How to understand the term $\frac{1}{-2E(\vec{p})} e^{-ip(x-y)}$ of Klein-Gordon propagator in Peskin & Schroeder's book? I am reading Peskin & Schroeder's book on Chapter 2. I have a question about how to get the term $\frac{1}{-2E(\vec{p})} e^{-ip(x-y)}$. The original equation for propagator is
$$
\langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} [ e^{-ip(x-y)} - e^{ip(x-y)} ].\tag{2.54}
$$
We separate this equation, namely,
$$
\langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} + \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{-2E(\vec{p})} e^{ip(x-y)}
$$
Peskin & Schroeder's Book says that the energy $p^{0}=-E(\vec{p})$ in the second term is less than 0. I tried to expand $e^{ip(x-y)}$ in the following:
$$
e^{ip(x-y)} = e^{i(p^{\mu}(x-y)_{\mu})} = e^{i(Et-\vec{p}\cdot(\vec{x}-\vec{y}))} = e^{i[-(-Et)-\vec{p}\cdot(\vec{x}-\vec{y})]}=e^{-i[p^{0}t+\vec{p}\cdot(\vec{x}-\vec{y})]}
$$
However, it seems that I can not write $e^{-i[p^{0}t+\vec{p}\cdot(\vec{x}-\vec{y})]}$ as $e^{-ip(x-y)}$, where $p^{0}=-E(\vec{p})$. Where is the problem?

 A: Yeah, basically as hft said in his comment, the main point is the integrals are only in the space/momentum components and not in time/energy components.
And because you are integrating over all momenta $\vec{p}$, for every $\vec{x}-\vec{y}$ you pass, you will also pass through a $\vec{y}-\vec{x}$. Meaning that effectively, and only inside the integral:
\begin{equation}
 \vec{x}-\vec{y}=\vec{y}-\vec{x}
\end{equation} and because this only happens in space components the time component of $p$ which is $E$ will gain a minus sign, telling you that you are dealing with "negative energy" particles (antiparticles):
$$
\langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} \Big|_{p_0=E} + \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2(-E(\vec{p}))} e^{-ip(x-y)} \Big|_{p_0=-E}
$$
Notice you could also have absorbed the minus in the time component of $x$ instead, telling you that those antiparticles can be thought as normal particles kind of travelling back in time:
$$
\langle 0 | [\phi(x), \phi(y)] | 0 \rangle = \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} \Big|_{p_0=E, \ t_+} - \int \frac{d^{3}p}{(2\pi)^3} \frac{1}{2E(\vec{p})} e^{-ip(x-y)} \Big|_{p_0=E, \ t_-}
$$
(This last paragraph has to be taken with caution, since in reality is more complex than that)
