Statement of the lemma
edit
Let
0
<
T
≤
∞
{\displaystyle 0<T\leq \infty }
φ
(
t
)
=
t
λ
g
(
t
)
{\displaystyle \varphi (t)=t^{\lambda }\,g(t)}
g
(
t
)
{\displaystyle g(t)}
t
=
0
{\displaystyle t=0}
g
(
0
)
≠
0
{\displaystyle g(0)\neq 0}
λ
>
−
1
{\displaystyle \lambda >-1}
Suppose, in addition, either that
|
φ
(
t
)
|
<
K
e
b
t
∀
t
>
0
,
{\displaystyle |\varphi (t)|<Ke^{bt}\ \forall t>0,}
where
K
,
b
{\displaystyle K,b}
t
{\displaystyle t}
∫
0
T
|
φ
(
t
)
|
d
t
<
∞
.
{\displaystyle \int _{0}^{T}|\varphi (t)|\,\mathrm {d} t<\infty .}
Then, it is true that for all positive
x
{\displaystyle x}
|
∫
0
T
e
−
x
t
φ
(
t
)
d
t
|
<
∞
{\displaystyle \left|\int _{0}^{T}e^{-xt}\varphi (t)\,\mathrm {d} t\right|<\infty }
and that the following asymptotic equivalence holds:
∫
0
T
e
−
x
t
φ
(
t
)
d
t
∼
∑
n
=
0
∞
g
(
n
)
(
0
)
Γ
(
λ
+
n
+
1
)
n
!
x
λ
+
n
+
1
,
(
x
>
0
,
x
→
∞
)
.
{\displaystyle \int _{0}^{T}e^{-xt}\varphi (t)\,\mathrm {d} t\sim \ \sum _{n=0}^{\infty }{\frac {g^{(n)}(0)\ \Gamma (\lambda +n+1)}{n!\ x^{\lambda +n+1}}},\ \ (x>0,\ x\rightarrow \infty ).}
See, for instance, Watson (1918) for the original proof or Miller (2006) for a more recent development.
We will prove the version of Watson's lemma which assumes that
|
φ
(
t
)
|
{\displaystyle |\varphi (t)|}
t
→
∞
{\displaystyle t\to \infty }
g
(
t
)
{\displaystyle g(t)}
Taylor series . Since the derivatives of
g
{\displaystyle g}
Taylor's theorem with remainder in the remaining small interval, then adding the tail back on in the end. At each step we will carefully estimate how much we are throwing away or adding on. This proof is a modification of the one found in Miller (2006) .
Let
0
<
T
≤
∞
{\displaystyle 0<T\leq \infty }
φ
{\displaystyle \varphi }
φ
(
t
)
=
t
λ
g
(
t
)
{\displaystyle \varphi (t)=t^{\lambda }g(t)}
λ
>
−
1
{\displaystyle \lambda >-1}
g
{\displaystyle g}
[
0
,
δ
]
{\displaystyle [0,\delta ]}
0
<
δ
<
T
{\displaystyle 0<\delta <T}
|
φ
(
t
)
|
≤
K
e
b
t
{\displaystyle |\varphi (t)|\leq Ke^{bt}}
δ
≤
t
≤
T
{\displaystyle \delta \leq t\leq T}
K
{\displaystyle K}
b
{\displaystyle b}
t
{\displaystyle t}
We can show that the integral is finite for
x
{\displaystyle x}
(
1
)
∫
0
T
e
−
x
t
φ
(
t
)
d
t
=
∫
0
δ
e
−
x
t
φ
(
t
)
d
t
+
∫
δ
T
e
−
x
t
φ
(
t
)
d
t
{\displaystyle (1)\quad \int _{0}^{T}e^{-xt}\varphi (t)\,\mathrm {d} t=\int _{0}^{\delta }e^{-xt}\varphi (t)\,\mathrm {d} t+\int _{\delta }^{T}e^{-xt}\varphi (t)\,\mathrm {d} t}
and estimating each term.
