The analytical result for free massless fermion propagator For massless fermion, the free propagator in quantum field theory
is
\begin{eqnarray*}
 &  & \langle0|T\psi(x)\bar{\psi}(y)|0\rangle=\int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}.
\end{eqnarray*}
In Peskin & Schroeder's book, An introduction to quantum field theory
(edition 1995, page 660, formula 19.40), they obtained the analytical
result for this propagator,
\begin{eqnarray*}
 &  & \int\frac{d^{4}k}{(2\pi)^{4}}\frac{i\gamma\cdot k}{k^{2}+i\epsilon}e^{-ik\cdot(x-y)}=-\frac{i}{2\pi^{2}}\frac{\gamma\cdot(x-y)}{(x-y)^{4}} .\tag{19.40}
\end{eqnarray*}
Question: Is this analytical result right? Actually I don't know
how to obtain it.

The calculation of the propagator in four dimensions is as follows. 
\begin{eqnarray*}
\int\frac{d^4 k}{(2\pi)^4}e^{-ik\cdot (x-y)}\frac{1}{k^2} 
&=&  i\int \frac{d^4 k_E}{(2\pi)^4}e^{ik_E\cdot (x_E-y_E)}\frac{1}{-k_E^2}  \\
&=& \frac{-i}{(2\pi)^4} \left( \int_0^{2\pi}d\theta_3 \int_0^{\pi}d\theta_2 \sin \theta_2 \right) \int_0^{\infty} dk_E k_E^3 \frac{1}{k_E^2} \int_0^{\pi}d\theta_1 \sin^2 \theta_1 e^{ik_E | x_E-y_E | \cos \theta_1} \\
&=& \frac{-i4\pi}{(2\pi)^4}  \int_0^{\infty} dk_E k_E \int_0^{\pi}d\theta_1 \frac{1-\cos 2\theta_1}{2} e^{ik_E | x_E-y_E | \cos \theta_1} \\
&=& \frac{-i}{4\pi^3} \frac{1}{| x_E-y_E |^2} \int_0^{\infty} ds s (\frac{\pi}{2} J_0(s)- \frac{\pi i^2}{2} J_2(s)) 
\end{eqnarray*}
where $s\equiv k_E\| x_E-y_E \| $, and $J_n(s)$'s are bessel functions and I made use of Hansen-Bessel Formula.
\begin{eqnarray*}
&=& \frac{-i}{4\pi^3} \frac{1}{| x_E-y_E |^2}  \int_0^{\infty} ds s \frac{\pi}{2} \frac{2}{s} J_1(s) \\
&=& -\frac{i}{4\pi^2} \frac{1}{| x_E-y_E|^2} \int_0^{\infty} ds \, J_1(s) \\
&=& -\frac{i}{4\pi^2} \frac{1}{| x_E-y_E |^2} \\
&=& \frac{i}{4\pi^2} \frac{1}{(x-y)^2}
\end{eqnarray*}
