The Laplace integral in one dimension

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For a scalar \theta, show that

\int_{-\infty}^{\infty}\exp\bigl\{-\tfrac12 h(\theta-\widetilde\theta)^2\bigr\}\,d\theta=\sqrt{2\pi/h},

and explain how this is the d_k=1 case of the Laplace integral (2\pi)^{d_k/2}|H|^{-1/2}. If h=n I_1 for a constant I_1, show that -\tfrac12\log h=-\tfrac12\log n+O(1).

Worked answer (click to show)

The integrand is \sqrt{2\pi/h} times the \operatorname{N}(\widetilde\theta,1/h) density, which integrates to 1. Hence

\int_{-\infty}^{\infty}\exp\bigl\{-\tfrac12 h(\theta-\widetilde\theta)^2\bigr\}\,d\theta=\sqrt{2\pi/h}.

The curvature h has to be positive for the integral to be finite. In the Laplace approximation with d_k=1 and H=h, the factor (2\pi)^{1/2}h^{-1/2} is the same quantity. If h=n I_1, then

-\tfrac12\log h=-\tfrac12\log n-\tfrac12\log I_1,

and the second term is a constant, that is O(1).