When proving that $$\biggl|\delta(y)- \frac{1}{2 \pi i}\int_{c-iT}^{c+iT}\frac{y^{s}}{s}\,ds\biggr| = y^{c}\min\{1, \frac{1}{T\log|y|} y \neq 1\}, \quad \min\left\{1, \frac{c}{T}, y=1\right\}$$ Where $\delta(y) = 0,\; 0< y<1, \frac{1}{2}, y=1, 1 y > 1$
In the $\; 0<y <1 $ case we have that $\frac{y^{s}}{s} \rightarrow 0, \Re(s) \rightarrow \infty$, applying the Cauchy residue theorem we obtain the following integral: $$ \biggl\lvert\int_{c-iT}^{\infty+iT}\frac{y^{s}}{s}\,ds\biggr\lvert$$ Applying the triangle inequality we obtain
$$ \biggl|\int_{c-iT}^{\infty+iT}\frac{y^{s}}{s}\,ds\biggr| \leq \frac{1}{T}\int_{c}^{\infty}y^{\sigma}d \sigma$$
Where does the 1/T appear from when we apply the triangle inequality above
Thanks.
Note that if $s= x+iT$ then $|s| \ge T$ and so ${1 \over |s|} \le {1 \over T}$.