Fourier transform of complex exponential with cubic power $\int_{-\infty}^{\infty}dx e^{iax-ibx^3}$

Question

How to evaluate the following integral, for real $a,b$? $$ \int_{-\infty}^{\infty}dx e^{iax-ibx^3} $$ If $b=0$, then the result is delta function $2\pi\delta(a)$ but how to do it for general b?

Answer 1

To me this question was screaming Airy function, so I looked up a few identities and found something pretty quick. From Abramowitz & Stegun (1965) we have $$ A_i(z) = \frac{1}{2\pi} \int_{-\infty}^{\infty} \exp \left[i\left(zx+\frac{x^3}{3}\right) \right]\,dx $$ or more generally, for all $\gamma \in \mathbb{R} \backslash\{0\}$, $$ A_i(\gamma z) = \frac{1}{2\pi\gamma} \int_{-\infty}^{\infty} \exp \left[i\left(zx+\frac{x^3}{3\gamma^3}\right) \right]\,dx.$$ Letting $z=a$ and $\frac{1}{3\gamma^3}=-b \implies \gamma = -(3b)^{-\frac13}$, the desired integral is given by $$ \int_{-\infty}^{\infty} \exp \left[i\left(ax-bx^3\right) \right]\,dx = 2\pi\gamma A_i(\gamma z) = -2\pi(3b)^{-\frac13}A_i\left(-\frac{a}{(3b)^\frac{1}{3}}\right). $$