Airy Bi
The Airy function $\mathrm{Ai}$ and "Bairy" function $\mathrm{Bi}$ are given by the formulas $$\mathrm{Ai}(x) = \dfrac{1}{\pi} \displaystyle\int_0^{\infty} \cos \left( \dfrac{t^3}{3} + xt \right) dt$$ and $$\mathrm{Bi}(x) = \dfrac{1}{\pi} \displaystyle\int_0^{\infty} \left[ e^{-\frac{t^3}{3} + xt} + \sin \left( \dfrac{t^3}{3}+xt \right) \right] dt.$$
Properties
Theorem: The function $\mathrm{Ai}$ is a solution to the differential equation $$y(z) - zy(z) = 0.$$
Proof: Suppose that $y$ has the form $$y(z) = \displaystyle\int_{\gamma} f(t)e^{zt} dt,$$ where $\gamma$ is an as-of-yet undefined contour in the complex plane. Assuming that we may differentiate under the integral it is clear that $$y(z)=\displaystyle\int_{\gamma} f(t)t^2 e^{zt} dt.$$ Thus we plug this representation into the differential equation to get $$(*) \hspace{35pt} y(z)-zy(z) = \displaystyle\int_{\gamma} (t^2-z)f(t)e^{zt} dt = 0.$$ Now we integrate by parts to see $$\begin{array}{ll} \displaystyle\int_{\gamma} zf(t)e^{zt} dt &= \displaystyle\int_{\gamma} f(t) \dfrac{d}{dt} e^{zt} dt \\ &= -f(t)e^{zt} \Bigg |_{\gamma} + \displaystyle\int_{\gamma} f'(t)e^{zt} dt. \end{array}$$ We will pick the contour $\gamma$ to enforce $f(t)e^{zt} \Bigg |_{\gamma}=0$. We will do this by first determining the function $f$. Plugging this back into the formula $(*)$ yields $$\begin{array}{ll} 0 &= y(z) - zy(z) \\ &= f(t)e^{zt} \Bigg |_{\gamma} + \displaystyle\int_{\gamma} (t^2f(t)-f'(t))e^{zt} dt. \end{array}$$ We have the freedom to choose $f$ and $\gamma$. We will choose $f$ so that $$t^2f(t)-f'(t)=0.$$ This is a simple differential equation whose solution is $$f(t)=c e^{-\frac{t^3}{3}}.$$ So we have derived $$y(z)=\displaystyle\int_{\gamma} e^{zt - \frac{t^3}{3}} dt.$$ To pick the contour $\gamma$ note that the integrand of $y$ is an entire function and hence if $\gamma$ is a simple closed curve we would have $y(z)=0$ for all $z \in \mathbb{C}$.
The variable of the integral defining $y$ is $t$ and for $t \in \mathbb{C}$ with $|t|$ very large, the cubic term in the exponent dominates. Hence consider polar form $t=|t|e^{i\theta}$ and compute $$e^{-\frac{t^3}{3}} = \exp\left( \frac{-|t|^3 e^{3i\theta}}{3} \right).$$ Notice that the inequality $\mathrm{Re} \hspace{2pt} e^{3i\theta} > 0$ forces $\cos(3\theta)>0$ yielding three sectors defined by $\theta$: $$-\dfrac{\pi}{6} < \theta < \dfrac{\pi}{6},$$ $$\dfrac{\pi}{2} < \theta < \dfrac{5\pi}{6},$$ $$\dfrac{7\pi}{6} < \theta < \dfrac{3\pi}{2}.$$ Picking the contour $\gamma$ so that it tends to complex infinity inside two of the sectors forces the condition $f(t)e^{zt} \Bigg |_{\gamma}=0$. █
