在數學 上，以皮埃爾-西蒙·拉普拉斯 命名的拉普拉斯方法 是用於得出下列積分 形式的近似解的方法：
∫
a
b
e
M
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}\!e^{Mf(x)}\,dx}
其中的 ƒ (x ) 是一個二次可微 函數 ， M 是一個很大的數，而積分邊界點 a 與 b 則允許為無限大。此外，函數 ƒ (x ) 在此積分範圍內的 全域極大值 所在處必須是唯一的並且不在邊界點上。則它的近似解可以寫為:
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a
b
e
M
f
(
x
)
d
x
≈
2
π
M
|
f
″
(
x
0
)
|
e
M
f
(
x
0
)
as
M
→
∞
.
{\displaystyle \int _{a}^{b}\!e^{Mf(x)}\,dx\approx {\sqrt {\frac {2\pi }{M\left|f''(x_{0})\right|}}}e^{Mf(x_{0})}{\text{ as }}M\to \infty .\,}
其中的 x 0 為極大值所在處。這方法最早是拉普拉斯在 (1774, pp. 366–367) 所發表。(待考查)
1. 相對誤差
首先，我們得先知道的是，這裏所謂的近似解是以 相對誤差 在講，而不是指 絕對誤差 。因此，令
s
=
2
π
M
|
f
″
(
x
0
)
|
{\displaystyle s={\sqrt {\frac {2\pi }{M\left|f''(x_{0})\right|}}}}
此 s 很顯然的當 M 很大時為一個微小的數，則上述的積分可以寫為
∫
a
b
e
M
f
(
x
)
d
x
=
s
e
M
f
(
x
0
)
1
s
∫
a
b
e
M
(
f
(
x
)
−
f
(
x
0
)
)
d
x
=
s
e
M
f
(
x
0
)
∫
(
a
−
x
0
)
/
s
(
b
−
x
0
)
/
s
e
M
(
f
(
s
y
+
x
0
)
−
f
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x
0
)
)
d
y
{\displaystyle {\begin{aligned}\int _{a}^{b}\!e^{Mf(x)}\,dx&=se^{Mf(x_{0})}{\frac {1}{s}}\int _{a}^{b}\!e^{M(f(x)-f(x_{0}))}\,dx\\&=se^{Mf(x_{0})}\int _{(a-x_{0})/s}^{(b-x_{0})/s}\!e^{M(f(sy+x_{0})-f(x_{0}))}\,dy\end{aligned}}}
因此，我們的相對誤差很顯然的為
|
∫
(
a
−
x
0
)
/
s
(
b
−
x
0
)
/
s
e
M
(
f
(
s
y
+
x
0
)
−
f
(
x
0
)
)
d
y
−
1
|
{\displaystyle \left|\int _{(a-x_{0})/s}^{(b-x_{0})/s}e^{M(f(sy+x_{0})-f(x_{0}))}dy-1\right|}
現在，讓我們把積分分為
y
∈
[
−
D
y
,
D
y
]
{\displaystyle y\in [-D_{y},D_{y}]}
基於上述四點，就有辦法證明拉普拉斯方法的可靠性。而 Fog(2008) 又將此方法推廣到任意精確。 ***待考查***
此方法的正式表述與證明：
假設
f
(
x
)
{\displaystyle f(x)}
x
0
∈
(
a
,
b
)
{\displaystyle x_{0}\in (a,b)}
f
(
x
0
)
=
max
[
a
,
b
]
f
(
x
)
{\displaystyle f(x_{0})=\max _{[a,b]}f(x)}
x
0
{\displaystyle x_{0}}
f
″
(
x
0
)
<
0
{\displaystyle f''(x_{0})<0}
則，
lim
n
→
+
∞
(
∫
a
b
e
n
f
(
x
)
d
x
(
e
n
f
(
x
0
)
2
π
n
(
−
f
″
(
x
0
)
)
)
)
=
1
{\displaystyle \lim _{n\to +\infty }\left({\frac {\int _{a}^{b}e^{nf(x)}\,dx}{\left(e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}\right)}}\right)=1}
證明：
下限 ：
在
f
″
{\displaystyle f''}
ε
{\displaystyle \varepsilon }
δ
>
0
{\displaystyle \delta >0}
|
x
0
−
x
|
<
δ
{\displaystyle |x_{0}-x|<\delta }
f
″
(
x
)
≥
f
″
(
x
0
)
−
ε
.
