25.8 APPROXIMATIONS TO INTEGRALS
From the observations contained in the two previous paragraphs, we deduce
that a path that follows the lines of steepest descent (or ascent) can never form
a closed loop. On such a path,φ, and hence|h(z)|, must continue to decrease
(increase) until the path meets a singularity off(z). It also follows that if a level
line ofh(z) forms a closed loop in the complex plane, then the loop must enclose
a singularity off(z). This may (ifφ→∞) or may not (ifφ→−∞) produce a
singularity inh(z).
We now turn to the study of the behaviour ofh(z) at a saddle point and how
this enables us to find an approximation to the integral ofh(z)alongacontour
that can be deformed to pass through the saddle point. At a saddle pointz 0 ,at
whichf′(z 0 ) = 0, both∇φand∇ψare zero, and consequently the magnitude and
phase ofh(z) are both stationary. The Taylor expansion off(z) at such a point
takes the form
f(z)=f(z 0 )+0+
1
2!
f′′(z 0 )(z−z 0 )^2 +O(z−z 0 )^3. (25.60)
We assume thatf′′(z 0 )= 0 and write it explicitly asf′′(z 0 )≡Aeiα, thus defining
the real quantitiesAandα.Ifithappensthatf′′(z 0 ) = 0, then two or more
saddle points coalesce and the Taylor expansion must be continued until the first
non-vanishing term is reached; we will not consider this case further, though the
general method of proceeding will be apparent from what follows. If we also
abbreviate the (in general) complex quantityf(z 0 )tof 0 , then (25.60) takes the
form
f(z)=f 0 +^12 Aeiα(z−z 0 )^2 +O(z−z 0 )^3. (25.61)
To study the implications of this approximation forh(z), we writez−z 0 as
ρeiθwithρandθboth real. Then
|h(z)|=|exp(f 0 )|exp[^12 Aρ^2 cos(2θ+α)+O(ρ^3 )]. (25.62)
This shows that there are four values ofθfor which|h(z)|is independent ofρ(to
second order). These therefore correspond to two crossing level lines given by
θ=^12
(
±^12 π−α
)
andθ=^12
(
±^32 π−α
)
. (25.63)
The two level lines cross at right angles to each other. It should be noted that
the continuations of the two level lines away from the saddle are not straight
in general. At the saddle they have to satisfy (25.63), but away from it the lines
must take whatever directions are needed to make∇φ= 0. In figure 25.12 one of
the level lines (|h|= 1) has a continuation (y= 0) that is straight; the other does
not and bends away from its initial directionx=1.
So far as the phase ofh(z) is concerned, we have
arg[h(z)] = arg(f 0 )+^12 Aρ^2 sin(2θ+α)+O(ρ^3 ),
which shows that there are four other directions (two lines crossing at right