Behaviour of Complex Airy function

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carlos g
carlos g 2017-7-6
回答： David Goodmanson 2017-7-7

I want to compute numerically the following integral (I am using MATLAB)

$$\int_{\eta_0}^\infty Ai(p)dp$$ where $\eta_0=-9.0311 - 5.2141i$

What should my $\infty$ be to get the right result?

I choose $\infty$ to obey:

$(0.332i)^{1/3}N+\eta_0$

where N is a real value. At the moment I choose it to be N=80 but I am not sure on how much it has to be to be considered "infinity".

So, I am using 38.9422 +22.4833i for the upper-bound (is it a good "infinite"?) and got 9.4214E+04 - 3.7640E+05i for the integral but I don't know if this result is right or not.

An additional question: is a branch cut needed?

  1 个评论
Walter Roberson
Walter Roberson 2017-7-6
Symbolic might be easier. In Maple notation,
(1/12)*(6*(I*A*B+(1/2)*A^2-(1/2)*B^2)*GAMMA(2/3)^2*3^(1/6)*hypergeom([2/3], [4/3, 5/3], (1/9)*(A+I*B)^3)-4*(3^(1/3)*hypergeom([1/3], [2/3, 4/3], (1/9)*(A+I*B)^3)*(A+I*B)-GAMMA(2/3))*Pi)/(Pi*GAMMA(2/3))
where A and B are your real and imaginary components respectively. Roughly 94229.49804-376355.0182*I

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回答（1 个）

David Goodmanson
David Goodmanson 2017-7-7
Hi Carlos,
For this integral to be meaningful, the airy function at the upper limit has to have a well defined value. As |p| --> infinity, only two values occur:
airy(p) --> 0 -60 deg < angle(p) < 60 deg
--> infinity angle(p) anything else
If you look at this plot
theta = linspace(-pi,pi,1000);
R = 20;
y = airy(R*exp(i*theta));
semilogy((180/pi)*theta,abs(y))
grid on
you can see that happening as a function of the angle. airy is tiniest on the real axis, so you can pick the 'infinity' point to be real and something like 20. Or any point with a small value.
airy(20)
ans = 1.6917e-27
C = 38.9+22.5i % your example
(180/pi)*angle(C)
ans = 30.0454 % ok
airy(C)
ans = -1.6139e-63 + 2.2011e-63i

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