I am trying to show that that the Airy functions defined below satisfy: $W[Ai(x),Bi(x)]=1/\pi$.
$$Ai(x)=\frac{1}{\pi} \int_0^\infty \cos(t^3/3+xt)dt$$
$$Bi(x)=\frac{1}{\pi}\int_0^\infty \bigg[ \exp(-t^3/3+xt)+\sin(t^3/3+xt)\bigg]dt $$
I tried to compute it directly but I got stuck, here's the last term I got:
$$Ai(x)Bi'(x)-Ai'(x)Bi(x) = \frac{1}{\pi^2}\bigg[ \int_0^\infty \cos(t^3/3+xt)dt \int_0^\infty \bigg( s\exp(-s^3/3+xs)+s\cos(s^3/3+xs)\bigg) ds + \int_0^\infty \sin(t^3/3+xt)tdt\int_0^\infty \bigg(\exp(-s^3/3+xs)+\sin(s^3/3+xs)\bigg)ds \bigg]$$
I don't see how to proceed from here, I guess I need complex integration contour but how exactly?
Thanks. I want also to show that $Bi(x),Bi'(x)>0 \forall x>0$, and to conclude the asymptotic identities: $$Bi(x) \sim \pi^{-1/2}x^{-1/4}\exp(2/3 x^{3/2})$$
$$Bi'(x)\sim \pi^{-1/2}x^{1/4}\exp(2/3 x^{3/2})$$
From the Airy differential equation it is straightforward to show : $$Ai''(x)=x\:Ai(x)\quad\text{and}\quad Bi''(x)=x\:Bi(x)$$ Let $\quad W(x)=Ai(x)Bi'(x)-Ai'(x)Bi(x)$ $$\frac{dW}{dx}=Ai'(x)Bi'(x)+Ai(x)Bi''(x)-Ai'(x)Bi'(x)-Ai''(x)Bi(x)$$ $$\frac{dW}{dx}=xAi(x)Bi(x)-xAi(x)Bi(x)=0$$ Thus $\quad W(x)=$constant.
Computing $\quad W(0)=Ai(0)Bi'(0)-Ai'(0)Bi(0)\quad$ leads to the constant=$\frac{1}{\pi}$.