On Perron's formula with remainder

Question

I am following Tenenbaum's Introduction to Analytic and Probabilistic Number Theory to prove Perron's formula with remainder. Particularly, in section II.2.1 the book states that

$$ \left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right|\le{x^\kappa\over T\log x} $$

where $k>\kappa>0$ and $x>1$, but when I perform integration by parts, I get

$$ \int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds={x^\kappa\over\log x}\left({x^{iT}\over\kappa+iT}-{x^{-k+iT}\over\kappa-k+iT}\right)+{1\over\log x}\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s^2}\mathrm ds $$

Now taking absolute value on both side, the equation becomes

$$ \left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right|\le{x^\kappa\over T\log x}+{1\over|\kappa-k+iT|}+{1\over\log x}\int_{\kappa-k+iT}^{\kappa+iT}{\mathrm ds\over|s|^2} $$

Apparently, my bound is much more loose than that of the book, and I wonder what I can do more to make the bounds more tight.

Answer 1

Using Greg Martin's comment, I get that

$$ \begin{aligned} \left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right| &=\left|\int_{\kappa-k}^\kappa{x^{t+iT}\over t+iT}\mathrm dt\right| \\ &\le\int_{\kappa-k}^\kappa{x^t\over|t+iT|}\mathrm dt \end{aligned} $$

By the property of right triangle, we see that $|t+iT|\ge T$, so

$$ \left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right|\le\frac1T\left|\int_{\kappa-k}^\kappa x^t\mathrm dt\right|\le{x^\kappa-x^{\kappa-k}\over T|\log x|} $$

Since $x>1$ and $\kappa>k$, we have $x^{\kappa-k}<x^\kappa$. Consequently, we obtain Tenenbaum's bound that

$$ \left|\int_{\kappa-k+iT}^{\kappa+iT}{x^s\over s}\mathrm ds\right|\le{x^\kappa\over T\log x} $$