Page 283 - Elements of Distribution Theory

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

9.4 Watson’s Lemma 269

satisfies
ˆ
+
h(t) = h m (t) + O(t a m+1 ) as t → 0 .
Consider the integral

T
I n = h(t)exp{−nt} dt, 0 < T ≤∞.
0
Then
m
c j (a j + 1) 1
I n = + O as n →∞.
a
n a j +1 n m+1 +1
j=0
Proof. Fix 0 < < T . There exists a constant K such that
|h(t)|≤ K exp{bt}, t ≥ .
Hence, for sufficiently large n,

T
∞ K
h(t)exp{−nt}dt ≤ K exp{−(n − b)} dt ≤ exp{−(n − b) }

n − b

exp{−n }
= O as n →∞.
n
Consider the integral

ˆ
h(t)exp{−nt} dt = h m (t)exp{−nt} dt + R m (t)exp{−nt} dt
0 0 0
+
where R m (t) = O(t a m+1 )as t → 0 .For any α> −1,
∞ ∞
α α α
t exp{−nt} dt = t exp{−nt} dt − t exp{−nt} dt.
0 0
Note that

∞ (α + 1)
α
t exp{−nt} dt = α+1
0 n
and

∞ ∞
α α
exp{n } t exp{−nt} dt = t exp{−n(t − )} dt

∞

α
= (t + ) exp{−nt} dt = O(1)
0
as n →∞.It follows that

α (α + 1)
t exp{−nt} dt = α+1 + O(exp{−n }).
0 n
Hence, for any > 0,
m
c j (a j + 1)
ˆ
h m (t)exp{−nt} dt = a j +1 + O(exp{−n }).
0 j=0 n