This is a detail from a proof in Ingham's Distribution of Prime Numbers, p. 91-92. He forms a Dirichlet integral and assumes for contradiction that the numerator $c(x)\geq 0.$ Then he bounds $f(s)$ in order to find an inequality for a constant $k$ that is inherently contradictory. My question is about the details of the complex algebra in the penultimate step. Note, $k\neq K.$
Assume RH. $k$ is a positive constant.
$$c(x) = \frac{\psi(x)- x + k x^{1/2}}{x},$$
then for $\sigma > 1$ we have $f(s)=\int_1^{\infty}\frac{c(x)}{x^s}dx$ in which
$$f(s) = -\frac{1\zeta'}{s\zeta}(s)-\frac{1}{s-1}+\frac{k}{s-1/2}$$
For $\sigma > 1/2$ there are no singularities of $f$ on the real line, and so the abscissa of convergence is $1/2.$ Suppose (for contradiction) $c(x) \geq 0$ for all $x>X (>1).$
Leaving out some detail we get
$ | f (\sigma+ t i) | \leq \int_1^X \frac{ | c(x |)}{x^{\sigma}}dx +\int_X^{\infty} \frac{c(x)}{x^{\sigma}} dx = ...\leq K+ f(\sigma),$
with K independent of $\sigma, t.$
Take $t = \gamma_1,$ where $1/2 + \gamma_1 i$ is the zero with the least positive $\gamma,$ multiply both sides by $(\sigma -1/2),$ and let $\sigma \to (1/2 + 0);$ this gives
$$ \frac{m_1}{|1/2 + \gamma_1|}\leq k$$
where $m_1$ is the multiplicity of the zero $1/2 + \gamma_1.$
The part in yellow is verbatim. Can someone explain the details of this step?
Let $z_1 = \frac{1}{2} + \gamma_1 i$ be the zero of $\zeta$ in question. We can then write
$$\zeta(s) = (s-z_1)^{m_1}\cdot h(s)$$
with a holomorphic function $h$ without zeros in a neighbourhood $U$ of $z_1$. On $U$, we thus have
$$\frac{\zeta'(s)}{\zeta(s)} = \frac{m_1(s-z_1)^{m_1-1}\cdot h(s) + (s-z_1)^{m_1}\cdot h'(s)}{(s-z_1)^{m_1}\cdot h(s)} = \frac{m_1}{s-z_1} + \frac{h'(s)}{h(s)}.$$
For $s = \sigma + \gamma_1 i$, we have $\sigma - \frac{1}{2} = s - z_1$, and so
$$\bigl(\sigma - \tfrac{1}{2}\bigr)\frac{\zeta'(s)}{s\zeta(s)} = \frac{m_1}{s} + (s-z_1)\frac{h'(s)}{sh(s)}.$$
Since $\frac{h'(s)}{sh(s)}$ is bounded near $z_1$, it follows that
$$\lim_{\sigma\to \frac{1}{2}+0} \bigl(\sigma - \tfrac{1}{2}\bigr)\frac{\zeta'(s)}{s\zeta(s)} = \frac{m_1}{z_1}.$$
The parts $\frac{1}{s-1}$ and $\frac{k}{s-1/2}$ are also bounded near $z_1$, so
$$\lim_{\sigma\to \frac{1}{2}+0} \bigl(\sigma - \tfrac{1}{2}\bigr) f(s) = - \frac{m_1}{\frac{1}{2} + \gamma_1 i}.$$
On the right hand side of the inequality, we note that $\frac{\zeta'(s)}{s\zeta(s)}$ and $\frac{1}{s-1}$ are bounded near $\frac{1}{2}$, so we have
$$\lim_{\sigma\to \frac{1}{2} + 0} \bigl(\sigma - \tfrac{1}{2}\bigr)\bigl(K + f(\sigma)\bigr) = k.$$