One integral, five moves

Lesson 3.1 told you which diagrams diverge. Lesson 3.2 told you what a continuation is. Neither of them evaluated anything.

Here is the integral this lesson does, once, in the abstract:

I \;=\; \int\!\frac{\mathrm{d}^d\ell}{(2\pi)^d}\;\frac{1}{\big(\ell^2 - m_1^2 + i\varepsilon\big)\big((\ell+p)^2 - m_2^2 + i\varepsilon\big)}

No process, no diagram, no external spinors. Just the shape every one-loop calculation in this course eventually reduces to, with the loop momentum $\ell$ running over everything and $p$ , $m_1$ , $m_2$ whatever the diagram happened to supply.

It takes five moves: combine the denominators, complete the square and shift, rotate to Euclidean space, do the angles and the radius, and expand in $\varepsilon$ . Each one is a step you are licensed to take, and the licences are the interesting part — two of them were issued in lesson 3.2 and one of them can fail.

Here is the thing worth watching for. This integral diverges — 3.1's power counting says so. But nothing goes wrong until the very last move. Every expression between here and the end is a finite, ordinary function of $d$ . The infinity does not appear because you did something illegal; it appears because at the end you asked what happens at $d = 4$ , and that is where a $\Gamma$ function has a pole.