Understanding the divergence of an asymptotic series

Question

In class, we recently developed an asymptotic expansion for a perturbed Gaussian integral and I was wondering why it was divergent.

First, the $e^{-gx^4}$ term was Taylor expanded, then integral and summation swapped, and finally the formula of the $n$-th momentum of a Gaussian used to obtain: $$ \int_{-\infty}^{+\infty} e^{-x^2-gx^4} dx \rightarrow \sum_{n=0}^\infty \frac{(-g)^n}{n!} \int_{-\infty}^{+\infty} e^{-x^2} x^{4n}=\sqrt{\pi}\sum_{n=0}^{\infty}\frac{(-g)^n\cdot(4n-1)!!}{n!\cdot2^{2n}} $$ By the Ratio Test this diverges for all $g\neq0$ because $\frac{a_{n+1}}{a_n}=\frac{(-g)\cdot(4n+3)(4n-1)}{(n+1)\cdot2^2}\rightarrow\infty$.

So far I had only came across a bunch of asymptotic expansions and developed the (probably unjustified) intuition that the divergence of the asymptotics came from expanding a function beyond its radius of convergence: $$ \int_0^\infty \frac{e^{-x}}{1+x}dx=\sum_{n=0}^\infty(-1)^n\int_0^\infty e^{-x}x^ndx=\textrm{divergent} $$

because the expansion $\frac{1}{1+x}=\sum (-x)^n$ is not justified for the integration domain $\int_1^\infty$.

However, in the cause of the perturbed Gaussian, the Taylor expansion of $e^{-gx^4}$ is well justified for $\forall x$ and for $g\ge0$ the integral converges. I was expecting the series to converge for $g\ge 0$.

Even though I am aware that limits cannot arbitrarily be interchanged ($\int\sum^\infty \neq \sum^\infty\int$), I was wondering whether the divergence of the Gaussian series would admit some "intuitive" explanation appart from that.

Answer 1

Not sure if anybody would be interested but just in case...

TL;DR Interchanging $\int^\infty\lim_{N\to\infty}\sum^N \Rightarrow \lim_{N\to\infty}\int^\infty\sum^N$ adds an extra term (the bump) to the integrand caused by the "explosion" of the Taylor series at large $x$. Given that the area of the bump is divergent as $N\to\infty$ the series diverges.

I would blame the Taylor expansion $e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$ for being not uniformly convergent. Roughly speaking, if you decide to cut off the Taylor expansion of the exponential at some order $N$, the series $1+x+\frac{x^2}{2!}+\dots$ will only approximate $e^x$ on the finite range $(-\sqrt[N]{N!\ },+\sqrt[N]{N!\ })$. Beyond this point it 'explodes' and it would go like $\sim\frac{x^N}{N!}$.

In fact, cutting the Taylor expansion $e^{-x^2}e^{-gx^4}\rightarrow e^{-x^2}\sum_{n=0}^N\frac{(-gx^4)^n}{n!}$ at order $N$ results in the desired function in the range $(-\left(\frac{\sqrt[N]{N!\ }}{g}\right)^{1/4},+\left(\frac{\sqrt[N]{N!\ }}{g}\right)^{1/4})$ and an extra BIG bump starting at $x=\left(\frac{\root{N}\of{N!\ }}{g}\right)^{1/4}$: Therefore, truncating the Taylor expansion inside the integral for large $N$ gives $$\int_0^\infty e^{-x^2}\sum_{n=0}^N\frac{(-gx^4)^n}{n!}=\int_0^\infty e^{-x^2}e^{-gx^4}+\textrm{\{area of the bump\}}$$ where the area of the bump is roughly $\sim \int_0^\infty e^{-x^2} \cdot\frac{(-gx^4)^{N}}{N!} = \frac{(-g)^N}{N!} \frac{(4N-1)!!}{2^{2N}}$ which $\to\infty$ for large $N$.

Consequently the divergence is caused by $$\text{area of }e^{-x^2-gx^4} = \int_0^\infty e^{-x^2}\lim_{N\to\infty}\sum_{n=0}^N\frac{(-gx^4)^n}{n!}$$ $$\neq \lim_{N\to\infty}\int_0^\infty e^{-x^2}\sum_{n=0}^N\frac{(-gx^4)^n}{n!}\ =\ \text{area of }e^{-x^2-gx^4}\ +\ \text{\{area of the bump\}}$$

Answer 2

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\begin{align} &\bbox[5px,#ffd]{\int_{-\infty}^{\infty} \expo{-x^{2}\ -\ gx^{4}}\,\,\dd x} \,\,\,\stackrel{\mrm{as}\ g\ \to \infty}{\sim}\,\,\, 2\int_{0}^{\infty} \expo{-\ gx^{4}}\pars{1 - x^{2}}\,\,\dd x \\[5mm] = &\ 2\Gamma\pars{5 \over 4}{1 \over g^{1/4}} - {1 \over 2}\,\Gamma\pars{3 \over 4}{1 \over g^{3/4}} \end{align}