30A (Asymptotic Methods)

f (t)

f (t) \sim \sum_{n \geq 0} a_{n} t^{λ_{n} - 1}

t ↓ 0

, with

0 < λ_{0} < λ_{1} < \dots

, 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 \geq 0} a_{n} Γ (λ_{n}) x^{- λ_{n}}

x \to \infty

, with truncation after any fixed number of terms justified by the corresponding remainder estimate.

b

Complete the square:

z^{2} - 2 i z = (z - i)^{2} + 1

. The pole of

(1 - i z)^{- 1}

is at

z = - i

, so the contour may be shifted upward through the saddle point

z = i

without crossing a singularity. Setting

z = t + i

gives the exact steepest-descent representation

I (x) = e^{- x} \int_{- \infty}^{\infty} \frac{e ^{- x t^{2}}}{2 - i t} d t .

Near the saddle,

\frac{1}{2 - i t} = \sum_{n \geq 0} \frac{( i t ) ^{n}}{2 ^{n + 1}} .

Odd terms integrate to zero, while

\int_{- \infty}^{\infty} t^{2 m} e^{- x t^{2}} d t = Γ (m + 1/2) x^{- m - 1/2} .

I (x) \sim e^{- x} m = 0 \sum \infty \frac{( - 1 ) ^{m} Γ ( m + 1/2 )}{2 ^{2 m + 1} x ^{m + 1/2}} .

There are no discussions about this article yet.

There are currently no matching articles.