For the first term we have
|
∫
0
δ
e
−
x
t
φ
(
t
)
d
t
|
≤
∫
0
δ
e
−
x
t
|
φ
(
t
)
|
d
t
≤
∫
0
δ
|
φ
(
t
)
|
d
t
{\displaystyle \left|\int _{0}^{\delta }e^{-xt}\varphi (t)\,\mathrm {d} t\right|\leq \int _{0}^{\delta }e^{-xt}|\varphi (t)|\,\mathrm {d} t\leq \int _{0}^{\delta }|\varphi (t)|\,\mathrm {d} t}
for
x
≥
0
{\displaystyle x\geq 0}
g
{\displaystyle g}
[
0
,
δ
]
{\displaystyle [0,\delta ]}
λ
>
−
1
{\displaystyle \lambda >-1}
φ
{\displaystyle \varphi }
x
>
b
{\displaystyle x>b}
|
∫
δ
T
e
−
x
t
φ
(
t
)
d
t
|
≤
∫
δ
T
e
−
x
t
|
φ
(
t
)
|
d
t
≤
K
∫
δ
T
e
(
b
−
x
)
t
d
t
≤
K
∫
δ
∞
e
(
b
−
x
)
t
d
t
=
K
e
(
b
−
x
)
δ
x
−
b
.
{\displaystyle {\begin{aligned}\left|\int _{\delta }^{T}e^{-xt}\varphi (t)\,\mathrm {d} t\right|&\leq \int _{\delta }^{T}e^{-xt}|\varphi (t)|\,\mathrm {d} t\\&\leq K\int _{\delta }^{T}e^{(b-x)t}\,\mathrm {d} t\\&\leq K\int _{\delta }^{\infty }e^{(b-x)t}\,\mathrm {d} t\\&=K\,{\frac {e^{(b-x)\delta }}{x-b}}.\end{aligned}}}
The finiteness of the original integral then follows from applying the triangle inequality to
(
1
)
{\displaystyle (1)}
We can deduce from the above calculation that
(
2
)
∫
0
T
e
−
x
t
φ
(
t
)
d
t
=
∫
0
δ
e
−
x
t
φ
(
t
)
d
t
+
O
(
x
−
1
e
−
δ
x
)
{\displaystyle (2)\quad \int _{0}^{T}e^{-xt}\varphi (t)\,\mathrm {d} t=\int _{0}^{\delta }e^{-xt}\varphi (t)\,\mathrm {d} t+O\left(x^{-1}e^{-\delta x}\right)}
as
x
→
∞
{\displaystyle x\to \infty }
By appealing to Taylor's theorem with remainder we know that, for each integer
N
≥
0
{\displaystyle N\geq 0}
g
(
t
)
=
∑
n
=
0
N
g
(
n
)
(
0
)
n
!
t
n
+
g
(
N
+
1
)
(
t
∗
)
(
N
+
1
)
!
t
N
+
1
{\displaystyle g(t)=\sum _{n=0}^{N}{\frac {g^{(n)}(0)}{n!}}\,t^{n}+{\frac {g^{(N+1)}(t^{*})}{(N+1)!}}\,t^{N+1}}
for
0
≤
t
≤
δ
{\displaystyle 0\leq t\leq \delta }
0
≤
t
∗
≤
t
{\displaystyle 0\leq t^{*}\leq t}
(
2
)
{\displaystyle (2)}
(
3
)
∫
0
δ
e
−
x
t
φ
(
t
)
d
t
=
∫
0
δ
e
−
x
t
t
λ
g
(
t
)
d
t
=
∑
n
=
0
N
g
(
n
)
(
0
)
n
!
∫
0
δ
t
λ
+
n
e
−
x
t
d
t
+
1
(
N
+
1
)
!