{\displaystyle f''(x)\geq f''(x_{0})-\varepsilon .}
泰勒展開 我們可以得知，在
x
∈
(
x
0
−
δ
,
x
0
+
δ
)
{\displaystyle x\in (x_{0}-\delta ,x_{0}+\delta )}
f
(
x
)
≥
f
(
x
0
)
+
1
2
(
f
″
(
x
0
)
−
ε
)
(
x
−
x
0
)
2
{\displaystyle f(x)\geq f(x_{0})+{\frac {1}{2}}(f''(x_{0})-\varepsilon )(x-x_{0})^{2}}
這樣，我們就得到本方法的積分下限了：
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a
b
e
n
f
(
x
)
d
x
≥
∫
x
0
−
δ
x
0
+
δ
e
n
f
(
x
)
d
x
≥
e
n
f
(
x
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)
∫
x
0
−
δ
x
0
+
δ
e
n
2
(
f
″
(
x
0
)
−
ε
)
(
x
−
x
0
)
2
d
x
=
e
n
f
(
x
0
)
1
n
(
−
f
″
(
x
0
)
+
ε
)
∫
−
δ
n
(
−
f
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x
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+
ε
)
δ
n
(
−
f
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(
x
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)
+
ε
)
e
−
1
2
y
2
d
y
{\displaystyle \int _{a}^{b}e^{nf(x)}\,dx\geq \int _{x_{0}-\delta }^{x_{0}+\delta }e^{nf(x)}\,dx\geq e^{nf(x_{0})}\int _{x_{0}-\delta }^{x_{0}+\delta }e^{{\frac {n}{2}}(f''(x_{0})-\varepsilon )(x-x_{0})^{2}}\,dx=e^{nf(x_{0})}{\sqrt {\frac {1}{n(-f''(x_{0})+\varepsilon )}}}\int _{-\delta {\sqrt {n(-f''(x_{0})+\varepsilon )}}}^{\delta {\sqrt {n(-f''(x_{0})+\varepsilon )}}}e^{-{\frac {1}{2}}y^{2}}\,dy}
其中最後一個等號來自於變數變換：
y
=
n
(
−
f
″
(
x
0
)
+
ε
)
(
x
−
x
0
)
{\displaystyle y={\sqrt {n(-f''(x_{0})+\varepsilon )}}(x-x_{0})}
f
″
(
x
0
)
<
0
{\displaystyle f''(x_{0})<0}
接着讓我們對上面的不等式兩邊同除以
e
n
f
(
x
0
)
2
π
n
(
−
f
″
(
x
0
)
)
{\displaystyle e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}
n
{\displaystyle n}
lim
n
→
+
∞
(
∫
a
b
e
n
f
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x
)
d
x
(
e
n
f
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x
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)
2
π
n
(
−
f
″
(
x
0
)
)
)
)
≥
lim
n
→
+
∞
1
2
π
∫
−
δ
n
(
−
f
″
(
x
0
)
+
ε
)
δ
n
(
−
f
″
(
x
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)
+
ε
)
e
−
1
2
y
2
d
y
−
f
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(
x
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−
f
″
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)
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ε
=
−
f
″
(
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−
f
″
(
x
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)
+
ε
