应用拉普拉斯方法验证积分渐近近似的技术问询

Let's walk through this step by step, starting with the suggested variable substitution and then applying Laplace's method (adjusted for endpoint maxima, since that's where our integral's dominant contribution comes from).

Step 1: Variable Substitution ( t = \tau^2 ) #

First, substitute ( t = \tau^2 ) into the original integral to transform it into a Laplace-friendly form:

• When ( t = 0 ), ( \tau = 0 ); when ( t = 1 ), ( \tau = 1 ).
• The differential ( dt = 2\tau d\tau ).
• Rewrite each term in the integrand:
  • ( t^{-1/2} = (\tau²⁾{-1/2} = \tau^{-1} )
  • ( \cos(t) = \cos(\tau^2) )
  • ( e^{-x t^{1/2}} = e^{-x \tau} )

Substituting all these into the original integral:
[
\int_0^1 t^{-1/2} \cos(t) e^{-x t^{1/2}} dt = \int_0^1 \tau^{-1} \cos(\tau^2) e^{-x \tau} \cdot 2\tau d\tau
]
The ( \tau^{-1} ) and ( \tau ) cancel out, simplifying to:
[
2 \int_0^1 \cos(\tau^2) e^{-x \tau} d\tau
]

Step 2: Fit to Laplace's Method Form #

Laplace's method applies to integrals of the form ( \int_\alpha^\beta g(\tau) e^{x h(\tau)} d\tau ). For our transformed integral:

• ( g(\tau) = \cos(\tau^2) )
• ( h(\tau) = -\tau ) (since ( e^{-x \tau} = e^{x(-\tau)} ))

Key Observation: Maximum of ( h(\tau) ) #

The function ( h(\tau) = -\tau ) is strictly decreasing on ( [0,1] ), so its maximum occurs at the left endpoint ( \tau = 0 ) (with ( h(0) = 0 )). This is an endpoint maximum, not an interior one—so we use the endpoint variant of Laplace's method (the formula you provided applies to interior maxima).

Step 3: Approximate the Integral for Large ( x ) #

For large ( x ), the exponential term ( e^{-x \tau} ) decays extremely rapidly as ( \tau ) moves away from 0. This means the dominant contribution to the integral comes from a tiny neighborhood around ( \tau = 0 ). We can make two safe simplifications:

1. Near ( \tau = 0 ), ( \cos(\tau^2) \approx \cos(0) = 1 ) (since ( \tau^2 ) is negligible for small ( \tau )).
2. Extend the upper limit of integration from 1 to ( \infty ): the integral from 1 to ( \infty ) is exponentially small compared to the integral near 0 (since ( e^{-x \cdot 1} ) is negligible for large ( x )).

Applying these simplifications:
[
\int_0^1 \cos(\tau^2) e^{-x \tau} d\tau \approx \int_0^\infty e^{-x \tau} d\tau
]

Compute the integral:
[
\int_0^\infty e^{-x \tau} d\tau = \left[ -\frac{e^{-x \tau}}{x} \right]_0^\infty = 0 - \left( -\frac{1}{x} \right) = \frac{1}{x}
]

Step 4: Combine Results #

Multiply back by the factor of 2 from the transformed integral:
[
2 \int_0^1 \cos(\tau^2) e^{-x \tau} d\tau \sim 2 \cdot \frac{1}{x} = \frac{2}{x}
]

This confirms the asymptotic approximation as ( x \rightarrow \infty ).

内容的提问来源于stack exchange，提问作者Evan