How to integrate $\int_{-\infty}^{\infty} e^{-x^{2}}e^{ix^{3}}dx$ I've been stuck on trying to integrate

$\int_{-\infty}^{\infty} e^{-x^{2}}e^{ix^{3}}dx$

I initially thought this could be solved in similar fashion to solving
$\int^{\infty}_{-\infty} e^{-x^{2}}e^{ix}dx$ where we define $F(t) = \int^{\infty}_{-\infty} e^{-x^{2}}e^{itx}dx$ and recognize is at as a Fourier transform of $f(x)=e^{-x^{2}}$ and then use the properties of the Fourier transform, namely that $\frac{d}{dt}(\mathcal{F}f)(t) = \mathcal{F}(ixf)(t)$, to show that $F$ satisfies the differential equation $F'(t)=\frac{t}{2}F(t)$. See here for more details.
So I tried to adjust this by defining a transform as $\mathcal{F}_{c}f(t)=\int^{\infty}_{-\infty} f(x)e^{itx^{3}}dx$ and I saw that I actually get a similar property that $\frac{d}{dt}(\mathcal{F}_{c}f)(t) = \mathcal{F}(3ix^{2}f)(t)$ but I got stuck trying to apply this property in similar fashion to see what $F'(t)$ (in this case $F(t)$ is defined with the new transform instead of Fourier transform) was but ended up with the integral
$$\int^{\infty}_{-\infty} (3ix^{2})e^{-x^{2}}e^{ix^{3}t}$$
and am stuck from here....
Anyways, I was looking for either help with my approach or another way entirely to evaluate the integral. I would be very happy either way!
 A: There is no antiderivative but, as mentioned by @K B Dave, we have
$$I_1=\int_{0}^{\infty} e^{-x^{2}}e^{ix^{3}}\,dx$$
$$I_1=\frac{1}{9} \left(3^{2/3} e^{2/27} \pi  \left(3 \text{Ai}\left(\frac{1}{3
   \sqrt[3]{3}}\right)+i \text{Bi}\left(\frac{1}{3 \sqrt[3]{3}}\right)\right)-3 i
   \, _2F_2\left(\frac{1}{2},1;\frac{2}{3},\frac{4}{3};\frac{4}{27}\right)\right)$$
$$I_2=\int^{0}_{-\infty} e^{-x^{2}}e^{ix^{3}}\,dx$$
$$I_2=\frac{1}{9} \left(3 i \,
   _2F_2\left(\frac{1}{2},1;\frac{2}{3},\frac{4}{3};\frac{4}{27}\right)+3^{2/3}
   e^{2/27} \pi  \left(3 \text{Ai}\left(\frac{1}{3 \sqrt[3]{3}}\right)-i
   \text{Bi}\left(\frac{1}{3 \sqrt[3]{3}}\right)\right)\right)$$ where appear Airy functions.
$$I_3=\int^{\infty}_{-\infty} e^{-x^{2}}e^{ix^{3}}\,dx=I_1+I_2=\frac{2 e^{2/27} \pi  \text{Ai}\left(\frac{1}{3 \sqrt[3]{3}}\right)}{\sqrt[3]{3}}=\frac{2 e^{2/27} K_{\frac{1}{3}}\left(\frac{2}{27}\right)}{3 \sqrt{3}}$$ where apperas the modified Bessel function of the second kind.
A: $\DeclareMathOperator{\Ai}{Ai}$
$\newcommand{\d}{\mathrm{d}}$
$\newcommand{\e}{\mathrm{e}}$
Consider the twice continuously-differentiable functions $f_0$, $f_1$ on
$\mathbb{R}^{\geq 0}$ given by
$$\begin{aligned}
f_0(x)
  &= \frac{1}{\pi}\int_0^{\infty}\cos(\tfrac{1}{3}t^3 + xt)\d t \\
f_1(x)
  &= \frac{\e^{-\tfrac{2}{3}x^{3/2}}}{\pi}\int_{0}^{\infty}\e^{-x^{1/2}t^2}\cos\left(\tfrac{1}{3}t^{3}\right)\d t
 \end{aligned}$$
(DLMF 9.5.1, 9.5.7). One can show directly, by
regulating the integral,
differentiation and integration by parts, that they
are solutions to the initial-value problem (Airy's differential equation)
$$\begin{aligned}
  f''(x) &= x f(x) \\
  f(0) &= \frac{1}{\pi}\int_0^{\infty}\cos (\tfrac{1}{3}t^3)\d t \\
  f'(0) &= -\frac{1}{\pi}\int_{0}^{\infty}
    t\sin\left(\tfrac{1}{3}t^{3}\right)\d t
\end{aligned}$$
(DLMF 9.2.1, 9.2.3, 9.2.4; cf. DLMF
5.9.8, 5.9.9 for the relation to the Gamma
function). But solutions to such problems are unique by the Picard–Lindelöf
theorem (Wikipedia), so
$$\Ai(x)\equiv f_0(x) = f_1(x) $$ on $\mathbb{R}^{\geq 0}$. Your integral is
related to $f_1$ by scaling (note the imaginary part vanishes by symmetry), and
$f_0$ is an expression often used for defining $\Ai$.

As when invoking any special function, the usual caveat applies. Assignment of a symbol to an expression that is not known to be elementary does not provide a justification for special consideration of the expression, which must come from elsewhere. For the Airy function, this justification comes from being a uniform approximation to

*

*one-parameter families of integrals with a coalescing saddle point and

*solutions of one-parameter families of linear second-order differential equations with a simple turning point (DLMF 9.15).

A: I hope that the following ideas may help to find the solution (I am still trying to glue the puzzles together).
Let $u(x)=e^{-x^2}$ and $v(x)=e^{i\cdot x^3}$.
Then we have:

*

*$\int{u(x)}=\frac{1}{2} \sqrt{\pi } \text{erf}(x)$

*$\int{v(x)}=-\frac{x \Gamma \left(\frac{1}{3},-i x^3\right)}{3 \sqrt[3]{-i x^3}}$

*$u'(x)=-2 e^{-x^2} x$

*$v'(x)=3 i e^{i x^3} x^2$
Note that "erf" is the Error Function. Now using Integration by parts one may come to a generic formula.
Moreover, when I use the following MatLab Code (that performs a numerical integration), then I obtain $1.3881 - 0.0000i$ (so I guess we can concentrate on the real part $1.3881$ by rounding off/truncating the very small imaginary part):
syms x;
fun = @(x) exp(-x.^2).*exp(i*x.^3);
q = integral(fun,-Inf,Inf)

