Some time ago, while studying Feynman’s path integral theory, I encountered the perturbation method for path integrals, which involves approximating the propagator step-by-step through expansion in terms of a small parameter. Such a technique has very clear physical significance. Readers interested in learning about path integrals and quantum mechanics are encouraged to read Feynman’s Quantum Mechanics and Path Integrals. However, from a mathematical perspective, this approximation technique is actually quite crude, and its range and speed of convergence are difficult to guarantee. In fact, various perturbation techniques have been developed in mathematics to handle different types 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 for path integrals is slightly more complex than this, but it remains in a similar form.
Simple and Direct Expansion
To calculate the aforementioned 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 example, in the case where 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 region of convergence is only at \varepsilon=0! 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 taken, the larger the effective radius of calculation, allowing for computation but not necessarily with high precision; the more terms taken, the more accurate it naturally becomes, but the effective radius decreases. When taking infinitely many terms, the radius of convergence is zero.
Expansion in the Exponent: A Closed-form Solution
When studying the perturbation expansion of differential equations, in order to obtain periodic solutions, we not only expand the solution in terms of a small parameter but also expand the period itself in terms of the small parameter. Inspired by this, I wondered: 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 (a-A)x^2+\varepsilon x^4 in the square brackets 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 next step is crucial: choose an appropriate A such that this term becomes 0, i.e., 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 excellent approximation. How good? Unbelievably good! 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}}} The formula above not only provides an approximate value for the integral but even the scaling is correct. That is to say, the exact value of \int_{-\infty}^{+\infty} e^{-\varepsilon x^4} dx is in the form of a constant multiplied by \frac{1}{\varepsilon^{1/4}}! For different combinations of a and \varepsilon, the calculated values are better than the first-order approximation of Eq. (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 Eq. (4) can actually yield this result, whereas Eq. (3) fails completely once again.
The table below shows the exact values, the results of the first-order approximation of Eq. (3) \sqrt{\frac{\pi}{a}}\left(1-\frac{3\varepsilon}{4a^2}\right), and the results of Eq. (4). As can be seen from the table, Eq. (4) is superior to the first-order approximation of Eq. (3) in all scenarios.
| Case | \varepsilon=1, a=10 | \varepsilon=0.1, a=1 | \varepsilon=1, a=1 | \varepsilon=1, a=0 |
|---|---|---|---|---|
| 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 | -\infty |
| \sqrt{\frac{2\pi}{a+\sqrt{a^2+6 \varepsilon}}} | 0.556402 | 1.66558 | 1.31279 | 1.60159 |
Although the stability and accuracy of Eq. (4) are superior, it is essentially still a first-order approximation. We will continue to discuss its higher-order approximations later.
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