Hard integral that standard CAS get totally wrong How to solve the following integral:
$$\int_{-\infty }^{\infty }\exp \left ( i\left ( ax^3+bx^2 \right ) \right )dx$$
Standard CAS seem to get it totally wrong, see: http://www.walkingrandomly.com/?p=5031
So what is the right ansatz and solution?
EDIT
There seems to be a problem with the way this question is posed... which I quite frankly don't get. To clarify I posted this follow-up question:
In which senses can an integral exist?

The integral is slightly undefined in much the same way that the much simpler integral
$$ \int_{-\infty}^{\infty} e^{i a x}\,dx = 2\pi\delta(a) $$
is defined only in the sense of being a distribution.
To evaluate it in some acceptable sense, denote it by $I$ and write
$$ I = \int e^{i a x^3+i b x^2}\,dx = \int_0^\infty e^{i b x^2}(e^{i a x^3} + e^{-i a x^3})\,dx. $$
Consider the integral
$$ J(a) = \int_0^\infty e^{i a x^3 + i b x^2}\,dx, $$
and let $a$ have a positive imaginary part, which makes $J$ converge. Now expanding $e^{i b x^2}$ in power series and using a CAS to evaluate the integrals we can get
$$ J(a) = \sum_{n\geq0} \frac{(i b)^n}{n!} \int_0^\infty x^{2n}e^{i a x^3}\,dx = \sum_{n\geq0}\frac{(ib)^n}{n!}\frac{(-i a)^{-(1+2n)/3}}{3}\Gamma\left(\frac{1+2n}{3}\right). $$
The second integral making up $I$ has $-a$, so we can evaluate $J(-a)$ by letting $a$ have a negative imaginary part. All this leads to
$$ I = \sum_{n\geq0} \frac{(i b)^n}{n!}\frac{2}{3|a|^{(1+2n)/3}}\Gamma((1+2n)/3)\cos\left(\frac{(1+2n)\pi}{6}\right), $$
which evaluates to
$$ I = \frac{2\pi}{3^{1/3}|a|^{1/3}}e^{\frac{2i}{27}b^3/a^2}\mathop{\text{Ai}}\left(-\frac{b^2}{3^{4/3}|a|^{4/3}}\right), $$
where $\text{Ai}$ is an Airy function.
