Let’s begin by going over the computation from the end of Lecture 3, where we were trying to compute the coefficients in the expansion

with

From the Exponential Formula, we get

so for each partition of the corresponding term in the sum is

where is the number of blocks in The product

is counting the number of ways to build a cyclic permutation of length at least on each block of so we conclude that

with the number of permutations in the symmetric group made up of exactly cycles, each of length at least We can write this a bit more precisely as

It remains to solve Problem 2 from Lecture 3, which was to evaluate the integral

Unlike Problem 1, Problem 2 cannot be solved exactly. In the case this was discussed in Lecture 2: we can approximate

with an exponentially small errror, and then evaluate the Gaussian integral

For any fixed we have an analogous approximation

with an error term that remains to be quantified precisely but is clearly rapidly decaying given the exponentially small tails of the Gaussian density. We now evaluate the total integral

In fact, we will evaluate all such integrals simultaneously by looking instead at the integral

which is the Laplace transform of the Gaussian density. On one hand, observe that the integral converges to define an entire function of whose derivatives at can be computed by differentiating under the integral sign, and are exactly the integrals we’re trying to evaluate:

(Differentiation under the integral sign is our first contact with the ghost of Richard Feynman, a congenial specter that will continue to haunt us). On the other hand, we can compute exactly by completing the square: since

we obtain

by translation invariance of Lebesgue measure, and hence

This is an important feature of the Gaussian density: up to minor details, it is a fixed point of the Laplace transform. In particular, this makes the Maclaurin series of is easy to compute directly:

Taking a closer look at the nonzero coefficients of this series, we observe that

the product of all odd numbers below Thus in terms of the double factorial which is the product of all numbers below of the same parity (and should not be confused with the much larger number ),$ we have

We thus obtain the integral formula

**Exercise 1: **Show that for even, is the number of permutations in all of whose cycles have length exactly i.e. the number of fixed point free involutions.

We have finally reached the point where we can legitimately write down Stirling’s formula. Here’s a step-by-step reconstruction of the argument we’ve been building.

**Step 1:** Obtain the integral representation

where and

**Step 2:** Demonstrate that the integral localizes in an -neighborhood of the minimizer of

where

**Step 3:** Write the integrand of in the form

**Step 4:** Expand the non-Gaussian part of the integrand,

with the coefficients being the “disconnected” version of the coefficients

**Step 5**: Integrate term-by-term:

**Step 6:** Replace each truncated Gaussian integral with a total Gaussian integral at the cost of an exponentially small error and evaluate:

The bracket in Step 6 is an Iverson bracket.

**Problem 1:** Carefully executing the above steps, show that we have obtained Stirling’s approximation: as we have

**Problem 2: **Develop a good bookkeeping system allowing the computation of subleading terms in this approximation.

**Problem 3:** Go through Steps 1 to 6, consider what features of were used, and examine to what extent these features were needed.

These are the problems which we will focus on in the next lectures. In particular, Problem 2 leads to the concept of Feynman diagrams. You can start thinking about these problems now, and see what you can see.