# About two integrals in QCD textbook by muta

1. ### Thor Shen

14

1.How to deal with the delta functions in eq.2.3.153 to obtain the eq.2.3.154 by integrating over q'?
2.How to caculate the integral from eq.2.3.154 to eq.2.3.156, especially the theta function?

2. ### Einj

443
You already have a $\delta^4(q'-q)$ so just put q=q' and remove the integral over dq'. Then in Eq. (2.3.154) you have $\delta(k'\cdot q)$ that, in the COM, becomes just $\delta(k_0\sqrt{s})$. Therefore, taking care of the Jacobian coming out of the delta function, this just tells you that $k'_0=0$. Hence you don't need to worry about the thetas since they only give you $\theta(q_0)=\theta(\sqrt{s})=1$ since its argument is positive.

1 person likes this.
3. ### Thor Shen

14
Actually, when I try to simplify the eq.2.3.153 for obtain the bracket in eq.2.3.154, I find we must use the two delta functions in eq.2.3.154, but I wonder whether we can use δ(1/4(q'+k')^2-m^2)=δ(k'^2+s-4m^2)δ(k'*q) directly, and the δ(k*q) is equivalent the δ(k'*q) because of the k' is integrated over total space?

4. ### Einj

443
No, you can't do that. You can't split one delta function in two. However, you already have two deltas:
$$\delta\left(q^2+k'^2+2q\cdot k'-4m^2\right)\delta\left(q^2+k'^2-2q\cdot k'-4m^2\right).$$
Now keep in mind that if you have some function $f(x)$ multiplied by a delta then $\delta(x-x_0)f(x)=\delta(x-x_0)f(x_0)$. This is true also if the function is a delta itself.

Now, the second delta tells you that $q^2+k'^2-4m^2=2q\cdot k'$. Using this is the first delta you obtain (I always omit the necessary Jacobian):
$$\delta\left(q\cdot k'\right)\delta\left(q^2+k'^2-2q\cdot k'-4m^2\right)=\delta\left(q\cdot k'\right)\delta\left(q^2+k'^2-4m^2\right).$$
Keep also in mind that $q^2=s$.