Mathematical Methods for Physics and Engineering : A Comprehensive Guide

APPLICATIONS OF COMPLEX VARIABLES

sign is determined by the directionθin the complex plane in which the distorted

contour passes through the saddle point. If−^12 π<θ≤^12 π, then the positive

sign is taken; if not, then the negative sign is appropriate. In broad terms, if the

integration path through the saddle is in the direction of an increasing real part

forz, then the overall sign is positive.

Formula (25.66) is the main result from a steepest descents approach to evalu-

ating a contour integral of the type considered, in the sense that it is the leading

term in any more refined calculation of the same integral. As can be seen, it

is as an ‘omnibus’ formula, the various components of which can be found by

considering a number of separate, less-complicated, calculations.

Before presenting a worked example that generates a substantial result, useful

in another connection, it is instructive to consider an integral that can be simply

and exactly evaluated by other means and then apply the saddle-point result to

it. Of course, the steepest descents method will appear heavy-handed, but our

purpose is to show it in action and to try to see why it works.

Consider the real integral

I=

∫∞

−∞

exp(10t−t^2 )dt.

This can be evaluated directly by making the substitutions=t−5 as follows:

I=

∫∞

−∞

exp(10t−t^2 )dt=

∫∞

−∞

exp(25−s^2 )ds=e^25

∫∞

−∞

exp(−s^2 )ds=

√

πe^25.

The saddle-point approach to the same problem is to consider the integral as a

contour integral in the complex plane, but one that lies along the real axis. The

saddle points of the integrand occur wheref′(t)=10− 2 t= 0; there is thus a

single saddle point att=t 0 = 5. This is on the real axis, and no distortion of the

contour is necessary. The valuef 0 of the exponent isf(5) = 50−25 = 25, whilst

its second derivative at the saddle point isf′′(5) =−2. Thus,A= 2 andα=π.

The contour clearly passes through the saddle point in the directionθ=0,i.e.in

the positive sense on the real axis, and so the overall sign must be +. Sinceg(t 0 )

is formally unity, we have all the ingredients needed for substitution in formula

(25.66), which reads

I=+

√

2 π

2

1 exp(25) exp[^12 i(π−π)] =

√

πe^25.

As it happens, this is exactly the same result as that obtained by accurate

calculation. This would not normally be the case, but here it is, because of the

quadratic nature of 10t−t^2 ; all of its derivatives beyond the second are identically

zero and no approximation of the exponent is involved.

Given the very large value of the integrand at the saddle point itself, the

reader may wonder whether there really is a saddle there. However, evaluating

the integrand at points lying on a line through the saddle point perpendicular