# Approximating rapidly divergent integrals

A while back I ran across a paper  giving a trick for evaluating integrals of the form where M is large and f is an increasing function. For large M, the integral is asymptotically That is, the ratio of A(M) to I(M) goes to 1 as M goes to infinity.

This looks like a strange variation on Laplace’s approximation. And although Laplace’s method is often useful in practice, no applications of the approximation above come to mind. Any ideas? I have a vague feeling I could have used something like this before.

There is one more requirement on the function f. In addition to being an increasing function, it must also satisfy In  the author gives several examples, including using f(x) = x². If we wanted to approximate the method above gives

exp(10000)/200 = 4.4034 × 104340

whereas the correct value to five significant figures is 4.4036 × 104340.

Even getting an estimate of the order of magnitude for such a large integral could be useful, and the approximation does better than that.

 Ira Rosenholtz. Estimating Large Integrals: The Bigger They Are, The Harder They Fall. The College Mathematics Journal, Vol. 32, No. 5 (Nov., 2001), pp. 322-329

## 2 thoughts on “Approximating rapidly divergent integrals”

1. GlennF

Just for fun… I wanted to see if I could derive this. I converted it to a differential equation for A and then used standard ways of developing an asymptotic series, which gets the same result. I glanced at the paper briefly.. it looks like the author uses a different route.

exp(f(M)) is the controlling factor… 1/f'(M) is the first term in the asymptotic series. The next term is just the same f”(M)/(f”(M)^2) that is required to approach zero (so that makes sense).

With the next term for the example you gave, I used bc to find a value of 4.40362928 x 10^4340, which agrees with the Mathematica result to the stated precision.

2. GlennF

Sorry, correction: The second term is f”(M)/ (f'(M)^3).