) \sim \frac{l}{x} \Rightarrow \displaystyle \int_1^\infty f(x) \, dx = \mathrm{sign}(l)\infty$. Therefore, $l$ must equal $0$. Let us prove directly that $xf(x) \rightarrow 0$.
We have for all $x$, still using $f(x) \rightarrow 0$, $$ |xf(x)| = x\left|\int_x^{+\infty} f'(t) \, dt\right| \leqslant \int_x^{+\infty} \frac{x}{t}|tf'(t)| \, dt \leqslant \int_x^{+\infty} |tf'(t)| \, dt \rightarrow 0, $$ because $t \mapsto tf'(t)$ is integrable. Therefore, $xf(x) \rightarrow 0$.