{\displaystyle \lim _{n\to +\infty }\left({\frac {\int _{a}^{b}e^{nf(x)}\,dx}{\left(e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}\right)}}\right)\geq \lim _{n\to +\infty }{\frac {1}{\sqrt {2\pi }}}\int _{-\delta {\sqrt {n(-f''(x_{0})+\varepsilon )}}}^{\delta {\sqrt {n(-f''(x_{0})+\varepsilon )}}}e^{-{\frac {1}{2}}y^{2}}\,dy{\sqrt {\frac {-f''(x_{0})}{-f''(x_{0})+\varepsilon }}}={\sqrt {\frac {-f''(x_{0})}{-f''(x_{0})+\varepsilon }}}}
既然上式是對任意大於0的
ε
{\displaystyle \varepsilon }
lim
n
→
+
∞
(
∫
a
b
e
n
f
(
x
)
d
x
(
e
n
f
(
x
0
)
2
π
n
(
−
f
″
(
x
0
)
)
)
)
≥
1
{\displaystyle \lim _{n\to +\infty }\left({\frac {\int _{a}^{b}e^{nf(x)}\,dx}{\left(e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}\right)}}\right)\geq 1}
請注意，上面的證明也適用於
a
=
−
∞
{\displaystyle a=-\infty }
b
=
∞
{\displaystyle b=\infty }
上限 ：
證明上限的部分其實和證明下限的部分很像，但是會較麻煩。再一次，我們取
ε
>
0
{\displaystyle \varepsilon >0}
ε
{\displaystyle \varepsilon }
f
″
(
x
0
)
+
ε
<
0
{\displaystyle f''(x_{0})+\varepsilon <0}
f
″
{\displaystyle f''}
泰勒定理 我們會發現總存在一個
δ
>
0
{\displaystyle \delta >0}
|
x
−
x
0
|
<
δ
{\displaystyle |x-x_{0}|<\delta }
f
(
x
)
≤
f
(
x
0
)
+
1
2
(
f
″
(
x
0
)
+
ε
)
(
x
−
x
0
)
2
{\displaystyle f(x)\leq f(x_{0})+{\frac {1}{2}}(f''(x_{0})+\varepsilon )(x-x_{0})^{2}}
a
,
b
{\displaystyle a,b}
x
0
{\displaystyle x_{0}}
η
>
0
{\displaystyle \eta >0}
|
x
−
x
0
|
≥
δ
{\displaystyle |x-x_{0}|\geq \delta }
f
(
x
)
≤
f
(
x
0
)
−
η
{\displaystyle f(x)\leq f(x_{0})-\eta }
現在，萬事俱備，東風就在下面啦：
∫
a
b
e
n
f
(
x
)
d
x
≤
∫
a
x
0
−
δ
e
n
f
(
x
)
d
x
+
∫
x
0
−
δ
x
0
+
δ
e
n
f
(
x
)
d
x
+
∫
x
0
+
δ
b
e
n
f
(
x
)
d
x
≤
(
b
−
a
)
e
n
(
f
(
x
0
)
−
η
)
+
∫
x
0
−
δ
x
0
+
δ
e
n
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}e^{nf(x)}\,dx\leq \int _{a}^{x_{0}-\delta }e^{nf(x)}\,dx+\int _{x_{0}-\delta }^{x_{0}+\delta }e^{nf(x)}\,dx+\int _{x_{0}+\delta }^{b}e^{nf(x)}\,dx\leq (b-a)e^{n(f(x_{0})-\eta )}+\int _{x_{0}-\delta }^{x_{0}+\delta }e^{nf(x)}\,dx}
≤
(
b
−
a
)
e
n
(
f
(
x
0
)
−
η
)
+
e
n
f
(
x
0
)
∫
x
0
−
δ
x
0
+
δ
e
n
2
(
f
″
(
x
0
)
+
ε
)
(
x
−
x
0
)
2
d
x
≤
(
b
−
a
)
e
n
(
f
(
x
0
)
−
η
)
+
e
n
f
(
x
0
)
∫
−
∞
+
∞
e
n
2
(
f
″
(
x
0
)
+
ε
)
(
x
−
x
0
)
2
d
x
{\displaystyle \leq (b-a)e^{n(f(x_{0})-\eta )}+e^{nf(x_{0})}\int _{x_{0}-\delta }^{x_{0}+\delta }e^{{\frac {n}{2}}(f''(x_{0})+\varepsilon )(x-x_{0})^{2}}\,dx\leq (b-a)e^{n(f(x_{0})-\eta )}+e^{nf(x_{0})}\int _{-\infty }^{+\infty }e^{{\frac {n}{2}}(f''(x_{0})+\varepsilon )(x-x_{0})^{2}}\,dx}
