Why is the second part so late?
In fact, before starting this series, I had already conceptualized the content for the next few articles. I originally intended to confidently introduce some integral expansion techniques I had thought of. Since I am quite familiar with perturbation methods, under normal circumstances, the second part would not have been this delayed. However, after finishing the first part and while preparing to write the second, I came across this response on Zhihu:
http://www.zhihu.com/question/24735673
This article greatly expanded my understanding of series. It discussed how the expansion of an integral is an asymptotic series. This made me hesitate, questioning whether this series had any value, because an asymptotic series implies that no matter what expansion technique is used, the resulting series has a radius of convergence of zero.
Later, upon further reflection, I realized that even for asymptotic series, there is room for improvement and methods to accelerate convergence. Therefore, I believe these articles still hold some significance, and I can take the opportunity to introduce theories related to asymptotic series and singularities. Well, let’s proceed with that plan.
Step-by-step Expansion in the Exponent
The technique mentioned in the previous article involved the variation of constant coefficients—adjusting variables to make the first-order term zero. The main issue with that method is that it cannot solve for higher-order approximations step-by-step. However, its underlying idea is excellent: introducing adjustable variables to make various orders of the expansion zero, rather than performing a pure power series expansion. Below, we will first try to place the expansion within the exponent to obtain a method that allows for step-by-step approximation. This will serve as a primitive model for the methods we intend to develop further.
In general, we seek to evaluate the integral: \int_{-\infty}^{+\infty} e^{-a x^2-\varepsilon x^4} dx
In the previous article, we hoped to approximate it using the integral \int_{-\infty}^{+\infty} e^{-A x^2} dx. The approach was to adjust A to make the first-order term zero. However, we could only go that far because there was only one adjustable variable; generally, one can only make one term zero. To achieve a step-by-step expansion—that is, to introduce multiple adjustable quantities—we set: A=a+a_1 \varepsilon + a_2\varepsilon^2 + a_3 \varepsilon^3 + \dots Then we 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 brackets 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^3+\dots \end{aligned} Integrating term by term, we obtain: \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, we get 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 these yields: a_1=\frac{3}{2 a},\,a_2=-\frac{39}{8 a^3},\,a_3=\frac{657}{16 a^5},\,\dots Thus, we have 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 another asymptotic expression for the original integral. Equation (5) performs better than Equation (3); compared to Equation (4), when \varepsilon/a^2 is very small, truncating the first few terms of (5) for calculation can yield more significant digits than Equation (4). Of course, it fails as a \to 0 or \varepsilon \to \infty—but even when it fails, it merely results in a finite zero rather than other indeterminate results. This is one of the advantages of this expansion method.
The Significance of Asymptotic Series
As mentioned, when taking infinitely many terms, the radius of convergence of an asymptotic series is actually zero. So, what is the significance of an asymptotic series? In fact, first, for approximate calculations, one can truncate the first few terms of an asymptotic series to obtain a relatively good approximation for a certain range of the independent variable. Second, for some asymptotic series, we can use resummation techniques to restore the original series and obtain an exact solution—there are numerous examples of this in quantum field theory!
Therefore, we need asymptotic series, and we can also consider ways to improve them (i.e., reducing the rate of divergence). There are various techniques to achieve this, provided you have enough creativity. In the next article, we will demonstrate this once more.
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