∫
0
δ
g
(
N
+
1
)
(
t
∗
)
t
λ
+
N
+
1
e
−
x
t
d
t
.
{\displaystyle {\begin{aligned}(3)\quad \int _{0}^{\delta }e^{-xt}\varphi (t)\,\mathrm {d} t&=\int _{0}^{\delta }e^{-xt}t^{\lambda }g(t)\,\mathrm {d} t\\&=\sum _{n=0}^{N}{\frac {g^{(n)}(0)}{n!}}\int _{0}^{\delta }t^{\lambda +n}e^{-xt}\,\mathrm {d} t+{\frac {1}{(N+1)!}}\int _{0}^{\delta }g^{(N+1)}(t^{*})\,t^{\lambda +N+1}e^{-xt}\,\mathrm {d} t.\end{aligned}}}
To bound the term involving the remainder we use the assumption that
g
(
N
+
1
)
{\displaystyle g^{(N+1)}}
[
0
,
δ
]
{\displaystyle [0,\delta ]}
|
∫
0
δ
g
(
N
+
1
)
(
t
∗
)
t
λ
+
N
+
1
e
−
x
t
d
t
|
≤
sup
t
∈
[
0
,
δ
]
|
g
(
N
+
1
)
(
t
)
|
∫
0
δ
t
λ
+
N
+
1
e
−
x
t
d
t
<
sup
t
∈
[
0
,
δ
]
|
g
(
N
+
1
)
(
t
)
|
∫
0
∞
t
λ
+
N
+
1
e
−
x
t
d
t
=
sup
t
∈
[
0
,
δ
]
|
g
(
N
+
1
)
(
t
)
|
Γ
(
λ
+
N
+
2
)
x
λ
+
N
+
2
.
{\displaystyle {\begin{aligned}\left|\int _{0}^{\delta }g^{(N+1)}(t^{*})\,t^{\lambda +N+1}e^{-xt}\,\mathrm {d} t\right|&\leq \sup _{t\in [0,\delta ]}\left|g^{(N+1)}(t)\right|\int _{0}^{\delta }t^{\lambda +N+1}e^{-xt}\,\mathrm {d} t\\&<\sup _{t\in [0,\delta ]}\left|g^{(N+1)}(t)\right|\int _{0}^{\infty }t^{\lambda +N+1}e^{-xt}\,\mathrm {d} t\\&=\sup _{t\in [0,\delta ]}\left|g^{(N+1)}(t)\right|\,{\frac {\Gamma (\lambda +N+2)}{x^{\lambda +N+2}}}.\end{aligned}}}
Here we have used the fact that
∫
0
∞
t
a
e
−
x
t
d
t
=
Γ
(
a
+
1
)
x
a
+
1
{\displaystyle \int _{0}^{\infty }t^{a}e^{-xt}\,\mathrm {d} t={\frac {\Gamma (a+1)}{x^{a+1}}}}
if
x
>
0
{\displaystyle x>0}
a
>
−
1
{\displaystyle a>-1}
Γ
{\displaystyle \Gamma }
gamma function .
From the above calculation we see from
(
3
)
{\displaystyle (3)}
(
4
)
∫
0
δ
e
−
x
t
φ
(
t
)
d
t
=
∑
n
=
0
N
g
(
n
)
(
0
)
n
!
