Asymptotics: Motivation

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To make progress in perturbation theory, we need to introduce the notion of ‘asymptotic series’. To introduce this concept, we’ll consider the following integral:

$\displaystyle I(x) = \int_{0}^{\infty} \frac{e^{-xt}}{1+t}\, dt \quad (\mathrm{Re}(x) > 0)$

Using

$\displaystyle \frac{1}{1+t} = 1-t+t^2-t^3+\cdots, \qquad |t|<1$

we obtain the asymptotic expansion

$\displaystyle I(x) \sim \frac{0!}{x}-\frac{1!}{x^2}+\frac{2!}{x^3}-\cdots$

$\displaystyle I(x) \sim \sum_{n=0}^{\infty}(-1)^n\frac{n!}{x^{n+1}}, \qquad x\to+\infty$

This series is divergent for all $x\neq 0$ , since its terms do not tend to zero. Let us evaluate the first 5 terms of the series for $x=10$ and compare the result to a numerical solution $I(10)\approx 0.09156$ :

$\displaystyle 0.1-0.01+0.002-0.0006+0.00024=0.09164$

We observe that the first 5 terms of the divergent series give a good approximation of the value obtained using numerical methods. We find ourselves in the case of astronomers’ series as presented by H. Poincaré.

We can ask ourselves how many terms we need to use to achieve a certain level of accuracy in our calculations. To do this, we reformulate the integral above using a partial sum and a remainder $R_n(x)$ :

$\displaystyle I(x)=\sum_{k=0}^{n-1}(-1)^k\frac{k!}{x^{k+1}}+R_n(x)$

where

$\displaystyle R_n(x)=I(x)-\sum_{k=0}^{n-1}(-1)^k\frac{k!}{x^{k+1}}$

The remainder term can be seen as the error made by stopping after $n$ terms. We would like to estimate the remainder:

$\displaystyle \frac{1}{1+t}=\sum_{k=0}^{n-1}(-1)^k t^k+\frac{(-1)^n t^n}{1+t}$

Therefore,

$\displaystyle I(x)=\int_0^\infty e^{-xt}\left(\sum_{k=0}^{n-1}(-1)^k t^k\right)\,dt+\int_0^\infty\frac{(-1)^n t^n e^{-xt}}{1+t}\,dt$

The remainder is

$\displaystyle R_n(x) = \int_{0}^{\infty} \frac{(-1)^n t^n e^{-xt}}{1+t}\, dt$

$\displaystyle |R_n(x)| = \int_{0}^{\infty} \frac{t^n e^{-xt}}{1+t}\, dt$

$\displaystyle |R_n(x)| < \int_{0}^{\infty} t^n e^{-xt}\, dt$

$\displaystyle \int_{0}^{\infty} t^n e^{-xt}\, dt = \frac{n!}{x^{n+1}}, \quad x > 0$

We observe that $\frac{n!}{x^{n+1}}$ is the first omitted term in the series, so that the remainder is smaller than the first omitted term. The term $\frac{n!}{x^{n+1}}$ decreases but then eventually gets bigger. We have to stop after some terms. For example, for $x=10$ , the terms of minimum magnitude occur at indices $n=9$ and $n=10$ , so we can truncate after either 10 or 11 terms.