Confusing integral in Zee's QFT

[waht]

This is probably really simple. In chapter I.4 the jump from (4) -> (5) is sort of eluding

$$W(J) = - \iint dx^0 dy^0 \int \frac{dk^0}{2\pi} e^{i k^0(x - y)^0} \int \frac{d^3k}{(2 \pi)^3} \frac{e^{i \vec{k}(\vec{x_1} - \vec{x_2})}} {k^2 - m^2 + i\epsilon}$$

and

$$\omega^2 = \vec{k}^2 + m^2$$


He got

$$W(J) = \int dx^0 \int \frac{d^3k}{(2\pi)^3} \frac{e^{i \vec{k} (\vec{x_1} - \vec{x_2})}}{\vec{k}^2 + m^2}$$


the way I see it - the middle term is the delta function

$$W(J) = - \iint dx^0 dy^0 \delta(x^0 - y^0) \int \frac{d^3k}{(2 \pi)^3} \frac{e^{i \vec{k}(\vec{x_1} - \vec{x_2})}} {k^2 - m^2 + i\epsilon}$$

but how does it disappear, and how does

$$k^2 - m^2 + i\epsilon$$ turn into

$$\vec{k}^2 + m^2$$

$$k^0$$ would be the $$\omega$$

but somehow this doesn't add up.

so just wondering if anyone could give a pointer on how to solve this

 

[RedX]

$$W(J) = - \iint dx^0 dy^0 \int \frac{dk^0}{2\pi} e^{i k^0(x - y)^0} \int \frac{d^3k}{(2 \pi)^3} \frac{e^{i \vec{k}(\vec{x_1} - \vec{x_2})}} {k^2 - m^2 + i\epsilon}$$

To do this integral, he integrated over y⁰ first. That produces a delta function $$\delta(k_0)$$. Then he integrated over k₀, and because of the delta function, this just sets k₀ equal to zero everywhere.

k^2-m^2 if written out is k₀^2-k^2-m^2, so if k₀ is zero, then that writes out to -(k^2+m^2), which cancels the negative sign.

 

[waht]

Took a while to convince myself, but yes it makes sense. Thanks.