# In $d=4$ $g\phi^3$-theory where does the divergence in the 3-point one-loop diagram come from?

$$g\phi^3$$ , $$d=4$$ , 3-point One-loop diagram (three external legs) Divergence

I am trying to find where the divergence factor/pole is on the following diagram in 4 dimensions so that I can use minimal subtraction...

I know it is proportional to...

$$\int d^dy_1 \ d^dy_2 \ d^dy_3 \ D_f(x_1,y_1)D_f(x_2,y_2)D_f(x_3,y_3)D_f(y_1,y_2)D_f(y_2,y_3) D_f(y_3,y_1)$$

where $$D_f$$ are Feynmann propagators. Which I can manipulate to get... $$\int d^dp_1 \cdots d^dp_3 \left(\frac{\textrm{Exp}[ip_1x_1]}{p_1^2+m^2}\right) \cdots \left(\frac{\textrm{Exp}[ip_3x_3]}{p_3^2+m^2}\right) \delta[p_1+p_2+p_3] \bullet \int d^dp_4 \left(\frac{1}{p_4^2+m^2}\right) \left(\frac{1}{(p_2+p_4)^2+m^2}\right) \left(\frac{1}{(p_2+p_3+p_4)^2+m^2}\right).$$

Which looks like a renormalization to the coupling constant $$g$$. So I only mess around with the $$p_4$$ integral, however it seems to be finite (no divergences) when I use Feynmanns trick... $$\frac{1}{abc}=2 \int_0^1 dx \int^{1-x}_0 dy \ (ax+by+c(1-x-y))^{-3}$$ and the following integral... $$\int \frac{d^dp}{(p^2+2p \cdot q +m^2)^n} \textrm{(Minkowski)} = i \pi^{d/2}(m^2-q^2)^{d/2-n} \frac{\Gamma[n-d/2] \Gamma[d/2]}{\Gamma[n]}.$$

Was just wondering if I am going in the right direction or if I did something wrong.

• You have 3 propagators, so the denominator of your loop momentum goes like $\frac{1}{k^6}$ dimensionally, while you're in 4 dimensions so you have $\int\; dk \, k^3$. The field is not massless and you needn't worry about IR divergences. So what do you think about the UV behaviour of the loop integral?
– Siva
Dec 23, 2013 at 3:06

## 1 Answer

In $d=4$ and $n=3$, you the following relation $w = 4 - E - V$, where:

• $w=$ supercifial degree of divergence
• $E=$ external legs
• $V=$ number of vertices

So, if you replace the values of $d$ and $n$, you will get $w<0$, therefore you won't have $UV$ divergences in that diagram.