Gaussian-type integrals: perturbative expansion (III)

高斯型积分的微扰展开（三） 苏剑林 · 2015-04-26 · Mathematics integral perturbation perturbation

Choosing a different small parameter

Comparing the two posts 《Gaussian-type integrals: perturbative expansion (I)》 and 《Gaussian-type integrals: perturbative expansion (II)》, we can draw two conclusions about the integral

$$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx\tag{1}$$

First, we found that approximations of the form $(4)$ have good properties, giving a reasonably reliable approximation for any $\varepsilon$. Second, we found that expanding order by order inside the exponent gives a better series than expanding directly as a power series. So can the two ideas be combined?

Let's rewrite $(4)$ as

$$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx\approx\sqrt{\frac{2\pi}{a+\sqrt{a^2+6 \varepsilon}}}=\sqrt{\frac{\pi}{a+\frac{1}{2}\left(\sqrt{a^2+6 \varepsilon}-a\right)}}\tag{6}$$more

where

$$\lambda=\frac{1}{2}\left(\sqrt{a^2+6 \varepsilon}-a\right)\tag{7}$$

is itself a small quantity (when $\varepsilon$ is small). So why not use $\lambda$ as the small parameter to expand inside the exponent? In fact, this amounts to letting

$$\varepsilon=\frac{2}{3}\left(\lambda^2+a\lambda\right)\tag{8}$$

Substituting into $(1)$ gives

$$\int_{-\infty}^{+\infty} e^{-ax^2-\frac{2}{3}\left(\lambda^2+a\lambda\right) x^4} dx\tag{9}$$

Expanding inside the exponent, i.e.

$$\int_{-\infty}^{+\infty} e^{-ax^2-\frac{2}{3}\left(\lambda^2+a\lambda\right) x^4} dx=\int_{-\infty}^{+\infty} e^{-\left(a+a_1 \lambda+a_2 \lambda^2+\dots \right)x^2} dx\tag{10}$$

expand as a power series in $\lambda$, and matching coefficients term by term gives

$$\left\{\begin{aligned}&a_1 - 1=0\\ &12 a a_2-9 a_1^2+23=0\\ &-24 a^2 a_3+36 a a_1 a_2-15 a_1^3+245=0\\ &\dots\end{aligned}\right.$$

Solving these we get

$$\left\{\begin{aligned}&a_1 = 1\\ &a_2=-\frac{7}{6 a}\\ &a_3=\frac{47}{6 a^2}\\ &\dots\end{aligned}\right.$$

and so we obtain

$$\begin{aligned}&\int_{-\infty}^{+\infty} e^{-ax^2-\frac{2}{3}\left(\lambda^2+a\lambda\right) x^4} dx\\ =&\int_{-\infty}^{+\infty} e^{-\left(a+\lambda-\frac{7}{6 a}\lambda^2+\frac{47}{6 a^2}\lambda^3+\dots \right)x^2} dx\\ =&\sqrt{\frac{\pi}{a+\lambda-\frac{7}{6 a}\lambda^2+\frac{47}{6 a^2}\lambda^3+\dots}}\\ =&\sqrt{\frac{\pi}{a+\frac{1}{2}\left(\sqrt{a^2+6 \varepsilon}-a\right)-\frac{7}{24 a}\left(\sqrt{a^2+6 \varepsilon}-a\right)^2+\dots}}\end{aligned}\tag{11}$$

which is a considerable improvement on the result of $(5)$. Of course, this does not change the fact that the final result still diverges when $a\to 0$; it is still an asymptotic series, just with a different order of divergence.

An even better parameterization

Looking back at the process above, all we really did was apply a quadratic transformation to the parameter:

$$\varepsilon=\frac{2}{3}\left(\lambda^2+a\lambda\right)\tag{8}$$

But are these two coefficients actually optimal? Let's consider a general quadratic transformation

$$\varepsilon=b_2 \lambda^2+b_1\lambda\tag{12}$$

Substituting into $(1)$ and repeating the above procedure, we obtain

$$\left\{\begin{aligned}&2 a a_1-3 b_1=0\\ &16 a^3 a_2-12 a^2 \left(a_1^2+2 b_2\right)+105 b_1^2=0\\ &64 a^5 a_3-96 a^4 a_1 a_2+40 a^3 a_1^3+840 a^2 b_1 b_2-3465 b_1^3=0\\ &\dots\end{aligned}\right.$$

Solving gives

$$\left\{\begin{aligned}&a_1 = \frac{3 b_1}{2 a}\\ &a_2=\frac{3 \left(4 a^2 b_2-13 b_1^2\right)}{8 a^3}\\ &a_3=-\frac{3 \left(52 a^2 b_1 b_2-219 b_1^3\right)}{16 a^5}\\ &\dots\end{aligned}\right.$$

Taking $b_1=2a,b_2=13$, we get $a_1=3,a_2=0$, meaning that if we make the transformation

$$\varepsilon=13 \lambda^2+2\lambda\tag{13}$$

or

$$\lambda=\frac{\sqrt{a^2+13\varepsilon}-a}{13}$$

then we obtain

$$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx\approx\int_{-\infty}^{+\infty} e^{-(a+3\lambda)x^2}dx=\sqrt{\frac{13\pi}{10a+3\sqrt{a^2+13\varepsilon}}}\tag{14}$$

This is a better result than $(4)$: it is exact up to second order, and it retains the good properties possessed by $(4)$.

$$\begin{array}{c|cccc} \hline \text{case} & \varepsilon=1,a=10 & \varepsilon=0.1,a=1 & \varepsilon=1,a=1 & \varepsilon=1,a=0 \\ \hline \text{exact value} & 0.556466 &1.67409 &1.36843 &1.8128 \\ (4) & 0.556402 &1.66558 & 1.31279 & 1.60159 \\ (14) & 0.556468 & 1.6754 & 1.38715 & 1.94312 \\ \hline \end{array}$$

In fact, the whole process is really a journey of creativity — apply every trick you can think of to the expansion, and the only limit is your imagination.