We have seen that the Ward identity can be violated by an improperly chosen regulator. Let us check the validity of the identity $Z_1=Z_2$, to order $\alpha$, for several choices of regulator. We have already verified that the relation holds for Pauli-Villars regularisation.
(a) First, let us verify the identity, $\delta Z_1 = \delta Z_2$, by simply replacing introducing an UV cutoff, $\Lambda$, on the integration over $l_E$. Recall \[\Gamma^\mu(q=0) = Z_1^{-1}\gamma^\mu \implies \delta Z_1 = -\delta F_1(q^2=0),\tag{1.1}\] and \[Z_2^{-1} = 1-\frac{d\Sigma_2}{d\slashed{p}}\bigg|_{\slashed{p}=m} \implies \delta Z_2 = \frac{d\Sigma_2}{d\slashed{p}}\bigg|_{\slashed{p}=m_0}.\tag{1.2}\] (We have used the fact that $m=m_0$ at tree-level.)
Let us start by evaluating $\delta Z_2$. From Peskin (7.17), we have \[\Sigma_2(p) = -ie^2\int^1_0 dx\int\frac{d^4l}{(2\pi)^4}\frac{-2x\slashed{p}+4m_0}{(l^2-\Delta+i\varepsilon)^2},\tag{1.3}\] where $\Delta = -x(1-x)p^2+(1-x)m_0^2$. Perform a Wick rotation on the integral, and we obtain \[\begin{aligned} \int\frac{d^4l}{(2\pi)^4}\frac{1}{(l^2-\Delta+i\varepsilon)^2} &= i\int\frac{d^4l_E}{(2\pi)^4}\frac{1}{(l_E^2+\Delta)^2}\\ &=\frac{i}{8\pi^2}\int_0^\infty dl_E\frac{l_E^3}{(l_E^2+\Delta)^2}. \end{aligned}\tag{1.4}\] Now, we can apply the UV cutoff to the integral, which yields \[\frac{i}{8\pi^2}\int_0^\Lambda dl_E\frac{l_E^3}{(l_E^2+\Delta)^2}=\frac{i}{16\pi^2}\left[\ln\left(1+\frac{\Lambda^2}{\Delta}\right)-\frac{\Lambda^2}{\Lambda^2+\Delta}\right].\tag{1.5}\]
Thus, by differentiating $\Sigma_2$ and noting that $p^2 = \slashed{p}^2$, we can obtain \[\begin{aligned} \delta Z_2 &= -\frac{e^2}{8\pi^2}\int^1_0dx \frac{\partial}{\partial\slashed{p}}\left\{(x\slashed{p}-2m_0)\left[\ln\left(1+\frac{\Lambda^2}{\Delta}\right)-\frac{\Lambda^2}{\Lambda^2+\Delta}\right]\right\}\bigg|_{\slashed{p}=m_0}\\ &=-\frac{\alpha}{2\pi}\int^1_0dx \left\{x\left[\ln\left(1+\frac{\Lambda^2}{\Delta}\right)-\frac{\Lambda^2}{\Lambda^2+\Delta}\right]-2(x\slashed{p}-2m_0)\slashed{p}x(1-x)\left[-\frac{\Lambda^2}{\Delta(\Lambda^2+\Delta)}+\frac{\Lambda^2}{(\Lambda^2+\Delta)^2}\right]\right\}\bigg|_{\slashed{p}=m_0}\\ &=-\frac{\alpha}{2\pi}\int^1_0dx\left\{x\left[\ln\left(1+\frac{\Lambda^2}{\Delta}\right)-\frac{\Lambda^2}{\Lambda^2+\Delta}\right]+2m_0^2(x-2)x(1-x)\frac{\Lambda^4}{\Delta(\Lambda^2+\Delta)^2}\right\}\bigg|_{\Delta=(1-x)^2m_0^2}\\ &=-\frac{\alpha}{2\pi}\int^1_0dx\left\{x\ln\left(1+\frac{\Lambda^2}{m_0^2(1-x)^2}\right)-x+\frac{2(x-2)x}{(1-x)}+ \order{\frac{m_0^2}{\Lambda^2}}\right\}. \end{aligned}\tag{1.6}\]
