How do physicists integrate? I've always thought that the integration notation in physics is weird, but I understood it nevertheless for a single variable, until I started reading Zee's QFT in a nutshell, where 4 dimensions are used. So I was wondering how the integration notation works.
For example, for a free theory with sources at $x_1$ and $x_2$ we have:
$$ 
W(J)= -\frac{1}{2}\int\frac{d^4k}{(2\pi)^4}J_2(k)^*\frac{1}{k^2-m^2+i\epsilon}J_1(k)
$$
Now we let:
$$
J_a(x)=\delta^{(3)}(\vec{x}-\vec{x_a})
$$
where $a=1,2$.
Using the Fourier transform into momentum space:
$$
J(k)=\int d^4xe^{-ik\cdot x}J(x)
$$
What I don't understand is that W(J) becomes:
$$
W(J)=-\int\int dx^0dy^0\int \frac{dk^0}{2\pi}e^{ik^0(x^0-y^0)}\int\frac{d^3k}{(2\pi)^3}\frac{e^{ik\cdot(x_1-x_2)}}{k^2-m^2+i\epsilon}
=\int dx^0\int\frac{d^3k}{(2\pi)^3}\frac{e^{ik\cdot(x_1-x_2)}}{k^2+m^2}
$$
I know that the $2$ comes from the two terms being equal so don't mind that.
It says integrating over $y^0$ will get a delta function setting $k^0$ to zero, and that I don't understand.
The only thing I know (and I'm guessing but I'm not sure yet) is that:
$$
J_a(k)=\int d^4xe^{-ik\cdot x}\delta^{(3)}(x-x_a)=\int dx^0e^{-ik^0\cdot x^0}\int d^3xe^{-ik\cdot x}\delta^{(3)}(x-x_a)= \int dx^0e^{-ik^0\cdot x^0}e^{-ik\cdot x_a}
$$
and the integral over $x^0$ (or $y^0$) is independent of the spatial $k$ so it's outside the integral of spatial $k$, but inside the integral of temporal $k$ in the $W(J)$. And the one with the $x_a$ part is inside the spatial integral $k$. Am I correct about this?
But yeah, my questions are:


*

*What's with the integral over $y^0$?

*Why is there even a $y$? Is it equal to $x_2$?

*How did the sign changed for the $m^2$ part?

*How do you actually know what's inside the integral using this notation? Or actually how to read integrals like this?
 A: $$ J_a(k) = \int dx^0 e^{-ik^0x^0} \int d^3x e^{i\vec{k}\cdot\vec{x}}\delta^3(\vec{x}-\vec{x}_a) = e^{i\vec{k}\cdot\vec{x}_a}\int dx^0 e^{-ik^0x^0} $$
$$ J^{*}_a(k) = \int dy^0 e^{ik^0y^0} \int d^3y e^{-i\vec{k}\cdot\vec{y}}\delta^3(\vec{y}-\vec{x}_a) = e^{-i\vec{k}\cdot\vec{x}_a}\int dy^0 e^{ik^0y^0} $$
So you have
$$ J^*_2(k)J_1(k) = e^{i\vec{k}\cdot(\vec{x}_1-\vec{x_2})} \int dx^0 \int dy^0 e^{ik^0(y^0-x^0)} $$
Now let me work with the term $k^2-m^2$:
$$k^2-m^2 = (k^0)^2-\vec{k}^2-m^2 = (k^0)^2-(\vec{k}^2+m^2) $$
Now let's talk about $\int \frac{dy^0}{2\pi} e^{ik^0(x^0-y^0)}$: by changing the integration variable $y^0-x^0 = \tilde{y}^0$ (so that $dy^0 = d\tilde{y}^0$), you get $\int \frac{d\tilde{y}^0}{2\pi} e^{i\tilde{y}^0k^0} $ which is the one dimensional Dirac Delta (you can quickly Google it): $\delta(k^0-0)$.
Now taking the integral over $dk^0$ simply means setting $k^0=0$ since you have this Dirac Delta function. The sign change comes as a consequence since you have $-(\vec{k}^2+m^2)$ at the denominator and you finally get:
$$ W[J] = \int dx^0 \int \frac{d^3k}{(2\pi)^3} \frac{e^{i\vec{k}\cdot(\vec{x}_1-\vec{x}_2)}}{\vec{k}^2+m^2}$$
I hope this helps! 