∫
0
δ
t
λ
+
n
e
−
x
t
d
t
+
O
(
x
−
λ
−
N
−
2
)
{\displaystyle (4)\quad \int _{0}^{\delta }e^{-xt}\varphi (t)\,\mathrm {d} t=\sum _{n=0}^{N}{\frac {g^{(n)}(0)}{n!}}\int _{0}^{\delta }t^{\lambda +n}e^{-xt}\,\mathrm {d} t+O\left(x^{-\lambda -N-2}\right)}
as
x
→
∞
{\displaystyle x\to \infty }
We will now add the tails on to each integral in
(
4
)
{\displaystyle (4)}
n
{\displaystyle n}
∫
0
δ
t
λ
+
n
e
−
x
t
d
t
=
∫
0
∞
t
λ
+
n
e
−
x
t
d
t
−
∫
δ
∞
t
λ
+
n
e
−
x
t
d
t
=
Γ
(
λ
+
n
+
1
)
x
λ
+
n
+
1
−
∫
δ
∞
t
λ
+
n
e
−
x
t
d
t
,
{\displaystyle {\begin{aligned}\int _{0}^{\delta }t^{\lambda +n}e^{-xt}\,\mathrm {d} t&=\int _{0}^{\infty }t^{\lambda +n}e^{-xt}\,\mathrm {d} t-\int _{\delta }^{\infty }t^{\lambda +n}e^{-xt}\,\mathrm {d} t\\[5pt]&={\frac {\Gamma (\lambda +n+1)}{x^{\lambda +n+1}}}-\int _{\delta }^{\infty }t^{\lambda +n}e^{-xt}\,\mathrm {d} t,\end{aligned}}}
and we will show that the remaining integrals are exponentially small. Indeed, if we make the change of variables
t
=
s
+
δ
{\displaystyle t=s+\delta }
∫
δ
∞
t
λ
+
n
e
−
x
t
d
t
=
∫
0
∞
(
s
+
δ
)
λ
+
n
e
−
x
(
s
+
δ
)
d
s
=
e
−
δ
x
∫
0
∞
(
s
+
δ
)
λ
+
n
e
−
x
s
d
s
≤
e
−
δ
x
∫
0
∞
(
s
+
δ
)
λ
+
n
e
−
s
d
s
{\displaystyle {\begin{aligned}\int _{\delta }^{\infty }t^{\lambda +n}e^{-xt}\,\mathrm {d} t&=\int _{0}^{\infty }(s+\delta )^{\lambda +n}e^{-x(s+\delta )}\,\mathrm {d} s\\[5pt]&=e^{-\delta x}\int _{0}^{\infty }(s+\delta )^{\lambda +n}e^{-xs}\,\mathrm {d} s\\[5pt]&\leq e^{-\delta x}\int _{0}^{\infty }(s+\delta )^{\lambda +n}e^{-s}\,\mathrm {d} s\end{aligned}}}
for
x
≥
1
{\displaystyle x\geq 1}
∫
0
δ
t
λ
+
n
e
−
x
t
d
t
=
Γ
(
λ
+
n
+
1
)
x
λ
+
n
+
1
+
O
(
e
−
δ
x
)
as
x
→
∞
.
{\displaystyle \int _{0}^{\delta }t^{\lambda +n}e^{-xt}\,\mathrm {d} t={\frac {\Gamma (\lambda +n+1)}{x^{\lambda +n+1}}}+O\left(e^{-\delta x}\right){\text{ as }}x\to \infty .}
If we substitute this last result into
(
4
)
{\displaystyle (4)}
∫
0
δ
e
−
x
t
φ
(
t
)
d
t
=
∑
n
=
0
N
g
(
n
)
(
0
)
Γ
(
λ
+
n
+
1
)
n
!
x
λ
+
n
+
1
+
O
(
e
−
δ
x
)
+
O
(
x
−
λ
−
N
−
2
)
=
∑
n
=
0
N
g
(
n
)
(
0
)
Γ
(
λ
+
n
+
1
)
n
!
