This is for $k=1$, for which your formulas are valid.
In short, they come from Fourier integral representations of solutions to the ODE. This is much easier to see for $Ai$ than $Bi$.
Applying the standard Fourier transform
$$
\tilde\Psi(\omega) = \mathcal F\{\Psi\} = \int_{-\infty}^\infty \Psi(x)e^{-i\omega x}dx
$$
to the ODE, leads to the first order ODE $-\omega^2\tilde\Psi - i \frac{d\tilde\Psi}{d\omega} = 0$, thus
$$
\tilde\Psi = Ae^{i\omega^3/3}.
$$
Substituting into the inverse transform results in the above form
$$
\Psi(x) = \mathcal F^{-1}\tilde\Psi = \frac{A}{2\pi}\int_{-\infty}^\infty e^{i\omega^3/3 + ix\omega} d\omega
$$
(the imaginary part cancels over the entire line leading to your formula, while the constant $A$ is chosen with respect to the far field behaviour).
This has only recoverd $Ai$, rather than $Bi$, since implicit in applying the Fourier transform is the idea that the solution is integrable on $(-\infty,\infty)$, which $Bi$ is not.
The following approach to find other solutions is due to Ablowitz and Fokas (Complex Variables: Introduction and Applications, 2nd ed). Consider a more general Fourier-like integral representation of a solution to Airy's equation
$$
\Psi(x) = \int_C V(\omega)e^{i\omega x} d\omega
$$
where $C$ is a complex contour whose properties are yet to be decided (but we think of going to infinity in both direction along certain rays). Using integration by parts we have
$$
\Psi'' - x\Psi = \int_C [-\omega^2 V - i V'(\omega)]e^{i\omega x} d\omega - i[V(\omega)e^{i\omega x}]_C.
$$
Here the latter term represents the contribution from each endpoint of $C$, and the integration by parts is valid since $V$ will end up being entire. To set this to zero we have, as before, $V = i\omega^3/3$, and need to take $C$ such that $[e^{i(\omega^3/3 + \omega x)}]_C = 0$. If $\arg\omega=\theta$, this works for $\omega \to \infty$ in the sectors
$$
\sin(3\theta) > 0
$$
or $0 < \theta < \pi/3$, $2\pi/3 < \theta < \pi$, $-2\pi/3 < \theta < -\pi/3$.
Now the Fourier integral solution $Ai$ can be thought of as the result when $C$ is the curve connecting $\theta = \pi$ to $\theta = 0$ (in a limiting sense, as these angles are on the border between exponential decay and growth, hence the oscillatory nature of the integrand). However, if one takes $C$ to connect $\theta = -\pi/2$ to $\theta = 0$, that is, $C$ comes up the imaginary axis from $-i\infty$ and then goes out along the real axis to $+\infty$, the integral ends up being (taking $\omega = -i\zeta$ on $\theta = -\pi/2$)
\begin{align*}
\Psi(x) &= i\int_0^\infty e^{i(i\zeta^3/3 - ix\zeta)} d\zeta + \int_0^\infty e^{i(\omega^3/3 + x\omega)} d\omega \\ &= \int_0^\infty \cos(\omega^3/3 + x\omega) d\omega + i\int_0^\infty \sin(\omega^3/3 + x\omega)d\omega + i\int_0^\infty e^{-\zeta^3/3 + x\zeta} d\zeta.
\end{align*}
This result is of course up to the arbitrary multiplicative constant in $V$. The first term is proportional to $Ai$, so if it is subtracted off and the arbitrary constant is chosen appropriately (again with respect to some far field behaviour), you end up with your integral formula for $Bi$.
Best Answer
Given the integral representation it follows that $\text{Ai}(x)$ fulfills the differential equation $y''=x y$.
In particular $\text{Ai}(x)$ is an entire function and $$\text{Ai}(x)=\frac{1}{\pi}\int_{0}^{+\infty}\cos\left(\frac{t^3}{3}+xt\right)\,dt \tag{1}$$ $$\begin{eqnarray*}(\mathcal{L}\text{Ai})(s)&=&\frac{1}{\pi}\int_{0}^{+\infty}\int_{0}^{+\infty}\cos\left(\frac{t^3}{3}+xt\right)e^{-sx}\,dt\,dx\\&=&\frac{1}{\pi}\int_{0}^{+\infty}\frac{s\cos\left(\frac{t^3}{3}\right)-t\sin\left(\frac{t^3}{3}\right)}{s^2+t^2}\,dt \tag{2}\end{eqnarray*}$$ It is not difficult to show that $$ \lim_{s\to 0^+}\int_{0}^{+\infty}\cos\left(\frac{t^3}{3}\right)\frac{s\,dt}{s^2+t^2} = \frac{\pi}{2}\tag{3}$$ $$ \lim_{s\to 0^+}\int_{0}^{+\infty}\sin\left(\frac{t^3}{3}\right)\frac{t\,dt}{s^2+t^2} = \frac{\pi}{6}\tag{4}$$ and it follows that $$ \int_{0}^{+\infty}\text{Ai}(x)\,dx = \lim_{s\to 0^+}\left(\mathcal{L}\text{Ai}\right)(s) = \frac{1}{\pi}\left(\frac{\pi}{2}-\frac{\pi}{6}\right)=\color{red}{\frac{1}{3}}\tag{5} $$ as wanted.