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


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).


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One Response to “Monday Math 54”

  1. Says:

    I’m Confused :s

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