I am trying to solve
$$\frac{\partial^3u}{\partial x^3}=\frac{\partial u}{\partial t}$$ $x\in\mathbb{R},\:t >0$. Subject to the conditions $u(x,0)=f(x),\:u,\:\frac{\partial u}{\partial x},\:\frac{\partial^2 u}{\partial x^2}\rightarrow0,\:$as $|x|\rightarrow\infty,\forall t>0.$ Use the method of Fourier transforms to show that the solution is $$u(x,t)=\frac{1}{(3t)^{1/3}}\int_{-\infty}^{\infty}A\left ( \frac{r-x}{(3t)^{1/3}}\right)f(r)\:dr$$ Where the Airy function $A$ is defined by $$A(w)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(w\xi+\frac{1}{3}\xi^3)}\:d\xi$$
My work so far is $$\frac{\partial^3u}{\partial x^3}=\frac{\partial u}{\partial t}$$ Applying Fourier, $$\frac{\partial \hat{u}}{\partial t}=i\xi^3\hat{u}\Rightarrow \hat{u}(\xi,t)=g(\xi)e^{i\xi^3t}$$ $$g(\xi)=\hat{u}(\xi,0)=\int_{-\infty}^{\infty}u(r,0)e^{i\xi r}\:dr=\int_{\infty}^{\infty}e^{i\xi r}f(r)\:dr=\hat{f}(\xi)$$$$\hat{u}(\xi,t)=e^{i\xi^3 t}\int_{-\infty}^{\infty}e^{i\xi r}f(r)\:dr$$ Now applying the inverse Fourier, $$\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{-i\xi x}\left ( e^{i\xi^3 t}\int_{-\infty}^{\infty}e^{i\xi r}f(r)\:dr\right )d\xi $$ Now I have no idea how to relate that to the answer given in the question.
Following through on the comment I gave above, note that we can rewrite the integral by reversing the order of summation (Fubini's theorem) and thereby write the inverse transform as $$u(x,t)=\int_{-\infty}^{\infty}\left(\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(r-x)\xi +i t\xi^3}\:d\xi\right)f(r)\:dr.$$ It remains to show that the integral inside is of the desired form. To that end, suppose we relabel $\xi\to \xi'$ and then define $\xi=\xi'/\lambda$. (This isn't problematic, since $\xi$ is just a dummy variable; effectively, we're rescaling $\xi\mapsto \xi \lambda$). Then
$$\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{i(r-x)\xi' +i\xi'^3 t}\:d\xi'=\frac{\lambda }{2\pi}\int_{-\infty}^{\infty}e^{i(r-x)\xi\lambda +i(\lambda ^3 t)\xi^3}\:d\xi.$$ Comparing this with the definition $$A(w)=\frac{1}{2\pi}\int_{-\infty}^{\infty}e^{iw\,\xi+\frac{1}{3}i\xi^3}\:d\xi$$
motivates the choice $\lambda=1/(3t)^{1/3}$. Thus we may write the integral as
$$\frac{(3t)^{-1/3}}{2\pi}\int_{-\infty}^{\infty}e^{i(r-x)/(3t)^{1/3}\xi +\frac{1}{3}i\xi^3}\:d\xi=\frac{1}{(3t)^{1/3}}A\left(\frac{r-x}{(3t)^{1/3}}\right),$$ which verifies the desired identity.