# Approximating rapidly divergent integrals

A while back I ran across a paper [1] 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 [1] 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.

[1] 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).