Now, let us turn to $\delta Z_1$. From Peskin (6.54), we have \[\delta F_1(q^2=0)=2ie^2\int^1_0dx\,dy\,dz\,\delta(x+y+z-1)\int\frac{d^4l}{(2\pi)^4}\frac{-l^2+2(1-4x+x^2)m_0^2}{(l^2-\Delta+i\varepsilon)^3},\tag{1.7}\] where $\Delta=(1-x)^2m_0^2$. Again, Wick rotate the interal with respect to $l$, we obtain \[\begin{aligned} \int\frac{d^4l}{(2\pi)^4}\frac{-l^2+2(1-4x+x^2)m_0^2}{(l^2-\Delta+i\varepsilon)^3} &= -i\int\frac{d^4l_E}{(2\pi)^4}\frac{l_E^2+2(1-4x+x^2)m_0^2}{(l_E^2+\Delta)^3}\\ &=\frac{-i}{8\pi^2}\int^\infty_0 dl_E\, l_E^3\cdot \frac{l_E^2+2(1-4x+x^2)m_0^2}{(l_E^2+\Delta)^3}. \end{aligned}\tag{1.8}\] Let us regulate the integral with a UV cutoff: \[\begin{aligned} \int^\infty_0\to\int^\Lambda_0. \end{aligned}\tag{1.9}\] Thus, \[\begin{aligned} &\frac{-i}{8\pi^2}\int^\Lambda_0 dl_E\, l_E^3\cdot \frac{l_E^2+2(1-4x+x^2)m_0^2}{(l_E^2+\Delta)^3}= \frac{-i}{8\pi^2}\int^{\sqrt{\Lambda^2+\Delta}}_{\sqrt{\Delta}} dk\left[\frac{(k^2-\Delta)^2}{k^5} + 2(1-4x+x^2)m_0^2\frac{k^2-\Delta}{k^5}\right]\\ =&\frac{-i}{8\pi^2}\left[\frac{1}{2}\ln\left(1+\frac{\Lambda^2}{\Delta}\right)+\frac{\Delta}{\Lambda^2+\Delta}-\frac{3}{4} - \frac{\Delta^2}{4( \Lambda^2+\Delta)^2} + 2(1-4x+x^2)m_0^2\left(-\frac{1}{2(\Lambda^2+\Delta)}+\frac{1}{4\Delta} + \frac{\Delta}{4(\Lambda^2+\Delta)}\right)\right]\\ =&\frac{-i}{8\pi^2}\left[\frac{1}{2}\ln\left(1+\frac{\Lambda^2}{\Delta}\right)-\frac{3}{4}+\frac{(1-4x+x^2)m_0^2}{2\Delta}+\order{\frac{m_0^2}{\Lambda^2}}\right]. \end{aligned}\tag{1.10}\]
Therefore, we have \[\delta Z_1 = -\frac{\alpha}{2\pi}\int^1_0dx\int^{1-x}_0dy\left[\ln\left(1+\frac{\Lambda^2}{\Delta}\right)-1+\order{\frac{m_0^2}{\Lambda^2}}\right].\tag{1.11}\]
Hence, we have shown that $\delta Z_1 \neq \delta Z_2$ under the UV cutoff regulator.
(b) In this section, let us see what happened with dimensional regularisation. Note that in $d$ dimensions, \[\gamma^\mu\gamma^\nu\gamma_\mu = -(2-\epsilon)\gamma^\nu;\tag{1.12}\] \[\gamma^\mu \gamma_\mu = d.\tag{1.13}\] Then, Peskin (7.17) is modified to be \[\begin{aligned} \Sigma_2(p) &= ie^2\int^1_0dx\int\frac{d^dl}{(2\pi)^d}\frac{(2-d)x\slashed{p}+d\cdot m_0}{(l^2-\Delta+i\varepsilon)^2}\\ &=-e^2\int^1_0dx\int\frac{d^dl_E}{(2\pi)^d}\frac{(2-d)x\slashed{p}+d\cdot m_0}{(l_E^2+\Delta)^2}\\ &= \frac{-e^2}{(4\pi)^{d/2}}\Gamma(2-\frac{d}{2})\int^1_0dx[(2-d)x\slashed{p}+d\cdot m_0]\cdot\frac{1}{\Delta^{2-d/2}}. \end{aligned}\tag{1.14}\] Thus, \[\delta Z_2 = \frac{d\Sigma_2}{d\slashed{p}}\bigg|_{\slashed{p}=m_0}=-\frac{2e^2}{(4\pi)^{d/2}}\Gamma(2-\frac{d}{2})\int^1_0dx\left[\frac{(2-d)x}{\Delta^{2-d/2}} + (2-\frac{d}{2})\frac{[(2-d)xm_0+d\cdot m_0]2x(1-x)m_0}{\Delta^{3-d/2}}\right]\bigg|_{\Delta=(1-x)^2m_0^2}.\tag{1.15}\] Now, take the limit $d=4-\epsilon$, where $\epsilon\to0$, we have \[\begin{aligned} \delta Z_2 &= -\frac{e^2}{(4\pi)^{2}}\left(1+\frac{\epsilon}{2}\ln4\pi\right)\left(\frac{2}{\epsilon}-\gamma+\order{\epsilon} \right)\int^1_0dx \\ &\quad \times\left[-2x+\epsilon[(\epsilon-2)xm_0+(4-\epsilon)m_0]2x(1-x)m_0\cdot\frac{1}{\Delta}\right]\left(1-\frac{1}{2}\epsilon\ln\Delta+\order{\epsilon^2}\right)\\ &=\frac{\alpha}{2\pi} \int_0^1\!dx\, (2x) \left[ \frac{2}{\epsilon} -\gamma -\ln 4\pi -1 -\ln m_0^2 -2\ln(1-x) -\frac{2(2-x)}{1-x} \right] +\mathcal{O}(\epsilon). \end{aligned}\tag{1.16}\]