I am studying the derivation of Ward Takahashi identity using Peskin and Schroeder (Page number 311) What I understand from his statements is as follows, for a change of variables \begin{equation} \psi(x) \to(1+ie\alpha(x))\psi(x). \tag{9.100} \end{equation} The QED Lagrangian density transforms to \begin{equation} \mathscr{L}\to \mathscr{L}-e\partial_\mu\alpha\overline{\psi}\gamma^\mu\psi. \tag{9.101} \end{equation} I agreed with his statements till here. Then he says
"This transformation leads to the following identity for the functional integral over two fermion fields $$\begin{align} 0=\int\mathscr{D}\overline{\psi}\mathscr{D}\psi\mathscr{D}Ae^{i\int d^4x\mathscr{L}}\Bigg\{-i\int d^4x\partial_\mu\alpha(x)\Bigg[j^\mu(x)\psi(x_1)\overline{\psi}(x_2)\Bigg]\\+\bigg(ie\alpha(x_1)\psi(x_1)\bigg)\overline{\psi}(x_2)+\psi(x_1)\big(-ie\alpha(x_2)\overline{\psi}(x_2)\big)\Bigg\} \end{align} \tag{9.102} $$ with $j^\mu=e\overline{\psi}\gamma^\mu\psi$. [...] Dividing this equation by $Z$ gives $$\begin{align} i\partial_\mu⟨0|Tj^\mu(x)\psi(x_1)\overline{\psi}(x_2)|0⟩=-ie\delta (x-x_1)⟩⟨0|\psi(x_1)\overline{\psi}(x_2)|0⟩ +ie\delta (x-x_2)⟨0|\psi(x_1)\overline{\psi}(x_2)|0⟩.\end{align} \tag{9.103} $$ To put this equation into a more familiar form, compute its Fourier transform by integrating \begin{equation} \int d^4xe^{-ik\cdot x}\int d^4x_1e^{iq\cdot x_1}\int d^4x_1e^{-ip\cdot x_2}. \tag{9.104} \end{equation} Then the above amplitudes converted as \begin{equation} -ik_\mu\mathscr{M}^\mu(k;p;q)=-ie\mathscr{M}_0(p;q-k)+ie\mathscr{M}_0(p+k;q). \tag{9.105} \end{equation} This is exactly the Ward-Takahashi identity for two external fermions."
My questions are
How did he get $$\begin{align} 0=\int\mathscr{D}\overline{\psi}\mathscr{D}\psi\mathscr{D}Ae^{i\int d^4x\mathscr{L}}\Bigg\{-i\int d^4x\partial_\mu\alpha(x)\Bigg[j^\mu(x)\psi(x_1)\overline{\psi}(x_2)\Bigg]\\+\bigg(ie\alpha(x_1)\psi(x_1)\bigg)\overline{\psi}(x_2)+\psi(x_1)\big(-ie\alpha(x_2)\overline{\psi}(x_2)\big)\Bigg\}~? \end{align} \tag{9.102}$$
On dividing this equation by $Z$ gives $$\begin{align} i\partial_\mu⟨0|Tj^\mu(x)\psi(x_1)\overline{\psi}(x_2)|0⟩=-ie\delta (x-x_1)⟩⟨0|\psi(x_1)\overline{\psi}(x_2)|0⟩ +ie\delta (x-x_2)⟨0|\psi(x_1)\overline{\psi}(x_2)|0⟩,\end{align}\tag{9.103}$$ how did he get this?
In taking Fourier transform, why does he take $(-k,q,-p)$ as the set of momentum in exponential instead of $(-k,-q,-p)$ ? Please help me to get an intuition.