Mathematical Methods for Physics and Engineering : A Comprehensive Guide

25.8 APPROXIMATIONS TO INTEGRALS

which we already know has the value

√

πwhenzis real. This choice of demon-

stration model is not accidental, but is motivated by the fact that, as we have

already shown, in the neighbourhood of a saddle point all exponential integrands

can be approximated by a Gaussian function of this form.

The same integral can also be thought of as an integral in the complex plane,

in which the integration contour happens to be along the real axis. Since the

integrand is analytic, the contour could be distorted into any other that had the

same end-points,z=−∞andz=+∞, both on the real axis.

As a particular possibility, we consider an arc of a circle of radiusRcentred

onz=0.Itiseasilyshownthatcos2θ≥1+4θ/πfor−π/ 4 <θ≤0, whereθis

measured from the positive realz-axis and−π<θ≤π. It follows from writing

z=Reiθon the arc that, if the arc is confined to the region−π/ 4 <θ≤ 0

(actually,|θ|<π/4 is sufficient), then the integral of exp(−z^2 ) tends to zero as

R→∞anywhere on the arc. A similar result holds for an arc confined to the

region||θ|−π|<π/4. We also note for future use that, forπ/ 4 <θ< 3 π/ 4

or−π/ 4 >θ>− 3 π/4, the integrand exp(−z^2 ) grows without limit asR→∞,

and that the largerRis, the more precipitous is the ‘drop or rise’ in its value on

crossing the four radial linesθ=±π/4andθ=± 3 π/4.

Now consider a contour that consists of an arc at infinity running fromθ=π

toθ=π−αjoined to a straight line,θ=−α, which passes throughz= 0 and

continues to infinity, where it in turn joins an arc at infinity running fromθ=−α

toθ= 0. This contour has the same start- and end-points as that used inI 0 ,

and so the integral of exp(−z^2 ) along it must also have the value

√

π.Asthe

contributions to the integral from the arcs vanish, providedα<π/4, it follows

that the integral of exp(−z^2 ) along the infinite lineθ=−αis

√

π.Ifwenowtake

αarbitrarily close toπ/4, we may substitutez=sexp(−iπ/4) into (25.67) and

obtain

√

π=

∫∞

−∞

exp(−z^2 )dz

= exp(−iπ/4)

∫∞

−∞

exp(is^2 )ds (25.68)

=

√

2 πexp(−iπ/4)

[∫∞

0

cos(^12 πu^2 )du+i

∫∞

0

sin(^12 πu^2 )du

]

. (25.69)

The final line was obtained by making a scale changes=

√

π/ 2 u. This enables

the two integrals to be identified with the Fresnel integralsC(x)andS(x),

C(x)=

∫x

0

cos(^12 πu^2 )duandS(x)=

∫x

0

sin(^12 πu^2 )du,

mentioned on page 645. Equation (25.69) can be rewritten as

(1 +i)

√

π

√

2

=

√

2 π[C(∞)+iS(∞)],