Perturbation Expansion of Gaussian-type Integrals (III)

Changing to a Small Parameter

By comparing the articles "Perturbation Expansion of Gaussian-type Integrals (I)" and "Perturbation Expansion of Gaussian-type Integrals (II)", we can draw two conclusions regarding the integral \int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx \tag{1} First, we found that approximate results similar to equation (4) possess good properties, providing a relatively reliable approximation for any \varepsilon. Second, we found that expanding order-by-order within the exponent yields better results than a direct expansion into a power series. So, can these two approaches be combined?

We rewrite equation (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} where \lambda = \frac{1}{2}\left(\sqrt{a^2+6 \varepsilon}-a\right) \tag{7} is actually also a small quantity (when \varepsilon is very small). Therefore, why not expand in the exponent using \lambda as the small parameter? In fact, this is equivalent to letting \varepsilon = \frac{2}{3}\left(\lambda^2+a\lambda\right) \tag{8} Substituting this into (1), we have \int_{-\infty}^{+\infty} e^{-ax^2-\frac{2}{3}\left(\lambda^2+a\lambda\right) x^4} dx \tag{9} Expanding in the exponent: \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} Expanding as a power series in \lambda and comparing the coefficients of each term, we obtain: \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 gives: \left\{\begin{aligned} &a_1 = 1\\ &a_2=-\frac{7}{6 a}\\ &a_3=\frac{47}{6 a^2}\\ &\dots \end{aligned}\right. Thus, we have: \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} This is much better than the result of equation (5). Of course, it does not change the fact that the final result diverges as a \to 0; it remains an asymptotic series, though the order of infinity has been modified.

Better Parameters

Reviewing the process above, we simply performed a quadratic transformation on the parameter: \varepsilon = \frac{2}{3}\left(\lambda^2+a\lambda\right) \tag{8} However, are these two coefficients optimal? Let us consider a general quadratic transformation: \varepsilon = b_2 \lambda^2 + b_1 \lambda \tag{12} Substituting this into equation (1) and repeating the process above, we get: \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 these 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. By choosing b_1=2a and b_2=13, we can obtain a_1=3 and a_2=0. That is to say, if we make the transformation \varepsilon = 13 \lambda^2 + 2a\lambda \tag{13} or \lambda = \frac{\sqrt{a^2+13\varepsilon}-a}{13} then we can 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 equation (4); it is accurate to the second order and retains the favorable properties of equation (4).

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

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