Asymptotic expansion of $\int_{-\infty}^{\infty} e^{ix\left(\frac{t^3}{3}+t\right)}dt $

Question

How do you find the asymptotic expansion of $\int_{-\infty}^{\infty} e^{ix\left(\frac{t^3}{3}+t\right)}dt $ as $x \to \infty$ ?

I know that I will eventually have to use the fact that $\int_{-\infty}^{\infty} e^{-xt^2}dt = \frac{\sqrt{\pi}}{\sqrt{x}}$ but not sure how to get to this point.

Any hints/help would be appreciated!

Answer 1

Let $f(x)$ denote the integral. Substituting $t = 2\sinh\left(\frac{1}{3}\operatorname{arsinh}z\right)$, the integral reduces to

\begin{align*} f(x) &= \frac{2}{3}\int_{-\infty}^{\infty}\exp\left(\frac{2xi}{3}z\right)\cosh\left(\frac{1}{3}\operatorname{arsinh}z\right) \frac{1}{\sqrt{1+z^2}} \, \mathrm{d}z \end{align*}

Noting that the integrand is $\mathcal{O}\bigl( e^{-\frac{2x}{3}\operatorname{Im}(z)} |z|^{-2/3} \bigr)$ as $|z| \to \infty$ with $\operatorname{Im}(z)>0$, we can deform the contour to a Hankel-type contour wrapping around the principal branch cut $[i,i\infty)$ of the integrand. The resulting integral is

\begin{align*} f(x) &= \frac{2}{3}\int_{i+0^+}^{i\infty+0^+}\exp\left(\frac{2xi}{3}z\right)\cosh\left(\frac{1}{3}\operatorname{arsinh}z\right) \frac{1}{\sqrt{1+z^2}} \, \mathrm{d}z \\ &\quad - \frac{2}{3}\int_{i-0^+}^{i\infty-0^+}\exp\left(\frac{2xi}{3}z\right)\cosh\left(\frac{1}{3}\operatorname{arsinh}z\right) \frac{1}{\sqrt{1+z^2}} \, \mathrm{d}z. \end{align*}

Substituting $z = it \pm 0^+$ to each of the integral, respectively,

\begin{align*} f(x) &= \frac{2}{3}\int_{1}^{\infty} \exp\left(-\frac{2x}{3}t\right)\cosh\left(\frac{i\pi}{6} + \frac{1}{3} \operatorname{arcosh}t\right) \frac{1}{\sqrt{t^2-1}} \, \mathrm{d}t \\ &\quad + \frac{2}{3}\int_{1}^{\infty} \exp\left(-\frac{2x}{3}t\right)\cosh\left(\frac{i\pi}{6} - \frac{1}{3} \operatorname{arcosh}t\right) \frac{1}{\sqrt{t^2-1}} \, \mathrm{d}t \\ &= \frac{2}{\sqrt{3}}\int_{1}^{\infty} \exp\left(-\frac{2x}{3}t\right)\cosh\left(\frac{1}{3} \operatorname{arcosh}t\right) \frac{1}{\sqrt{t^2-1}} \, \mathrm{d}t \\ &= \frac{1}{\sqrt{3}}\int_{-\infty}^{\infty} \exp\left(-\frac{2x}{3}\cosh s\right)\cosh\left(\frac{1}{3}s\right) \, \mathrm{d}s. \tag{$t=\cosh s$} \end{align*}

The asymptotic behavior of the last integral as $x \to \infty$ can be studied using Laplace's method, yielding

$$ f(x) \sim e^{-2x/3} \sqrt{\frac{\pi}{x}} \quad \text{as } x \to \infty. $$

Answer 2

If you make the problem more general, if $a>0$, $b>0$ and $x>0$, $$I=\int_{-\infty}^{+\infty}e^{i x \left(a t^3+b t\right)}\,dt=\frac{2 \pi }{ \sqrt[3]{3a x}}\text{Ai}\left(b\sqrt[3]{\frac{x^2}{3 a}}\right)$$ which, as said in comments, reduces to a Bessel K function.

Expanded for large $x$ $$I=\frac {\sqrt \pi}{\sqrt[4]{3ab}}\frac 1{\sqrt x}\exp\left(-\frac 23\sqrt{ \frac {b^3} {3a}} \,x\right)\Bigg[1+\sum_{n=1}^\infty (-1)^n\,{c_n}{t^n}\Bigg]\quad \text{with}\quad t=\frac 1 x{\sqrt{\frac{a}{3b^3}}}$$ where the first coefficients are $$\left\{\frac{5}{16},\frac{385}{512},\frac{85085}{24576},\frac{3 7182145}{1572864},\cdots\right\}$$

For your specific case, this is a quite good approximation, the relative error being $0.10$% for $x=3.40$ and $0.01$% for $x=5.60$.

Edit

It is interesting to notice if that provided that $(a,b) \in \mathbb{R}$ $$9 | a| ^{3/2} | b| ^{9/2}\,I=3 \left(\left| a b^5\right| +a b^5\right) K_{\frac{1}{3}}\left(\frac{2 | b| ^{3/2}}{3 \sqrt{3| a| }}x\right)-$$ $$\sqrt{3} \pi \left(a b^5-| a| | b| ^5\right) \left(J_{-\frac{1}{3}}\left(\frac{2 | b| ^{3/2}}{3 \sqrt{3| a| }}x\right)+J_{\frac{1}{3}}\left(\frac{2 | b| ^{3/2}}{3 \sqrt{3| a| }}x\right)\right)$$

If $ab\gt 0$ only remains the Bessel K function (which disappear if $ab \lt 0$) and $$I=\alpha K_{\frac{1}{3}}(\beta\,x)$$ with $$\alpha=\frac{2 a b^5}{3 | a| ^{3/2} | b| ^{9/2}}\qquad \qquad \beta=\frac{2 | b| ^{3/2}}{3 \sqrt{|3 a| }}$$ $$K_{\frac{1}{3}}(\beta\,x)=\sqrt{\frac{\pi }{2 \beta x}}\,e^{-\beta x}\,\left(1-\frac{5}{72 \beta x}+\frac{385}{10368 \beta ^2 x^2}+O\left(\frac{1}{x^3}\right)\right)$$