homework and exercises - The computation of the propagator in two dimensions

I did the computation of the propagator in two dimensions at (19.26) in Peskin & Shroeder as follows.
First I performed a Wick rotation.

\begin{alignat}{2} \int\frac{d^2 k}{(2\pi)^2}e^{-ik\cdot (y-z)}\frac{ik^{\mu}\gamma_{\mu}}{k^2} &=&& -\partial^{\mu}\gamma_{\mu} \left( i\int \frac{d^2 k_E}{(2\pi)^2}e^{ik_E\cdot (y_E-z_E)}\frac{1}{-k_E^2} \right) \\ &=&& \frac{i}{4\pi^2} \partial^{\mu}\gamma_{\mu} \int_0^{\infty}dk_E k_E \frac{1}{k_E^2}\int_0^{2\pi}d\theta e^{ik_E|y_E-z_E|\cos \theta} \\ &=&& \frac{i}{4\pi^2} \partial^{\mu}\gamma_{\mu} \int_0^{\infty}dk_E k_E \frac{1}{k_E^2} 2\pi J_0(k_E|y_E-z_E|) \\ \end{alignat}

where $J_0(s)$ is a bessel function and I made use of Hansen-Bessel Formula.

Setting $s\equiv k_E|y_E-z_E|$

\begin{alignat}{2} &=&& \frac{i}{2\pi} \partial^{\mu}\gamma_{\mu} \int_0^{\infty} ds\frac{1}{s} J_0(s) \\ &=&& 0 \end{alignat} But in the book, $$ \int\frac{d^2 k}{(2\pi)^2}e^{-ik\cdot (y-z)}\frac{ik^{\mu}\gamma_{\mu}}{k^2}= -\partial^{\mu}\gamma_{\mu} \left( \frac{i}{4\pi}\log (y-z)^2 \right) \tag{19.26}$$ Where did I make a mistake?

Answer

After you take $\gamma_\mu \partial^\mu$ out of the integral sign, the integral becomes divergent, so this step is not guaranteed by theorems from calculus and apparently illegal. But I think your calculation can be continued in the following (heuristic) way.

Let $\vec{k}$ denote the momentum in Euclidean space and $k$ its norm, and $r=|\vec{y}-\vec{z}|$. In order to take $\partial^\mu$ out, we can introduce a cut off $\epsilon$, which is a positive small number, then the integration becomes $$ \lim_{\epsilon\to 0}\quad\frac{i}{2\pi}\int^\infty_\epsilon dk (\gamma_\mu \partial^\mu)\frac{J_0(kr)}{k}= \lim_{\epsilon\to 0}\quad\frac{i}{2\pi}(\gamma_\mu \partial^\mu)\int^\infty_{\epsilon\cdot r} ds \frac{J_0(s)}{s} $$

In the expansion of $J_0(s)$: $$ J_0(s)=\frac{sin(s)}{s}=1-\frac{1}{3!}s^2+... $$ the terms of order larger than 1 will contribute to the integral terms like $\epsilon^2 r^2$ which vanishes as $\epsilon$ goes to zero. So these terms as well as other constants are irrelevant to the result. The only relevant one is the zeroth order term: $$ \lim_{\epsilon\to 0}\quad\frac{i}{2\pi}(\gamma_\mu \partial^\mu)\int^\infty_{\epsilon\cdot r} ds \frac{1}{s}= \lim_{\epsilon\to 0}\quad -\frac{i}{2\pi}(\gamma_\mu \partial^\mu)\left[ln(r)+ln(\epsilon)+constant\right] $$ which turns out to be the right answer:$$ -\frac{i}{2\pi}(\gamma_\mu \partial^\mu)ln(r) $$