Perturbative Expansion of Gaussian-Type Integrals (I)

高斯型积分的微扰展开（一） 苏剑林 · 2015-02-14 · Mathematics integral perturbation perturbation asymptotic series

Some time ago I was studying Feynman's path integral theory, and came across the perturbative method used in path integrals — that is, the technique of gradually approximating the propagator by expanding in a small parameter. This kind of technique has a very clear physical meaning; readers interested in learning about path integrals and quantum mechanics are encouraged to read Feynman's Quantum Mechanics and Path Integrals. From a mathematical point of view, however, this approximation technique is actually quite crude, and the range and rate of convergence are hard to guarantee. In fact, mathematics has developed all sorts of perturbation techniques to deal with different kinds of perturbations. Below we study the integral

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

or, more generally,

$$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon V(x)} dx\tag{2}$$

The series expansion in path integrals is slightly more complicated than this, but still takes a similar form. more

A simple, direct expansion

The simplest, most direct way to compute the integral above is to expand $e^{-\varepsilon V(x)}$ as a series

$$e^{-\varepsilon V(x)}=\sum_{n=0}^{\infty} (-1)^n \frac{1}{n!}\varepsilon^n V^n (x)$$

giving

$$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon V(x)} dx=\sum_{n=0}^{\infty} (-1)^n\frac{1}{n!}\varepsilon^n \int_{-\infty}^{+\infty} e^{-ax^2} V^n (x)dx$$

If the definite integral of $e^{-ax^2}V^n (x)$ is easy to compute, then we get an approximation to the original integral. For instance, in the case of $V(x)=x^4$ we have

$$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx=\sqrt{\frac{\pi}{a}}\left(1-\frac{3\varepsilon}{4a^2}+\frac{105\varepsilon^2}{32a^4}-\frac{3465\varepsilon^3}{128a^6}+\dots\right)\tag{3}$$

This series completely breaks down once $\varepsilon=a^2$, and even when $\varepsilon < a^2$, the convergence rate is very slow. In fact, this is an asymptotic series; strictly speaking, its region of convergence is exactly $\varepsilon=0$! Of course, we can take its first few terms to approximate the integral for relatively large (compared to 0) values of $\varepsilon$. The "asymptotic" nature of an asymptotic series means: the fewer terms you take, the larger the radius over which the computation is effective — it will compute something, but the precision isn't necessarily high; the more terms you take, the more accurate it naturally becomes, but the effective radius shrinks. Taking infinitely many terms, the radius of convergence is reduced to 0.

Expanding inside the exponent: a closed-form solution

When studying perturbative expansions of differential equations, in order to obtain periodic solutions we not only expand the solution in the small parameter, but also expand the period itself in the small parameter. This inspired me to wonder: could we instead expand the original coefficient as a parameter? Concretely, let us introduce $A$

$$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx=\int_{-\infty}^{+\infty} e^{-Ax^2-[(a-A)x^2+\varepsilon x^4]} dx$$

Now let's expand the term $(a-A)x^2+\varepsilon x^2$ inside the square brackets to first order in $V(x)$, giving

$$\int_{-\infty}^{+\infty} e^{-Ax^2} \left\{1-[(a-A)x^2+\varepsilon x^4]\right\}dx$$

Compute

$$\int_{-\infty}^{+\infty} e^{-Ax^2}[(a-A)x^2+\varepsilon x^4]dx=\sqrt{\frac{\pi}{A}}\frac{2 A (a-A)+3 \varepsilon}{4 A^2}$$

The key step comes next: choose an appropriate $A$ so that this term equals $0$, that is,

$$2 A (a-A)+3 \varepsilon=0\,\to\, A=\frac{1}{2} \left(a+\sqrt{a^2+6 \varepsilon}\right)$$

so that

$$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx\approx\int_{-\infty}^{+\infty} e^{-Ax^2} =\sqrt{\frac{\pi}{A}}=\sqrt{\frac{2\pi}{a+\sqrt{a^2+6 \varepsilon}}}\tag{4}$$

This is an excellent approximation. How good is it? Almost unbelievably good! Consider the most extreme case: $a=0$, for which the formula above gives

$$\int_{-\infty}^{+\infty} e^{-\varepsilon x^4} dx\approx\sqrt{\frac{2\pi}{\sqrt{6 \varepsilon}}}$$

This result not only gives an approximation to the integral, but even gets the scaling right — that is, the exact value of $\int_{-\infty}^{+\infty} e^{-\varepsilon x^4} dx$ is indeed a constant times $\frac{1}{\varepsilon^{1/4}}$! For various combinations of $a,\varepsilon$, the value computed this way is better than the first-order approximation given by formula $(3)$ (sometimes even better than its higher-order approximations). Another extreme case is $\varepsilon\to+\infty$, where the exact value of the integral is 0, and $(4)$ manages to reproduce this result as well, whereas $(3)$ once again completely fails.

The table below shows the exact value, the first-order approximation $\sqrt{\frac{\pi}{a}}\left(1-\frac{3\varepsilon}{4a^2}\right)$ obtained by formula $(3)$, and the result computed via $(4)$. From the table we can see that formula $(4)$ outperforms the first-order approximation of formula $(3)$ in every case.

$$\begin{array}{c|cccc} \hline \text{case} & \varepsilon=1,a=10 & \varepsilon=0.1,a=1 & \varepsilon=1,a=1 & \varepsilon=1,a=0 \\ \hline \text{exact value} & 0.556466 &1.67409 &1.36843 &1.8128 \\ \sqrt{\frac{\pi}{a}}\left(1-\frac{3\varepsilon}{4a^2}\right) & 0.556295 &1.63952 & 0.443113 & \text{negative infinity} \\ \sqrt{\frac{2\pi}{a+\sqrt{a^2+6 \varepsilon}}} & 0.556402 &1.66558 & 1.31279 & 1.60159 \\ \hline \end{array}$$

Although formula $(4)$ is superior in both stability and precision, it is still essentially a first-order approximation. Later we will continue on to discuss its higher-order approximations.