Background. The Airy transform of $f$ is defined as $$\int_{-\infty}^\infty dx\, f(x)\,\text{Ai}(y-x)\;.$$ $\text{Ai}$ denotes Airy function, $$\text{Ai}(x)=\frac{1}{\pi}\int_{-\infty}^\infty dt\,\cos(xt+t^3/3)\;.$$ It satisfies Airy equation $\text{Ai}(x)''=x\,\text{Ai}(x)$.
The Airy transform of a gaussian $e^{-x^2}$ can be found explicitly, and it equals $$\sqrt\pi\, e^{\frac{1}{4}\left(y+\frac{1}{24}\right)}\,\text{Ai}\left(y+\frac{1}{16}\right)\;.$$ For a reference see pp. 77-78 of "Airy Functions and Applications to Physics". You can find a preview here.
My question. Is there some way to use this result to compute the similar integral $$\int_{0}^\infty dx\, e^{-x^2}\text{Ai}(y-x)\;,$$ where now integration is only on the half-line? Or, anyway, can this integral be given a closed-form expression?