We continue with the estimation of for large via Euler’s integral,

As at the end of Lecture 1, we make the substitution thereby obtaining

Now, one way to characterize an algebraic combinatorialist is to say that such a person loathes this being some horrible transcendental thing, but loves this being an exponential generating function for cyclic permutations:

Accordingly, we make the further change of variables which gets us to the formula

We now focus on the problem of determining the asymptotics of

where the “action” is a smooth function on the Maclaurin series of which is

**Exercise 1:** What is the radius of convergence of this series?

Having solved Exercise 1 (which is singularly easy), you know that the series expansion of is only valid in a neighborhood of so ostensibly is not of any help in the estimation of since the integrand involves with arbitrarily large inputs On the other hand, we’re trying to work out an approximation, not perform an exact evaluation, so perhaps we can get away with something local. Indeed, the shape of indicates that this might well be the case. Since

our action has a unique stationary point at and the value is the global minimum of on its domain of definition,

This means that is the global maximum of the integrand of : we have

In particular the integrand of is exponentially smaller than for all This means that the integrand of is sharply peaked at the unique minimizer of the action and the larger is, the more exaggerated this effect becomes, as in the plots below.

Accordingly, we expect that as the integral “localizes” at the stationary point of the action meaning that

might not be such a bad approximation. Let us investigate this more closely: we fix and seek to estimate the integral

Being combinatorialists, we will count the complement, meaning that we will try to estimate the above integral by instead estimating

To estimate the left integral, observe that left of zero we have

and hence

where the rightmost integral is convergent (say, by comparison with ). To control the rightmost integral, we can observe that

for so that we have the bound

where

satisfies

by the Bounded Convergence Theorem. In particular, we can conclude that

Now we estimate the -integral using a similar strategy Since is strictly increasing for we have

for Consequently, we have the bound

where

is a convergent integral, and moreover

by the Bounded Convergence Theorem. So we similarly have that

We thus conclude that

Just to remind ourselves where we are vis-a-vis our principal goal of approximation , we have arrived at the estimate

It remains to estimate the integral. This is ostensibly more difficult than what we did above, since it apparently requires some actual insight into the behavior of On the other hand, we are now reduced to considering the behavior of the action on a small interval where we can represent it as a series: we have that

for all sufficiently small. Now, since

we expect a further approximation of the form

Of course, such an approximation is not exactly true, and we have to understand how accurate it actually is in order to proceed rigorously. For now, however, let us see what would happen if we had the exact identity

where the quotes are meant to indicate that we are considering a hypothetical truth which we know to be false, but might still be conceptually useful.

Consider the integral

which we would like to approximate. An ideal outcome would be that we could exactly evaluate it. Unfortunately, even though the integrand here doesn’t look so bad, trying to compute the indefinite integral

is a fool’s errand, being in some sense analogous to trying to solve the quintic. On the other hand, by the same sort of rough estimates we have been using so far in this lecture, we have

The integral

which is known as the Gaussian integral, can in fact be exactly evaluated — not using the Fundamental Theorem of Calculus, but by exploiting symmetry. The answer is

a remarkable formula which is the basis of various amusing proclamations, including Lord Kelvin‘s assertion that “a mathematician is one to whom this is as obvious as that twice two makes four is to you.” I leave you to either try the derivation of this formula on your own, or look it up somewhere.

Changing variables in the Gaussian integral, we get the evaluation we actually need:

Following through on our thought experiment, this would imply the approximation

The meat of this formula is correct, but the error term is the result of our wishful thinking, and it is false — we will correct it in Lecture 3.

“… that twice two makes fo[u]r is to you.”

What an ignoramus that Lord Kelvin was. You’re right, I should correct his error.

Indeed. Solid beard though.