# 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?

• 1, 2) You need to Fourier transform both factors of $J$ in $W$. For the other one, use integration variable $y$. 3) Note that $k^2 = (k^0)^2 - \vec{k}^2$, causing a sign flip. 4) I'm not sure what you're asking here. What's inside an integral is pretty clear, it's everything to the right of the integral sign. Commented Mar 30, 2018 at 22:19

$$J_a(k) = \int dx^0 e^{-ik^0x^0} \int d^3x e^{i\vec{k}\cdot\vec{x}}\delta^3(\vec{x}-\vec{x}_a) = e^{i\vec{k}\cdot\vec{x}_a}\int dx^0 e^{-ik^0x^0}$$

$$J^{*}_a(k) = \int dy^0 e^{ik^0y^0} \int d^3y e^{-i\vec{k}\cdot\vec{y}}\delta^3(\vec{y}-\vec{x}_a) = e^{-i\vec{k}\cdot\vec{x}_a}\int dy^0 e^{ik^0y^0}$$

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!