The airy function $\mathrm{Ai}(x)$ can be defined by an improper integral: $$ \mathrm{Ai}(x) \equiv \frac{1}{\pi} \int^\infty_0\mathrm{d}t \cos\left(\frac{t^3}{3}+tx\right) \, . $$ I want to prove the convergence of this integral.
Wikipedia says this can be proven with the help of Dirichlet's test for convergence, which says that $$ \int fg $$ converges if (1) $f$ is uniformly bounded over all intervals and (2) $g$ is non-negative and monotonically decreasing.
In my attempt, I first make the substitution $u=\frac{t^3}{3}+tx$, so that $\mathrm{d}u = (t^2+x)\mathrm{d}t$. Now we need to write $\mathrm{d}t$ solely in terms of u, which we can expect to be messy but solvable.
Solving for $t$ gives the real solution as $$ t = \frac{\sqrt[3]{3u+\sqrt{9u^2+4x^3}}}{\sqrt[3]{2}} - \frac{\sqrt[3]{2}x}{\sqrt[3]{3u+\sqrt{9u^2+4x^3}}} \, , $$ and the integral becomes $$ \mathrm{Ai}(x) = \frac{1}{\pi}\int^\infty_0 \mathrm{d}u \cos{u} \left[\left( \frac{\sqrt[3]{3u+\sqrt{9u^2+4x^3}}}{\sqrt[3]{2}} - \frac{\sqrt[3]{2}x}{\sqrt[3]{3u+\sqrt{9u^2+4x^3}}} \right)^2+x\right]^{-1} $$
Now let $\cos{u}$ be $f$ and the other term be $g$, since $\cos{u}$ is obviously bounded, and the other term has a denominator that increase as $u^{2/3}$ and so should be decreasing, we can apply Dirichlet's test and conclude that the integral is convergent.
Is my reasoning correct? And are there more elegant ways of proving this result?