I wanna show that having Airy function defined as: $$ \mathrm {Ai} (x)={\dfrac {1}{\pi }}\int _{0}^{\infty }\cos \left({\dfrac {t^{3}}{3}}+xt\right)dt $$
Solves equation: $$y''=xy.$$
Edit: After clearing out that $k^3$ is not $x^3$, which i have misread: I am stuck at the integral: $$ \int_0^\infty (k^2+x)\cos(kx+k^3) $$ which is clearly not converging but some how wiki says it right solution?
Hints:
Rewrite the Airy function as $$\begin{align} \mathrm {Ai} (x)~:=~& \int_{\mathbb{R}_+}\frac{\mathrm{d}k}{\pi} \cos \left({\frac{k^{3}}{3}}+ xk\right) ~=~\ldots \cr ~=~& \int_{\mathbb{R}}\frac{\mathrm{d}k}{2\pi} \exp \left({\frac {\mathrm{i}k^{3}}{3}}+\mathrm{i} xk\right)\cr ~=~&\lim_{\epsilon\searrow 0^+} \int_{\mathbb{R}+\mathrm{i}\epsilon}\frac{\mathrm{d}k}{2\pi} \exp \left({\frac {\mathrm{i}k^{3}}{3}}+\mathrm{i} xk\right) .\end{align}\tag{1}$$
Therefore the Airy differential equation (DE) is satisfied: $$\begin{align} \mathrm {Ai}^{\prime\prime}(x)-x\mathrm {Ai} (x) ~\stackrel{(1)}{=}~&\lim_{\epsilon\searrow 0^+} \int_{\mathbb{R}+\mathrm{i}\epsilon}\frac{\mathrm{d}k}{2\pi} (-k^2-x)\exp \left({\frac {\mathrm{i}k^{3}}{3}}+\mathrm{i} xk\right)\cr ~=~&\lim_{\epsilon\searrow 0^+} \int_{\mathbb{R}+\mathrm{i}\epsilon}\frac{\mathrm{d}k}{2\pi} \mathrm{i} \frac{d}{dk}\exp \left({\frac {\mathrm{i}k^{3}}{3}}+\mathrm{i} xk\right)\cr ~=~&\ldots ~=~0. \end{align}\tag{2}$$
