Solving @Peskin Eqn 2.54 Homework

[SuperStringboy]

Homework Statement

I am facing problem to derive the 2nd expression from the first one. My problem is the 2nd term of the 2nd expression.

Homework Equations

$$\int\frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}[\exp(-ip\cdot(x - y)) - \exp(ip\cdot (x - y))]=\int\ \frac{d^3p} {(2\pi)^3}\ \{ \frac {1}{2E_p}\ e^{-ip.(x-y)}\left|_{p^0 = E_p}\ +\ \frac {1}{-2E_p}\ e^{-ip.(x-y)}\left|_{p^0 = -E_p}\ \}$$

The Attempt at a Solution

$$p\cdot (x - y)= p^0(x^0 - y^0) - \textbf p\cdot(x-y)$$

For Po = - Ep we can take
$$p\cdot (x - y)= - p^0(x^0 - y^0) - \textbf p\cdot(x-y)$$
If i am not wrong yet, then what now?
should i change the dummy variable as p = - p? But if do it then i think another change comes d³p becomes -d³p for the 2nd term and i loose the minus sign before the 2nd term.

i don't know how much wrong i am but i am expecting good solution from you guys.

 

[Ben Niehoff]

If i am not wrong yet, then what now?
should i change the dummy variable as p = - p? But if do it then i thing another change comes d³p becomes -d³p for the 2nd term and i loose the minus sign before the 2nd term.

The trick is that the measure d³p actually does not change sign under p -> -p.

 

[SuperStringboy]

Thanks Ben . But will you please explain that why the sign of d³ does not change. Is it because of spherical co-ordinates : d³p = p²sin(theta)d(theta)d(phi)dp , where p = |p| ?