Perturbative Expansion of Gaussian-type Integrals (I)

By 苏剑林 | February 14, 2015

Some time ago, while researching Feynman's path integral theory, I encountered perturbative methods for path integrals—approximating the propagator step-by-step through expansions in small parameters. Such techniques carry very clear physical meaning; readers interested in learning about path integrals and quantum mechanics should read Feynman's Quantum Mechanics and Path Integrals. However, from a mathematical perspective, this approximation technique is actually quite crude, and its convergence range and speed are often difficult to guarantee. In fact, mathematics has developed various perturbation (perturbation) techniques to handle different scenarios. 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 for path integrals is slightly more complex than this, but it follows a similar form.

Simple and Direct Expansion

To calculate the above integral, the simplest and most direct method is to perform a series expansion of $e^{-\varepsilon V(x)}$: $$e^{-\varepsilon V(x)}=\sum_{n=0}^{\infty} (-1)^n \frac{1}{n!}\varepsilon^n V^n (x)$$ Thus: $$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon V(x)} dx=\sum_{n=0}^{\infty} (-1)^n\frac{1}{n!}\varepsilon^n \int_{-\infty}^{+\infty} e^{-ax^2} V^n (x)dx$$ If the definite integral of $e^{-ax^2}V^n (x)$ is easy to compute, we can obtain an approximation of 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 fails completely when $\varepsilon=a^2$, and even if $\varepsilon < a^2$, the convergence speed is very slow. In fact, this is an asymptotic series; strictly speaking, its convergence radius is zero! Of course, we can take the first few terms to perform approximate calculations for relatively large $\varepsilon$ (relative to 0). The "asymptotic" nature of an asymptotic series means that the fewer terms you take, the larger the effective calculation radius, but the accuracy may not be high; the more terms you take, the more accurate it becomes locally, but the effective radius shrinks. If you take infinitely many terms, the convergence radius is exactly 0.

Expansion in the Exponent: A Closed-form Solution

In studying the perturbative expansion of differential equations, in order to obtain periodic solutions, we not only expand the solution in terms of a small parameter but often also expand the period itself in terms of the small parameter. This inspired me: can we expand the original coefficients as parameters? Specifically, we introduce a parameter $A$: $$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx=\int_{-\infty}^{+\infty} e^{-Ax^2-[(a-A)x^2+\varepsilon x^4]} dx$$ Now, treating the term in the brackets $(a-A)x^2+\varepsilon x^4$ as $V(x)$ and performing a first-order expansion, we get: $$\int_{-\infty}^{+\infty} e^{-Ax^2} \left\{1-[(a-A)x^2+\varepsilon x^4]\right\}dx$$ Calculating the integral: $$\int_{-\infty}^{+\infty} e^{-Ax^2}[(a-A)x^2+\varepsilon x^4]dx=\sqrt{\frac{\pi}{A}}\frac{2 A (a-A)+3 \varepsilon}{4 A^2}$$ The key next step is to choose an appropriate $A$ such that this term becomes $0$, namely: $$2 A (a-A)+3 \varepsilon=0\,\to\, A=\frac{1}{2} \left(a+\sqrt{a^2+6 \varepsilon}\right)$$ Therefore: $$\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx\approx\int_{-\infty}^{+\infty} e^{-Ax^2} =\sqrt{\frac{\pi}{A}}=\sqrt{\frac{2\pi}{a+\sqrt{a^2+6 \varepsilon}}}\tag{4}$$ This is an exceptionally good approximation. How good? Incredibly so! Consider the most extreme case: $a=0$. The above formula gives: $$\int_{-\infty}^{+\infty} e^{-\varepsilon x^4} dx\approx\sqrt{\frac{2\pi}{\sqrt{6 \varepsilon}}}$$ This equation not only provides an approximate value for the integral but even the scaling is correct—the exact value of $\int_{-\infty}^{+\infty} e^{-\varepsilon x^4} dx$ is indeed form of a constant multiplied by $\frac{1}{\varepsilon^{1/4}}$! For various combinations of $a, \varepsilon$, the values calculated are better than the first-order approximation of equation $(3)$ (sometimes they are even better than its higher-order approximations). Another extreme case is $\varepsilon\to+\infty$, where the exact value of the integral is 0, and $(4)$ can actually yield this result, whereas $(3)$ fails completely again.

The following table compares the exact values, the first-order approximation from equation $(3)$ $\sqrt{\frac{\pi}{a}}\left(1-\frac{3\varepsilon}{4a^2}\right)$, and the calculation results from equation $(4)$. From the table, it is evident that formula $(4)$ is superior to the first-order approximation of formula $(3)$ in all scenarios.

Although formula $(4)$ is superior in stability and accuracy, it remains fundamentally a first-order approximation. We will discuss its higher-order approximations in the future.