x
λ
+
n
+
1
+
O
(
x
−
λ
−
N
−
2
)
{\displaystyle {\begin{aligned}\int _{0}^{\delta }e^{-xt}\varphi (t)\,\mathrm {d} t&=\sum _{n=0}^{N}{\frac {g^{(n)}(0)\ \Gamma (\lambda +n+1)}{n!\ x^{\lambda +n+1}}}+O\left(e^{-\delta x}\right)+O\left(x^{-\lambda -N-2}\right)\\&=\sum _{n=0}^{N}{\frac {g^{(n)}(0)\ \Gamma (\lambda +n+1)}{n!\ x^{\lambda +n+1}}}+O\left(x^{-\lambda -N-2}\right)\end{aligned}}}
as
x
→
∞
{\displaystyle x\to \infty }
(
2
)
{\displaystyle (2)}
∫
0
T
e
−
x
t
φ
(
t
)
d
t
=
∑
n
=
0
N
g
(
n
)
(
0
)
Γ
(
λ
+
n
+
1
)
n
!
x
λ
+
n
+
1
+
O
(
x
−
λ
−
N
−
2
)
+
O
(
x
−
1
e
−
δ
x
)
=
∑
n
=
0
N
g
(
n
)
(
0
)
Γ
(
λ
+
n
+
1
)
n
!
x
λ
+
n
+
1
+
O
(
x
−
λ
−
N
−
2
)
{\displaystyle {\begin{aligned}\int _{0}^{T}e^{-xt}\varphi (t)\,\mathrm {d} t&=\sum _{n=0}^{N}{\frac {g^{(n)}(0)\ \Gamma (\lambda +n+1)}{n!\ x^{\lambda +n+1}}}+O\left(x^{-\lambda -N-2}\right)+O\left(x^{-1}e^{-\delta x}\right)\\&=\sum _{n=0}^{N}{\frac {g^{(n)}(0)\ \Gamma (\lambda +n+1)}{n!\ x^{\lambda +n+1}}}+O\left(x^{-\lambda -N-2}\right)\end{aligned}}}
as
x
→
∞
{\displaystyle x\to \infty }
Since this last expression is true for each integer
N
≥
0
{\displaystyle N\geq 0}
∫
0
T
e
−
x
t
φ
(
t
)
d
t
∼
∑
n
=
0
∞
g
(
n
)
(
0
)
Γ
(
λ
+
n
+
1
)
n
!
x
λ
+
n
+
1
{\displaystyle \int _{0}^{T}e^{-xt}\varphi (t)\,\mathrm {d} t\sim \sum _{n=0}^{\infty }{\frac {g^{(n)}(0)\ \Gamma (\lambda +n+1)}{n!\ x^{\lambda +n+1}}}}
as
x
→
∞
{\displaystyle x\to \infty }
asymptotic expansion of the integral in question.
When
0
<
a
<
b
{\displaystyle 0<a<b}
confluent hypergeometric function of the first kind has the integral representation
1
F
1
(
a
,
b
,
x
)
=
Γ
(
b
)
Γ
(
a
)
Γ
(
b
−
a
)
∫
0
1
e
x
t
t
a
−
1
(
1
−
t
)
b
−
a
−
1
d
t
,
{\displaystyle {}_{1}F_{1}(a,b,x)={\frac {\Gamma (b)}{\Gamma (a)\Gamma (b-a)}}\int _{0}^{1}e^{xt}t^{a-1}(1-t)^{b-a-1}\,\mathrm {d} t,}
where
Γ
{\displaystyle \Gamma }
gamma function . The change of variables
t
=
1
−
s
{\displaystyle t=1-s}
1
F
1
(
a
,
b
,
x
)
=
Γ
(
b
)
Γ
(
a
)
Γ
(
b
−
a
)
e
x
∫
0
1
e
−
x
s
(
1
−
s
)
a
−
1
s
b
−
a
−
1
d
s
,
{\displaystyle {}_{1}F_{1}(a,b,x)={\frac {\Gamma (b)}{\Gamma (a)\Gamma (b-a)}}\,e^{x}\int _{0}^{1}e^{-xs}(1-s)^{a-1}s^{b-a-1}\,ds,}
which is now amenable to the use of Watson's lemma. Taking
λ
=
b
−
a
−
1
{\displaystyle \lambda =b-a-1}
g
(
s
)
=
(
1
−
s
)
a
−
1
{\displaystyle g(s)=(1-s)^{a-1}}
∫
0
1
e
−
x
s
(
1
−
s
)
a
−
1
s
b
−
a
−
1
d
s
∼
Γ
(
b
−
a
)
x
a
−
b
as
x
→
∞
with
x
>
0
,
{\displaystyle \int _{0}^{1}e^{-xs}(1-s)^{a-1}s^{b-a-1}\,ds\sim \Gamma (b-a)x^{a-b}\quad {\text{as }}x\to \infty {\text{ with }}x>0,}
which allows us to conclude that
1
F
1
(
a
,
b
,
x
)
∼
Γ
(
b
)
Γ
(
a
)
x
a
−
b
e
x
as
x
→
∞
with
x
>
0.
