And so $$\int_1^{1/x} \frac{\{y\}}{y^2}\,\mathrm{d}y
= \left(
\int_1^{\lfloor1/x\rfloor}
+
\int_{\lfloor1/x\rfloor}^{1/x} \right) \frac{\{y\}}{y^2}\,\mathrm{d}y
=
\left[ \sum_{n=1}^{\lfloor 1/x \rfloor-1} \int_{n}^{n+1} \frac{y-n}{y^2}\,\mathrm{d}y\right ]+\int_{\lfloor1/x\rfloor}^{1/x}\frac{y-\lfloor 1/x\rfloor}{y^2}\,\mathrm{d}y
$$