30A (Asymptotic Methods)

Watson lemma states that if

f (t) \sim \sum_{n = 0}^{\infty} a_{n} t^{λ_{n} - 1} (t ↓ 0), 0 < λ_{0} < λ_{1} < \dots,

(1)

and the Laplace integral has suitable growth control away from zero, then

\int_{0}^{b} e^{- x t} f (t) d t \sim \sum_{n = 0}^{\infty} a_{n} Γ (λ_{n}) x^{- λ_{n}} (x \to + \infty) .

(2)

The Taylor series of the amplitude is

sin (t^{2}) = \sum_{k = 0}^{\infty} \frac{( - 1 ) ^{k}}{( 2 k + 1 )!} t^{4 k + 2} .

(3)

Termwise application of Watson lemma therefore gives the asymptotic expansion

I (x) \sim k = 0 \sum \infty \frac{( - 1 ) ^{k} Γ ( 4 k + 3 )}{( 2 k + 1 )! x ^{4 k + 3}} = \frac{2}{x ^{3}} - \frac{120}{x ^{7}} + \frac{30240}{x ^{11}} - \dots .

(4)

In particular, the leading asymptotic approximation is

I (x) \sim 2 x^{- 3}

b

 0  0

Solution

 0  0

Put

α = ν - \frac{1}{2}, χ = z - \frac{π ν}{2} - \frac{π}{4} .

(1)

The two endpoints

t = 1

and

t = - 1

both contribute. Near

t = 1

, set

u = 1 - t

. Then

(1 - t^{2})^{ν - 1/2} = 2^{α} u^{α} (1 - \frac{u}{2})^{α}, e^{i z t} = e^{i z} e^{- i z u} .

(2)

Expanding the last amplitude in powers of

u

and applying the Complex Watson lemma after rotating onto a decaying endpoint ray gives one exponential series. Repeating at

t = - 1

gives the other. After multiplication by the prefactor in the integral representation, the two endpoint contributions combine as

J_{ν} (z) \sim \frac{1}{2 π z} [e^{i χ} \sum_{k = 0}^{\infty} \frac{i ^{k} a _{k} ( ν )}{z ^{k}} + e^{- i χ} \sum_{k = 0}^{\infty} \frac{( - i ) ^{k} a _{k} ( ν )}{z ^{k}}],

(3)

where

a_{k} (ν) = \frac{\prod _{j = 1}^{k} ( 4 ν ^{2} - ( 2 j - 1 ) ^{2} )}{k ! 8 ^{k}}, a_{0} = 1.

(4)

Separating the even and odd powers gives the large-argument asymptotic expansion of the Bessel function of the first kind:

J_{ν} (z) \sim \frac{2}{π z} [cos χ m = 0 \sum \infty \frac{( - 1 ) ^{m} a _{2 m} ( ν )}{z ^{2 m}} - sin χ m = 0 \sum \infty \frac{( - 1 ) ^{m} a _{2 m + 1} ( ν )}{z ^{2 m + 1}}] .

(5)

Its first terms are

J_{ν} (z) \sim \frac{2}{π z} [cos χ - \frac{4 ν ^{2} - 1}{8 z} sin χ - \frac{( 4 ν ^{2} - 1 ) ( 4 ν ^{2} - 9 )}{2 ! ( 8 z ) ^{2}} cos χ + \dots] .

(6)

For fixed

ν

, the expansion is uniform in every closed sector

∣ ar g z ∣ \leq π - δ

(7)

with

δ > 0

. The exclusion of the negative real axis fixes the branch of

z^{1/2}

and stays away from its Stokes boundary.

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Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 30A (Asymptotic Methods)

a

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