## Monday Math 54

Let us consider an integral of the form , with parameter λ>0, and where f(x) has a single local minimum in the interval (a,b). The method of steepest decent, also known as saddle-point integration, is a useful method for approximating such an integral in the small λ limit. Namely, if the local minimum occurs at x=x0, then we can expand f(x) in the Taylor series about this point; as it is a local minimum, f‘(x0)=0 and f”(x0)≥0. Here, we assume that this second derivative is nonzero. Thus, our Taylor series is
.
Now, we note that as λ→0+, the term in the exponent becomes ever more negative, and the integrand approaches zero; the local minimum at x0 is the “slowest” to approach, and thus the function near that point comes to dominate the rest of the integral; and so

and

Now, we see that the last integral is part of a gaussian integral; the peak of our gaussian is in the region of integration, and as λ→0+, the gaussian’s width goes to zero as well, so that the tails become negligible in the limit, and
.
Now, recall that the integral of the gaussian over all real numbers is given by
.
Thus ,
and so
.

Let us perform an example: . This does not fit the form as it is, but can be transformed to do so. First, let us make the substitution . Then we have:
. Second, we note that for positive u, .
Thus, we have
,
and the integral above fits our form with  and ; thus our approximation will be for n→∞. Now, , which is zero for u=1. , so , and . Thus, our first saddle-point approximation says

and so

which you may recognize as Stirling’s approximation (see here).