≤
(
b
−
a
)
e
n
(
f
(
x
0
)
−
η
)
+
e
n
f
(
x
0
)
2
π
n
(
−
f
″
(
x
0
)
−
ε
)
{\displaystyle \leq (b-a)e^{n(f(x_{0})-\eta )}+e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0})-\varepsilon )}}}}
如果我們對上面的不等式兩邊皆除以
e
n
f
(
x
0
)
2
π
n
(
−
f
″
(
x
0
)
)
{\displaystyle e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}
lim
n
→
+
∞
(
∫
a
b
e
n
f
(
x
)
d
x
(
e
n
f
(
x
0
)
2
π
n
(
−
f
″
(
x
0
)
)
)
)
≤
lim
n
→
+
∞
(
(
b
−
a
)
e
−
η
n
n
(
−
f
″
(
x
0
)
)
2
π
+
−
f
″
(
x
0
)
−
f
″
(
x
0
)
−
ε
)
=
−
f
″
(
x
0
)
−
f
″
(
x
0
)
−
ε
{\displaystyle \lim _{n\to +\infty }\left({\frac {\int _{a}^{b}e^{nf(x)}\,dx}{\left(e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}\right)}}\right)\leq \lim _{n\to +\infty }\left((b-a)e^{-\eta n}{\sqrt {\frac {n(-f''(x_{0}))}{2\pi }}}+{\sqrt {\frac {-f''(x_{0})}{-f''(x_{0})-\varepsilon }}}\right)={\sqrt {\frac {-f''(x_{0})}{-f''(x_{0})-\varepsilon }}}}
再一次，因為
ε
{\displaystyle \varepsilon }
lim
n
→
+
∞
(
∫
a
b
e
n
f
(
x
)
d
x
(
e
n
f
(
x
0
)
2
π
n
(
−
f
″
(
x
0
)
)
)
)
≤
1
{\displaystyle \lim _{n\to +\infty }\left({\frac {\int _{a}^{b}e^{nf(x)}\,dx}{\left(e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}\right)}}\right)\leq 1}
把上限與下限兩個證明同時考慮，整個證明就完成了。
注意，關於上限的證明，很明顯的當我們把它應用在
a
{\displaystyle a}
b
{\displaystyle b}
n
=
1
{\displaystyle n=1}
∫
a
b
e
n
f
(
x
)
d
x
{\displaystyle \int _{a}^{b}e^{nf(x)}\,dx}
η
{\displaystyle \eta }
[
a
,
b
]
{\displaystyle [a,b]}
∫
a
x
0
−
δ
e
n
f
(
x
)
d
x
+
∫
x
0
+
δ
b
e
n
f
(
x
)
d
x
{\displaystyle \int _{a}^{x_{0}-\delta }e^{nf(x)}\,dx+\int _{x_{0}+\delta }^{b}e^{nf(x)}\,dx}
必須利用上述的假設，而改為：
∫
a
x
0
−
δ
e
n
f
(
x
)
d
x
+
∫
x
0
+
δ
b
e
n
f
(
x
)
d
x
≤
∫
a
b
e
f
(
x
)
e
(
n
−
1
)
(
f
(
x
0
)
−
η
)
d
x
=
e
(
n
−
1
)
(
f
(
x
0
)
−
η
)
∫
a
b
e
f
(
x
)
d
x
{\displaystyle \int _{a}^{x_{0}-\delta }e^{nf(x)}\,dx+\int _{x_{0}+\delta }^{b}e^{nf(x)}\,dx\leq \int _{a}^{b}e^{f(x)}e^{(n-1)(f(x_{0})-\eta )}\,dx=e^{(n-1)(f(x_{0})-\eta )}\int _{a}^{b}e^{f(x)}\,dx}
以取代先前會得到的
(
b
−
a
)
e
n
(
f
(
x
0
)
−
η
)
{\displaystyle (b-a)e^{n(f(x_{0})-\eta )}}
e
n
f
(
x
0
)
2
π
n
(
−
f
″
(
x
0
)
)
{\displaystyle e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}
e
(
n
−
1
)
(
f
(
x
0
)
−
η
)
∫
a
b
e
f
(
x
)
d
x
e
n
f
(
x
0
)
2
π
n
(
−
f
″
(
x
0
)
)
=
e
−
(
n
−
1
)
η
n
e
−
f
(
x
0
)
∫
a
b
e
f
(
x
)
d
x
−
f
″
(
x
0
)
2
π
