Why are Airy like Integrals zero when the start and endpoints of integration are in the same $\frac{1}{3}$ slice of convergence?

Question

Background: This is Bence et al. Mathematical methods:

The version of the equation being solved is $$\frac{d^2y}{ dz^2}=zy\tag{1}$$

If z is a parameter, a solution as a contour integral is $$y(z)=\int_a^bf(t)e^{zt}dt\tag{2}$$ which yields when plugged into (1) $$\int_a^bt^2f(t)e^{zt}dt=\int_a^bzf(t)e^{zt}dt$$ $$=f(t)e^{zt}\mid_a^b-\int_a^b\frac{df(t)}{dt}e^{zt}dt$$ If a and b are chosen so that the middle term is zero the equation can be solved as $$\frac{df(t)}{dt}+t^2f(t)=0$$ $$f(t)=Ae^{-t^3/3}\tag{3}$$ This occurs when $arg(t)=\pm\pi/6+\frac{2n\pi}{3}$ because the real part of t is positive and the exponent goes to zero. Plugging (3) into (1) and normalizing yields the $$Ai(z)=\frac{1}{2\pi i}\int_{C1} e^{-t^3/3+zt}dt$$ As long as the endpoints are in the shaded regions, the equation is valid:

QuestionAccording to the text, the contour integral will be zero if the start and endpoints are in the same shaded region. Why would the integral be zero when the path is completely in one shaded region but not when the path is in two different shaded regions? Hypothesis/uneducated guess: Equation (1) has a singular point at infinity. Perhaps this means that the contour which crosses a branch cut encompasses points at infinity? Could this be used to solve the integral by using the residue at infinity?

References: Bence, Hobson, Riley, Mathematical methods pg 891

Answer 1

Basically, when the contour has its ends in the same region, it can be "shrunk towards \infty" by using Cauchy's theorem: because the integrand decays exponentially in the shaded regions as $\lvert t \rvert \to \infty$, one can bound the integral above by something that decays to zero as the contour is deformed to move it further from the origin.

On the other hand, if the contour has its ends in different shaded regions, it loops around the unshaded region where the integrand blows up as $\lvert t \rvert \to \infty$. The end can't be deformed through this region because the integral then would not converge. One can check that the integral is nonzero by deforming the contour to two straight rays meeting at the origin and evaluating it (numerically, for example, although it has a closed form in terms of the Gamma-function for $z=0$).

Answer 2

Consider a curve $C$ that stays in the segment $|\arg(t)|\leq\theta<\pi/6$ and goes to infinity in both directions. I will argue that the curve can be deformed so that the integral is arbitrarily small. A similar argument applies for the other segments.

Pick points $z_0,z_1,\dots$ going along $C$ in the same direction, with $|z_i| = N+i$, where $N$ will be chosen later. Choose $z_{-1},z_{-2}$ going along $C$ in the other direction from $z_0$, with $|z_i| = N+|i|.$ We can replace $C$ by the piecewise-linear contour $C'$ that goes from each $z_i$ to $z_{i+1}$ by a straight line. Since the integrand is entire, by Cauchy's integral theorem, replacing $C$ by $C'$ does not change the value of the integral. Using a crude bound $|z_i-z_{i+1}|\leq 5n$, the integral along $C'$ can be bounded by $\sum_{n\geq N} 10ne^{-\cos(3\theta)n^3/3+(n+1)|z|}$ which tends to zero as $N\to\infty$. By choosing $N$ large enough we can therefore give an arbitrarily small bound on the integral.

Regarding "Could this be used to solve the integral by using the residue at infinity?": No. Try substituting $\tau=1/t$; you get an essential singularity at the origin, so can't use the residue theorem.