Counterterms in on-shell subtraction scheme I’m reading Schwartz’s QFT book, on page 332, I get the conditions for the on-shell subtraction scheme, but I don’t know how they imply equation (18.45):
$$\Sigma_2(m_p)=-\frac{\alpha}{2\pi}m_p(\frac{3}{2}m_p\ln \frac{\Lambda^2}{m_p^2} +\frac{3}{4}).\tag{18.45}$$
My attempt:
From (18.11) $$\Sigma_2(\not p)=-\frac{\alpha}{\pi}(m \ln \Lambda^2 -\frac{1}{4}\not p \ln \Lambda^2 +finite)\tag{18.11}$$ i get
$$\Sigma_2(m_p)=-\frac{\alpha}{\pi}(m_R ln \Lambda^2 -\frac{1}{4}m_pln \Lambda^2 +finite)$$
$$\Sigma_2(m_p)=-\frac{\alpha}{2\pi}(\frac{3}{4}m_pln \Lambda^2 +finite).$$
Substitute this into (18.44) $$\delta_m=\Sigma_2(m_p),\tag{18.44}$$
$$\delta_m=-\frac{\alpha}{2\pi}(\frac{3}{4}m_pln \Lambda^2 +finite).$$
 A: I would recommend rereading the chapter to make sure you really "get the conditions for the on-shell subtraction scheme". There are several issues here.
First of all, the sum of all the 1PI graphs is given by:
 $$\Sigma_R({p\!\!\!/})=\Sigma_2({p\!\!\!/})+\delta_2 {p\!\!\!/}-(\delta_m+\delta_2)m_R$$
For the on-shell subtraction scheme, we want the mass appearing in the Lagrangian to match the physical mass. That means setting $m_R=m_P$. To be completely clear, we now have:
 $$\Sigma_R({p\!\!\!/})=\Sigma_2({p\!\!\!/})+\delta_2 {p\!\!\!/}-(\delta_m+\delta_2)m_P$$
Keeping in mind that the mass that appears in the calculation of $\Sigma_2({p\!\!\!/})$ is now $m_P$.
The two conditions for this subtraction scheme are:
$$
\begin{align}
\Sigma_R(m_P) &=0, \\
\frac{d}{d{p\!\!\!/}}\Sigma_R({p\!\!\!/})\Big|_{{p\!\!\!/}=m_P} &=0,
\end{align}
$$
which lead to equations (18.43) and (18.44):
$$
\begin{align}
\delta_2&=-\frac{d}{d{p\!\!\!/}}\Sigma_2({p\!\!\!/})\Big|_{{p\!\!\!/}=m_P}, \\
\delta_m m_P&=\Sigma_2(m_P).
\end{align}
$$
Now to actually compute $\delta_2$ and $\delta_m$, we need to evaluate $\Sigma_2({p\!\!\!/})$. 
From Eqn. (18.11), setting $m\rightarrow m_P$ and evaluate the integral at ${p\!\!\!/}=m_P$ (which means $p^2=m_P^2$):
$$
\begin{align}
\Sigma_2(m_P)&=-\frac{\alpha}{2\pi} \intop_0^1 dx\;(2-x)m_P\ln\frac{x\Lambda^2}{(1-x)(m_P^2-m_P^2x)} \\
&=-\frac{\alpha}{2\pi} \intop_0^1 dx\;(2-x)m_P\Big(\ln\frac{\Lambda^2}{m_P^2}+\ln\frac{x}{(1-x)^2}\Big) \\
&=-\frac{\alpha}{2\pi} m_P\Big(\frac{3}{2}\ln\frac{\Lambda^2}{m_P^2}+\frac{3}{4}\Big) \\
\end{align}
$$
That is the expression in Eqn. (18.45). Note that the "finite" term in (18.11) is not a constant but momentum-dependent. You can't just plug in ${p\!\!\!/}=m_P$ while ignoring that term.
Hope this is clear.
