By 苏剑林 | April 26, 2015
Comparing the two articles "Perturbation Expansion of Gaussian-type Integrals (Part I)" and "Perturbation Expansion of Gaussian-type Integrals (Part II)", we can derive two conclusions regarding the integral:
\[\int_{-\infty}^{+\infty} e^{-ax^2-\varepsilon x^4} dx\tag{1}\]First, we found that approximation results similar to equation (4) possess good properties, yielding a relatively reliable approximation for any $\varepsilon$. Second, we found that expanding term by term within the exponent produces better results than directly expanding as a power series. So, can we combine these two approaches?
Let us 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 small). Therefore, why not use $\lambda$ as a small parameter to expand within the exponent? 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 inside 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 then 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 yields:
\[\left\{\begin{aligned}&a_1 = 1\\ &a_2=-\frac{7}{6 a}\\ &a_3=\frac{47}{6 a^2}\\ &\dots\end{aligned}\right.\]So 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 result is much better than that of equation (5). Of course, it does not change the fact that the final result diverges as $a \to 0$. At the same time, it remains an asymptotic series, merely with a change in the order of the divergence.
Looking back at 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 the most 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:
\[\left\{\begin{aligned}&a_1 = \frac{3 b_1}{2 a}\\ &a_2=\frac{3 \left(4 a^2 b_2-13 b_1^2\right)}{8 a^3}\\ &a_3=-\frac{3 \left(52 a^2 b_1 b_2-219 b_1^3\right)}{16 a^5}\\ &\dots\end{aligned}\right.\]Setting $b_1=2a$ 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 same good properties that equation (4) possessed.
| 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 |
| (4) | 0.556402 | 1.66558 | 1.31279 | 1.60159 |
| (14) | 0.556468 | 1.6754 | 1.38715 | 1.94312 |
In fact, the entire process is a voyage of creativity—apply every technique you can think of to the expansion, and the limit is only your imagination.