This is from the book A Course in Analytic Number Theory from Marius Overholt involving the sawtooth function:
$$S(x)=x-\lfloor x \rfloor-{1\over{2}}$$
Proposition 1.6.(page 11)
If $1 \leq a\leq b$ the estimate
$$\left | \int_a^b{S(x)\over{x^s}}dx\right|\leq \left( {1\over{8}} +{\left|s\right|\over{16\sigma}} \right)a^{-\sigma}$$ holds for any complex number $s=\sigma+it$ with positive real part $\sigma$
Proof (taken from the book): If $$\beta(x)={1\over{16}}+\int_0^xS(u)du$$ then $\left|\beta(x)\right|\leq{1\over{16}}$. Integration by parts gives:
$$\left|\int_a^b{S(x)\over{x^s}}dx\right|=\left| {\beta(x)\over{x^s}}\Bigl|_a^b-\int_a^b(-s){\beta(x)\over{x^{s+1}}}dx \right| \leq2.{1/16\over{a^{\sigma}}} +\int_a^b\left|s\right|{1/16\over {x^{\sigma+1}}}={1\over{8a^{\sigma}}}-{\left|s\right|\over{16\sigma}}{1\over{x^{\sigma}}}\Bigl|_a^b \leq {1\over{8a^{\sigma}}}+{\left|s\right|\over{16\sigma}}{1\over{a^{\sigma}}}$$ The integration by parts part seems baffling to me. Could someone smart explain the equality: $$\left|\int_a^b{S(x)\over{x^s}}dx\right|=\left| {\beta(x)\over{x^s}}\Bigl|_a^b-\int_a^b(-s){\beta(x)\over{x^{s+1}}}dx \right| $$
One may recall that, integrating by parts, $$ \int_a^b u(x)\cdot v'(x)\:dx =\left[ u(x)\frac{}{} v(x)\right]_a^b-\int_a^b u'(x)\cdot v(x)\:dx $$ then applying it with $$ u(x):=\frac1{x^s},\qquad u'(x)=-\frac{s}{x^{s+1}}, $$$$ v'(x):=S(x)=\beta'(x),\qquad v(x)=\beta(x)={1\over{16}}+\int_0^xS(u)du $$ one gets $$ \int_a^b \frac{S(x)}{x^s}\:dx =\left[ \frac1{x^s}\cdot \beta(x)\right]_a^b-\int_a^b(-s){\beta(x)\over{x^{s+1}}}\:dx $$ giving the desired equality with absolute values.