I wonder if the convolution


\begin{equation}
f(y)=\int\_{-\infty}^{+\infty} \mathrm{Airy}(a\cdot x)\cdot e^{-b(y-x)^2} dx
\end{equation}


can be solved analytically. Or in case not, if there is an analytic expression for the zeros of $f(y) = 0$ and for which values of $a$ and $b$ they exist.


edit: $a>0$ and $b>0$



Convolution of an Airy function with a Gaussian
Here's a proof sketch of
$$(\*) \quad \int\_{-\infty}^\infty e^{-b \ u^2}\text{Ai}(a(u+y))du = \frac{\sqrt{\pi}}{\sqrt{b}} 
\exp{\big(\frac{a^6}{96b^3} + \frac{a^3y}{4b}}\big)\text{Ai}\big(a(y+\frac{a^3}{16b})\big).$$


The following formula will be used twice:
$$ (1) \quad \frac{1}{2\pi} \int\_{-\infty}^\infty \exp{\big(iu^3/3-u^2\ t/2+i\ u \ x\big) } = \exp{\big(t^3/12 + t \ x/2 \big)} \text{Ai}\big( x+t^2/4 \big)$$
Also used used is the well-known
$$ (2) \quad \int\_{-\infty}^\infty \exp{\big(-iz \ t - \frac{z^2}{4b} \big) }dz = 2 \sqrt{b \pi } e^{-b t^2} $$
and the Fourier integral representation of the Dirac delta function,
$$ (3) \quad \frac{1}{2 \pi} \int\_{-\infty}^\infty \exp{\big(i x (u-z) \big)} dx = \delta(u-z) .$$


Starting with (1), apply $\int\_{-\infty}^\infty \exp{(-ix \ z)}dx/(2\pi) .$ Use (3), then (1) to get $ \exp{(-iz^3/3-z^2 \ t/2)} $= $\exp(t^3/3)\int\_{-\infty}^\infty \exp{(-ix \ z + t\,x/2)} Ai(x+t^2/4)dx .$ Shift $z \to z-it/2$ to get a simpler looking integral. Let $t/2 \to y.$ The parameters can be rescaled to get the attractive intermediate form


$$ (4) \quad a \int\_{-\infty}^\infty \exp{(-iu \ z)}\text{Ai}(a(u+y)) du = \exp{(i\frac{z^3}{3a^3}+iy \ z)}$$


Apply (2) to (4), use (1) again, and simplify to get (\*).


The proof uses interchange of integrations that have not been justified, and I don't want to do it, hence I call it a proof *sketch*. Also, I expected something with a nice closed form as $(\*)$ to have been discovered before. It has, in arXive:0906.3666v4 29 Sept 2009, 'Zeros of Airy Function and Relaxation Process,' M, Katori and H. Tanemura. See formula 3.10. Their proof is different, but it also involves an interchange of $\int$ about which nothing is said. If starting from their 3.10, you'll need to work in a scaling parameter. That formula is part of an appendix that has some references that I have not read, so I don't know if their equivalent to $(\*)$ is the first time it appears in the literature.