{\displaystyle {}_{1}F_{1}(a,b,x)\sim {\frac {\Gamma (b)}{\Gamma (a)}}\,x^{a-b}e^{x}\quad {\text{as }}x\to \infty {\text{ with }}x>0.}
References
edit
Miller, P.D. (2006), Applied Asymptotic Analysis , Providence, RI: American Mathematical Society, p. 467, ISBN 978-0-8218-4078-8 Watson, G. N. (1918), "The harmonic functions associated with the parabolic cylinder" , Proceedings of the London Mathematical Society , vol. 2, no. 17, pp. 116–148, doi :10.1112/plms/s2-17.1.116 Ablowitz, M. J., Fokas, A. S. (2003). Complex variables: introduction and applications. Cambridge University Press . 
![{\displaystyle [0,\delta ]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e470c2978f96e03afad4bc73d198c8dd6ade046a)
![{\displaystyle {\begin{aligned}\left|\int _{0}^{\delta }g^{(N+1)}(t^{*})\,t^{\lambda +N+1}e^{-xt}\,\mathrm {d} t\right|&\leq \sup _{t\in [0,\delta ]}\left|g^{(N+1)}(t)\right|\int _{0}^{\delta }t^{\lambda +N+1}e^{-xt}\,\mathrm {d} t\\&<\sup _{t\in [0,\delta ]}\left|g^{(N+1)}(t)\right|\int _{0}^{\infty }t^{\lambda +N+1}e^{-xt}\,\mathrm {d} t\\&=\sup _{t\in [0,\delta ]}\left|g^{(N+1)}(t)\right|\,{\frac {\Gamma (\lambda +N+2)}{x^{\lambda +N+2}}}.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/088c6d8aead67a29697d944b25ece6f6740c5714)
![{\displaystyle {\begin{aligned}\int _{0}^{\delta }t^{\lambda +n}e^{-xt}\,\mathrm {d} t&=\int _{0}^{\infty }t^{\lambda +n}e^{-xt}\,\mathrm {d} t-\int _{\delta }^{\infty }t^{\lambda +n}e^{-xt}\,\mathrm {d} t\\[5pt]&={\frac {\Gamma (\lambda +n+1)}{x^{\lambda +n+1}}}-\int _{\delta }^{\infty }t^{\lambda +n}e^{-xt}\,\mathrm {d} t,\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5a636c582e462d650f8bb182df4a3967f0c9f59d)
![{\displaystyle {\begin{aligned}\int _{\delta }^{\infty }t^{\lambda +n}e^{-xt}\,\mathrm {d} t&=\int _{0}^{\infty }(s+\delta )^{\lambda +n}e^{-x(s+\delta )}\,\mathrm {d} s\\[5pt]&=e^{-\delta x}\int _{0}^{\infty }(s+\delta )^{\lambda +n}e^{-xs}\,\mathrm {d} s\\[5pt]&\leq e^{-\delta x}\int _{0}^{\infty }(s+\delta )^{\lambda +n}e^{-s}\,\mathrm {d} s\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ff7014de6e6869b74a2c78b7ab408d78cf4ad82a)
