homework and exercises - The contraction of fermion field in 1+1-dimensional massless QED

My question comes from the textbook by Peskin & Schroeder, the integral (19.26): $$\begin{align} \int \frac{d^2 k}{(2\pi)^2}\! e^{- i k\cdot (y-z)}\frac{i \not{k}}{k^2} = -\not\partial \left(\frac{i}{4\pi}\log (y-z)^2\right) \end{align}$$

Question: how to derive the formula from the left hand side to the right hand side ?

If considering the identity (3.117) and set $m=0$, I have $$\begin{align} \int \frac{d^2k}{(2\pi)^2}\! \frac{i k\cdot\gamma}{k^2} e^{-i k\cdot (y-z)} = i \not\partial \left(D_R(y-z)\right) \end{align}$$ here $$\begin{align} D_R(y-z)= \int \frac{d^2 k}{(2\pi)^2}\frac{i}{k^2}e^{-i k\cdot (y-z)} \end{align}$$ the 2-vector: $k^\mu=(k^0,k^1)$ and owing to the massless condition:$(k^0)^2 = (k^1)^2$. set $\kappa \equiv k^1$.therefore I got $$\begin{align} &\int_{-\infty}^{+\infty}\frac{d k^1}{(2\pi)}\! \bigg[\frac{1}{2 k^0} e^{-i [k^0(y-z)_0 - k^1(y-z)_1]} + \frac{1}{-2 k^0}e^{-i[-k^0(y-z)_0 - k^1(y-z)_1]}\bigg] \\ &= - \frac{i}{4\pi}\ 2 \int_{-\infty}^{+\infty} \frac{\sin\left(\kappa (y-z)_0\right)}{\kappa}e^{i \kappa(y-z)_1}\; d\kappa \end{align}$$

But I failed to get the log-term from the above formula.

NOTE I found a related answer A four-dimensional integral in Peskin & Schroeder