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282 Approximation of Integrals

the maximizer of g,is less than or greater than z.For instance, suppose that g is strictly
decreasing on (ˆ y, ∞). If z < ˆ y, then the approximation is based on Theorem 9.14; if z ≥ ˆ y,
so that over the interval [z, ∞) g is maximized at z, then the approximation is based on
Theorem 9.15. Furthermore, in the case z < ˆ y, the approximation does not depend on the
value of z.
This is illustrated in the following example.

Example 9.12. Consider approximation of the integral
∞

√ √
exp(x) nφ( nx) dx
z
for large n, where φ(·) denotes the standard normal density function. It is straightforward
to show that the exact value of this integral is given by
√ √
exp{1/(2n)}[1 − ( nz − 1/ n)].
If z < 0 then, by Theorem 9.14,
√
∞ n (2π)

−1
exp(x)exp − x 2 dx = √ [1 + O(n )]
z 2 n
so that

∞
√ √ −1
exp(x) nφ( nx) dx = 1 + O(n )as n →∞.
z
If z > 0, then Theorem 9.15 must be used, leading to the approximation

∞ n exp − z exp(z)
n 2
−1
exp(x)exp − x 2 dx = 2 [1 + O(n )]
z 2 z n
so that

∞
√ √ √ exp(z) −1
exp(x) nφ( nx) dx = φ( nz) √ [1 + O(n )] as n →∞.
z nz
Hence,
∞ −1

√ √ 1 + O(n ) if z < 0
exp(x) nφ( nx) dx = √ √ −1 . (9.3)
z [φ( nz)exp(z)/( nz)][1 + O(n )] if z > 0
2
Since z has derivative 0 at z = 0, neither Theorem 9.14 nor Theorem 9.15 can be used
when z = 0.
In addition to the fact that the form of the approximation depends on the sign of z, the
approach based on Laplace’s method also has disadvantage that the approximations are not
−1
valid uniformly in z. That is, the O(n ) terms in (9.3) refer to asymptotic properties for
each fixed z, not to the maximum error over a range of z values.
√
For instance, suppose that z n = z 0 / n, where z 0 > 0isa fixed constant. If the approx-
imation in (9.3) is valid uniformly for all z in a neighborbood of 0 then
√ √
∞

z exp(x) nφ( nx) dx −1
sup √ √ = 1 + O(n )
0≤z≤ φ( nz)exp(z)/( nz)