# Superficial degree of divergence in $\lambda\phi^4$

Ryder at the beginning of the chapter about renormalization defines the "superficial degree of divergence" of diagrams in $$\lambda \phi^4$$ theory. I'll recap the derivation.

A diagram in $$\lambda\phi^4$$ theory has $$E$$ external lines, $$I$$ internal lines of which $$L$$ are loops and $$n$$ vertices. Each internal line contributes with a factor $$\frac{1}{p^2}$$ and each undetermined loop momentum with an integral $$\int d^d p$$ in $$d$$ dimensions, so we define the superficial degree of divergence as $$D=dL-2I$$

and we say that if $$D\geq0$$, the diagram must diverge.

But consider the simplest diagram in $$\lambda\phi^4$$ with 4 external lines and no loops, just crossing in the center like an $$X$$. This diagram has no internal lines and no loops, so $$D=0$$ and the diagram should diverge logarithmically. Nevertheless, the value of the diagram is just $$-i\lambda(2\pi)^4\delta^{(4)}(p_1+p_2-p_3-p_4)$$

which is finite. What happened?