Page 284 - Elements of Distribution Theory

P1: JZP
052184472Xc09 CUNY148/Severini June 2, 2005 12:8

270 Approximation of Integrals

+
Note that, since R m (t) = O(t a m+1 )as t → 0 , there exists a constant K m such that for
sufficiently small ,
|R m (t)|≤ K m t a m+1 .

Hence,
(a m+1 + 1)
t a m+1 .
|R m (t)| exp{−nt} dt ≤ K m exp{−nt} dt ≤ K m a
0 0 n m+1 +1
It follows that
T
I n = h(t)exp{−nt} dt
0
T
ˆ
= h m (t)exp{−nt} dt + R m (t)exp{−nt} dt + h(t)exp{−nt} dt
0 0
m
c j (a j + 1) 1 exp{−n }
= + O(exp{−n }) + O + O ,
a
n a j +1 n m+1 +1 n
j=0
proving the theorem.
Example 9.4 (Exponential integral). Consider the exponential integral function defined
by
1
∞

E 1 (z) = exp(−u) du;
z u
this function arises in a number of different contexts. See, for example, Barndorff-Nielsen
and Cox (1989, Example 3.3) for a discussion of a distribution with density function given
by E 1 .
Consider an asymptotic expansion for E 1 (z) for large z.In order to use Watson’s lemma,
the integral used to define E 1 must be rewritten. Note that, using the change-of-variable
t = u/z − 1,
1 1
∞ ∞

exp(−u) du = exp(−z) exp(−zt) dt.
u 1 + t
z 0
The function h(t) = 1/(1 + t) satisfies the conditions of Watson’s lemma with a j = j and
j
c j = (−1) , j = 0, 1,.... Hence, for any m = 0, 1,...,

m
j j! 1
E 1 (z) = exp(−z) (−1) + O .
z j+1 z m+1
j=0
Since this holds for all m we may write
∞
j j!
E 1 (z) ∼ exp(−z) (−1) as n →∞.
z j+1
j=0
Example 9.5 (Ratio of gamma functions). Consider an asymptotic expansion of the ratio
of gamma functions (z)/ (z + x) for large values of z and fixed x.By Theorem 9.2 we
can write
(z) 1 1 z−1 x−1
= u (1 − u) dx.
(z + x) (x) 0