I am looking to find an explicit integral formula for $\operatorname{Bi}(x)$ via solving Airy's equation:
$${\mathrm d^2y\over\mathrm dx^2}-xy=0\tag1$$
Currently I am able to solve for $\operatorname{Ai}(x)$ using the means of Fourier transform:
Let $F(\omega)$ denote the Fourier transform of $y(x)$. If we take Fourier transform on both side of (1), then
$$ -\omega^2F(\omega)-iF'(\omega)=0\tag2 $$
Solving (2) yields $$ F(\omega)=F(0)e^{i\omega^3/3}\tag3 $$
Take inverse Fourier transform on (3) gives
$$ y={F(0)\over2\pi}\int_{-\infty}^\infty e^{i(\omega x+\omega^3/3)}\mathrm d\omega $$
Due to the fact that $\int_{-\infty}^\infty f(x)\mathrm dx=\int_0^\infty[f(x)+f(-x)]\mathrm dx$, we are able to get rid of the complex exponential:
$$ y={F(0)\over\pi}\int_0^\infty\cos\left(\omega x+{\omega^3\over3}\right)\mathrm d\omega\tag4 $$
and (4) is identical to $y=F(0)\operatorname{Ai}(x)$. Fourier transform is only valid for square-integrable funcntion, so the particular solution $\operatorname{Ai}(x)$ is only the square-integrable branch of the more general solution:
$$ y=C_1\operatorname{Ai}(x)+C_2\operatorname{Bi}(x) $$
However, I would like to derive an explicit integral formula for $\operatorname{Bi}(x)$, so I plug $y=f(x)\operatorname{Ai}(x)$ back into (1) and obtain
$$ f''(x)\operatorname{Ai}(x)+2f'(x)\operatorname{Ai}'(x)+f(x)\operatorname{Ai}''(x)-xf(x)\operatorname{Ai}(x)=0 \\ f(x)[\operatorname{Ai}''(x)-x\operatorname{Ai}(x)]+f''(x)\operatorname{Ai}(x)+f'(x)\operatorname{Ai}'(x)=0 $$ $$ f''(x)\operatorname{Ai}(x)+2f'(x)\operatorname{Ai}'(x)=0\tag5 $$
Via some basic algebraic operation, I arrived at
$$ {f''(x)\over f'(x)}=-2{\operatorname{Ai}'(x)\over\operatorname{Ai}(x)}\tag6 $$
Integrate and exponentiate on both side of (6) gives
$$ f'(x)={C\over\operatorname{Ai}^2(x)} $$
and eventually I got stuck on
$$ f(x)=C\int{\mathrm dx\over\operatorname{Ai}^2(x)} $$
Although this integral became a simple business if we made use of the Wronskian of $\operatorname{Ai}(x)$ and $\operatorname{Bi}(x)$, but I wonder if it is possible for me to obtain an explicit integral formula $\operatorname{Bi}(x)$ via this integral. Alternatively, could somebody provide some other ways for me to find this integral formula
