Mathematical Methods for Physics and Engineering : A Comprehensive Guide

25.8 APPROXIMATIONS TO INTEGRALS

In the worked example in subsection 25.8.2 the function

F(x)=

1

π

∫∞

0

cos(^13 s^3 +xs)ds (∗)

was shown to have the properties associated with the Airy function,Ai(x),whenx> 0 .Use

the stationary phase method to show that, forx< 0 and−xsufficiently large,

F(x)∼

1

√

π(−x)^1 /^4

sin

[

2

3

(−x)^3 /^2 +

π

4

]

,

in accordance with equation (25.53) forAi(z).

Since the cosine function is an even function and its argument in (∗) is purely real, we
may considerF(x) as the real part of

G(x)=

1

2 π

∫∞

−∞

exp[i(^13 s^3 +xs)].

This is of the standard form for a saddle-point approach withg(s)=1/ 2 πandf(s)=
i(^13 s^3 +xs). The latter hasf′(s)=0whens^2 =−x.Sincex<0 there are two saddle points
ats=+

√

−xands=−

√

−x. These are both on the real axis separated by a distance
2

√

−x.
If−xis sufficiently large, the Gaussian-like stationary phase integrals can be treated
separately and their contributions simply added. In terms of a phase–amplitude diagram,
the Cornu spiral from the first saddle will have effectively reached its final winding point
before the spiral from the second saddle begins. The second spiral therefore takes the final
point of the first as its starting point; the vector representing its net contribution need not
be in the same direction as that arising from the first spiral, and in general it will not be.
Nearthesaddleats=+

√

−xthe form off(s) is, in the usual notation,

f(s)=f 0 +^12 Aeiα(s−s 0 )^2

=−

2 i

3

(−x)^3 /^2 +

1

2

2

√

−xeiπ/^2 (ρeiθ)^2

=−

2 i

3

(−x)^3 /^2 +

√

−xeiπ/^2 ρ^2 (cos 2θ+isin 2θ).

For the exponent to be purely imaginary requires sin 2θ= 0, implying that the level lines
are given byθ=0,π/2,πor 3π/2. The same conclusions hold at the saddle ats=−

√

−x,
which differs only in that the sign off 0 is reversed andα=3π/2 rather thanπ/2; exp(iα)
is imaginary in both cases. Thus the obvious path is one that approaches both saddles
from the directionθ=πand leaves them in the directionθ=0.As− 3 π/ 4 < 0 <π/4,
the±choice is resolved as positive at both saddles.
Next we calculate the approximate valuesof the integrals from equation (25.73). At
s=+

√

−xit is

+

√

2 π

2

√

−x

1

2 π

exp

[

−

2 i

3

(−x)^3 /^2

]

exp

[

i

2

(

π−

π

2

)]

=+

1

2

√

π(−x)^1 /^4

exp

[

−i

(

2

3

(−x)^3 /^2 −

π

4

)]

.

The corresponding contribution from the saddle ats=−

√

−xis

+

√

2 π

2

√

−x

1

2 π

exp

[

2 i

3

(−x)^3 /^2

]

exp

[

i

2

(

π−

3 π

2

)]

,