In Edwards' very nice book ``Riemann's zeta function'' the following integral comes up in section 1.14. Suppose $\beta = \sigma + i\tau$ with $\sigma > 0$. Suppose $x > 1$. Fix some real number $a > \sigma$ and let

$$I(\beta) := \int_{a-i\infty}^{a+i\infty} \frac{\log(1-\frac s{\beta})}{s^2} x^s ds.$$

[Here, the logarithm is taken as $\log(s-\beta)-\log(-\beta)$ for $\tau \ne 0$ using the branch of logarithm defined away from the negative real axis, which is real on the positive real axis.] It's easy to see that this integral is absolutely convergent (note that $|x^s| = x^a$). Now Edwards claims that $\lim_{\tau \to \infty} I(\beta) = 0$ because ``it is not difficult to show, using the Lebesgue bounded convergence theorem ... that the limit of this integral is the integral of the limit, namely zero''. I didn't find a good way to bound $\log(1-\frac s{\beta})$ for fixed $s$. Is there a solution that avoids long and messy calculations?