Perturbative Expansion of Gaussian-type Integrals (II)

高斯型积分的微扰展开（二） 苏剑林 · 2015-03-07 · Mathematics integral perturbation perturbation asymptotic series

Why has the second installment been so long in coming?

Actually, before starting this series I had already sketched out the content of the next few posts; I wanted to confidently present some integral-expansion techniques I'd come up with, and since I'm reasonably familiar with perturbation methods myself, under normal circumstances the second post wouldn't have taken this long. However, in the interval between finishing the first post and preparing to write the second, I came across this answer on Zhihu:

http://www.zhihu.com/question/24735673

That piece greatly expanded my understanding of series. It discusses how the expansion of an integral is in fact an asymptotic series. This gave me pause, and made me doubt whether this series of posts had any value at all, since an asymptotic series means that no matter what expansion technique you use, the resulting series always has a radius of convergence of zero.

But thinking it over some more, even for an asymptotic series there's still room for improvement — there are methods to accelerate convergence — so I think these posts of mine still have a bit of merit. And along the way I can introduce some of the theory concerning asymptotic series and singularities. All right, let's do that. more

Term-by-term expansion in the exponent

The technique already mentioned in the previous post was varying a constant coefficient, adjusting a variable so that the first-order term vanishes; the main issue with that method is that it doesn't let us obtain higher-order approximations term by term. Still, the underlying idea is a good one: introduce adjustable variables so that each order of the expansion vanishes in turn, rather than simply performing a plain power-series expansion. Below, we'll try putting the expansion inside the exponent, obtaining a method that can produce successive-order approximations — this will serve as the prototype model for a method we'll develop further later on.

In general, we want to compute the integral

$$\int_{-\infty}^{+\infty} e^{-a x^2-\varepsilon x^4} dx$$

In the previous post, we hoped to approximate it using the integral $\int_{-\infty}^{+\infty} e^{-A x^2} dx$, with the idea of adjusting $A$ so that the first-order term vanishes. But that's as far as we could go, because with only one adjustable variable we can generally make just one term vanish. To achieve a term-by-term expansion — i.e., to introduce multiple adjustable quantities — let us set

$$A=a+a_1 \varepsilon + a_2\varepsilon^2 + a_3 \varepsilon^3 + \dots$$

Then consider

$$\int_{-\infty}^{+\infty} \left[e^{-a x^2-\varepsilon x^4}-e^{-(a+a_1 \varepsilon + a_2\varepsilon^2 + a_3 \varepsilon^3 + \dots)x^2}\right] dx$$

Expanding the integrand inside the parentheses in powers of $\varepsilon$, we get

$$\begin{aligned}x^2 e^{-a x^2} \left(a_1-x^2\right)\varepsilon &+ \frac{1}{2} x^2 e^{-a x^2} \left(-a_1^2 x^2+2 a_2+x^6\right)\varepsilon^2 \\&- \frac{1}{6} x^2 e^{-a x^2} \left(-a_1^3 x^4+6 a_1 a_2 x^2-6 a_3+x^{10}\right)\varepsilon+\dots\end{aligned}$$

Integrating term by term gives

$$\begin{aligned}\sqrt{\frac{\pi}{a} }&\left[\frac{1}{4}(2 a a_1 -3)\left(\frac{\varepsilon}{a^2} \right)+\frac{1}{32}\left(4 a^2 \left(4 a a_2-3 a_1^2\right)+105\right)\left(\frac{\varepsilon}{a^2} \right)^2\right.\\ &\left.+\frac{1}{128}\left(3465-8 a^3 \left(8 a^2 a_3-12 a a_1 a_2+5 a_1^3\right)\right)\left(\frac{\varepsilon}{a^2} \right)^3+\dots\right]\end{aligned}$$

Setting each term to zero yields the system of equations

$$\left\{\begin{aligned}&2 a a_1 -3=0\\ &4 a^2 \left(4 a a_2-3 a_1^2\right)+105=0\\ &3465-8 a^3 \left(8 a^2 a_3-12 a a_1 a_2+5 a_1^3\right)=0\\ &\dots\end{aligned}\right.$$

Solving it gives

$$a_1=\frac{3}{2 a},\,a_2=-\frac{39}{8 a^3},\,a_3=\frac{657}{16 a^5},\,\dots$$

so we obtain the series

$$\begin{aligned}&\int_{-\infty}^{+\infty} e^{-a x^2-\varepsilon x^4} dx \\ =&\int_{-\infty}^{+\infty} e^{-\left(a+\frac{3}{2}\frac{\varepsilon}{a}-\frac{39}{8}\frac{\varepsilon^2}{a^3}+\frac{657}{16}\frac{\varepsilon^3}{a^5}\dots\right)x^2} dx\\=&\sqrt{\frac{\pi}{a+\frac{3}{2}\frac{\varepsilon}{a}-\frac{39}{8}\frac{\varepsilon^2}{a^3}+\frac{657}{16}\frac{\varepsilon^3}{a^5}\dots}}\end{aligned}\tag{5}$$

This is an asymptotic expression for the original integral. Formula $(5)$ performs somewhat better than formula $(3)$; compared with $(4)$, when $\varepsilon/a^2$ is small, truncating the first few terms of $(5)$ and computing gives more significant digits of accuracy than formula $(4)$. Of course, it fails when $a\to 0$ or $\varepsilon\to \infty$ — but even when it fails, it simply gives a finite zero rather than some other indeterminate result. That's one of the nice features of this kind of expansion.

The significance of asymptotic series

As already mentioned, when you take infinitely many terms, the radius of convergence of an asymptotic series is actually zero — so what use is such a series? In fact, first of all, for numerical approximation purposes, you can truncate the first few terms of the asymptotic series to get a fairly good approximate value over some range of the independent variable. Second, for certain asymptotic series, we can use resummation techniques to recover the original series and thereby obtain an exact solution — there are plenty of examples of this in quantum field theory!

So we do need asymptotic series, and at the same time we can think about ways to improve them (i.e., slow down their rate of divergence). There are all sorts of techniques for doing this, limited only by how much creativity you can bring to bear. In the next post, we'll demonstrate this once again.