Prove this limit of ratios of divergent integrals Define
$$ m_{n}(a)\equiv\int_{-\infty}^{\infty}x^{n}e^{x^{2}-ax^{4}}dx $$
I would like to prove that
$$ \lim_{a\to 0^+} \frac{m_{2}^{2}(a)}{m_{0}(a)m_{4}(a)} = 1. $$
This limit bears out numerically, but I can't figure out how to prove it analytically. All of the integrals individually diverge. $a$ is real and positive.
OP Edit:
I don't know if this helps, but since both limits of the integrand are zero when $a>0$, we can deduce that:
$$\int_{-\infty}^{\infty}\frac{d}{dx}\left(x^{n}e^{x^{2}-ax^{4}}\right)dx=0$$
$$\Rightarrow2m_{n+1}(a)-4am_{n+3}(a)+nm_{n-1}(a)=0$$
$$\Rightarrow2m_{2}(a)-4am_{4}(a)+m_{0}(a)=0$$
This relation felt like progress.
 A: Let $n = 2k$ be even and we first make a simple change of variables $x \mapsto \sqrt{x}$ to write
$$ m_{2k}(a) = \int_{0}^{\infty} x^{k-\frac{1}{2}} e^{-(ax^2 - x)} \, dx. $$
Now set $u^2 = 4a(ax^2 - x) + 1 = (2ax - 1)^2$. When $u \geq 0$, this relation gives two inverses
$$ x_{+}(u) = \frac{1 + u}{2a} \quad \text{for } u \in [0,\infty), \qquad x_{-}(u) = \frac{1 - u}{2a} \quad \text{for } u \in [0,1]. $$
Now splitting the integral into two parts, one on $[0, \frac{1}{2a}]$ and the other on $[\frac{1}{2a}, \infty)$, and applying the substitution $x = x_{\pm}(u)$ appropriately,
\begin{align*}
m_{2k}(a)
&= \int_{0}^{\frac{1}{2a}} x^{k-\frac{1}{2}} e^{-(ax^2 - x)} \, dx + \int_{\frac{1}{2a}}^{\infty} x^{k-\frac{1}{2}} e^{-(ax^2 - x)} \, dx \\
&= \int_{0}^{1} \left(\frac{1-u}{2a}\right)^{k-\frac{1}{2}} e^{-\frac{u^2-1}{4a}} \, \frac{du}{2a} + \int_{0}^{\infty} \left(\frac{1+u}{2a}\right)^{k-\frac{1}{2}} e^{-\frac{u^2-1}{4a}} \, \frac{du}{2a} \\
&= \frac{e^{1/4a}}{(2a)^{k+\frac{1}{2}}} \int_{0}^{\infty} \left( (1-u)_{+}^{k-\frac{1}{2}} + (1+u)^{k-\frac{1}{2}} \right) e^{-\frac{u^2}{4a}} \, du.
\end{align*}
Here $x_+ = \max\{0, x\}$ denotes the positive part of $x$. Symmetrizing the last integral and rewriting a little bit, we finally obtain the expression
$$ m_{2k}(a) = \frac{\sqrt{2\pi} e^{1/4a}}{2^{k+1} a^k} \int_{-\infty}^{\infty} \left( (1-|u|)_{+}^{k-\frac{1}{2}} + (1+|u|)^{k-\frac{1}{2}} \right) \frac{1}{\sqrt{4\pi a}} e^{-\frac{u^2}{4a}} \, du. $$
Notice that the last integral involves the Gaussian kernel which 'converges' to $\delta_0(x)$ as $a \to 0^+$. This tells that
$$ m_{2k}(a) = \frac{\sqrt{2\pi} e^{1/4a}}{(2a)^k} (1 + o(1)) $$
as $a \to 0^+$. Plugging this back, all the term cancels out to prove that the limit is indeed $1$.
I know little about physics theory, but somehow the exponent reminds me of the $\phi^4$-theory. If it is indeed related to the theory, I guess they might have a fancy way of explaining the limit.
