This is in reference to baby Rudin's Ch 6 exer 16:
He defines there $\zeta(s)=\sum_1^\infty\frac1{n^s}$ for $1<s<\infty$, and then asks to show that $$\zeta(s)=\frac s{s-1}-s\int_1^\infty \frac{x-[x]}{x^{s+1}}\,dx,$$ but he also asks to
show that $$\int_1^\infty \frac{x-[x]}{x^{s+1}}\,dx$$ converges for $0<s<1$ as well.
My try:
$$\begin{align} \int_i^{i+1}\frac{x-[x]}{x^{s+1}}\,dx &= \int_i^{i+1}\frac{x-i}{x^{s+1}}\,dx\\ &= \int_i^{i+1}\frac{x}{x^{s+1}}\,dx-i\int_i^{i+1}\frac{1}{x^{s+1}}\,dx\\ &=\frac{(i+1)^{1-s}}{1-s}-\frac{i^{1-s}}{1-s}-i\left(\frac1{si^s}-\frac1{s(i+1)^s}\right). \end{align}$$
How to show that $$\sum_1^\infty\left\{\frac{(i+1)^{1-s}}{1-s}-\frac{i^{1-s}}{1-s}-i\left(\frac1{si^s}-\frac1{s(i+1)^s}\right)\right\}$$ indeed converges for $0<s<1$?
The simple fact that $0\leq x-[x]\leq1$ immediately leads to absolute convergence: $$ \int_1^N\frac{x-[x]}{x^{s+1}}dx \leq \int_1^N\frac{dx}{x^{s+1}} =\frac1{sx^{s}}\Big|_N^1\leq\frac1s. $$