{\displaystyle {\frac {e^{(n-1)(f(x_{0})-\eta )}\int _{a}^{b}e^{f(x)}\,dx}{e^{nf(x_{0})}{\sqrt {\frac {2\pi }{n(-f''(x_{0}))}}}}}=e^{-(n-1)\eta }{\sqrt {n}}e^{-f(x_{0})}\int _{a}^{b}e^{f(x)}\,dx{\sqrt {\frac {-f''(x_{0})}{2\pi }}}}
這樣的話，當我們取
n
→
∞
{\displaystyle n\rightarrow \infty }
0
{\displaystyle 0}
再強調一次，這裏我們多加給無限大積分範圍的情形的條件，是充分，但非必要。不過，這樣的條件已經可以適用在許多情形了(但非全部)。 這考慮條件簡單來講就是積分區間得是被良好定義的(即不能是無限大的)，並且被積函數在
x
0
{\displaystyle x_{0}}
η
>
0
{\displaystyle \eta >0}
n
=
1
{\displaystyle n=1}
n
{\displaystyle n}
有時拉普拉斯方法也會被寫成其他形式，如：
∫
a
b
h
(
x
)
e
M
g
(
x
)
d
x
≈
2
π
M
|
g
″
(
x
0
)
|
h
(
x
0
)
e
M
g
(
x
0
)
as
M
→
∞
{\displaystyle \int _{a}^{b}\!h(x)e^{Mg(x)}\,dx\approx {\sqrt {\frac {2\pi }{M|g''(x_{0})|}}}h(x_{0})e^{Mg(x_{0})}{\text{ as }}M\to \infty \,}
其中
h
{\displaystyle h}
重要的是，這方法精確度與函數
g
(
x
)
{\displaystyle g(x)}
h
(
x
)
{\displaystyle h(x)}
[ 1]
相對誤差的推導
首先，由於極大值所在設為
x
0
=
0
{\displaystyle x_{0}=0}
|
R
|
{\displaystyle \left|R\right|}
∫
a
b
h
(
x
)
e
M
g
(
x
)
d
x
=
h
(
0
)
e
M
g
(
0
)
s
∫
a
/
s
b
/
s
h
(
x
)
h
(
0
)
e
M
[
g
(
s
y
)
−
g
(
0
)
]
d
y
⏟
1
+
R
{\displaystyle \int _{a}^{b}\!h(x)e^{Mg(x)}\,dx=h(0)e^{Mg(0)}s\underbrace {\int _{a/s}^{b/s}{\frac {h(x)}{h(0)}}e^{M\left[g(sy)-g(0)\right]}dy} _{1+R}}
其中
s
≡
2
π
M
|
g
″
(
0
)
|
{\displaystyle s\equiv {\sqrt {\frac {2\pi }{M\left|g''(0)\right|}}}}
A
≡
h
(
s
y
)
h
(
0
)
e
M
[
g
(
s
y
)
−
g
(
0
)
]
{\displaystyle A\equiv {\frac {h(sy)}{h(0)}}e^{M\left[g(sy)-g(0)\right]}}
A
0
≡
e
−
π
y
2
{\displaystyle A_{0}\equiv e^{-\pi y^{2}}}
|
R
|
=
|
∫
a
/
s
b
/
s
A
d
y
−
∫
−
∞
∞
A
0
d
y
|
{\displaystyle \left|R\right|=\left|\int _{a/s}^{b/s}A\,dy-\int _{-\infty }^{\infty }A_{0}\,dy\right|}
所以，只要我們求得上式的上限為何，即可得相對誤差。注意，這裏的推導還可以再優化，不過，由於此處我只想顯示它會收斂到0，因此，不再細推。
由於
|
A
+
B
|
≤
|
A
|
+
|
B
|
{\displaystyle \left|A+B\right|\leq |A|+|B|}
|
R
|
<
|
∫
D
y
∞
A
0
d
y
|
⏟
(
a
1
)
+
|
∫
D
y
b
/
s
A
d
y
|
⏟
(
b
1
)
+
|
∫
−
D
y
D
y
(
A
−
A
0
)
d
y
|
⏟
(
c
)
+
|
∫
a
/
s
−
D
y
A
d
y
|
⏟
(
b
2
)
+
|
∫
−
∞
−
D
y
A
0
d
y
|
⏟
(
a
2
)
{\displaystyle |R|<\underbrace {\left|\int _{D_{y}}^{\infty }A_{0}dy\right|} _{(a_{1})}+\underbrace {\left|\int _{D_{y}}^{b/s}Ady\right|} _{(b_{1})}+\underbrace {\left|\int _{-D_{y}}^{D_{y}}\left(A-A_{0}\right)dy\right|} _{(c)}+\underbrace {\left|\int _{a/s}^{-D_{y}}Ady\right|} _{(b_{2})}+\underbrace {\left|\int _{-\infty }^{-D_{y}}A_{0}dy\right|} _{(a_{2})}}
其中
(
a
1
)
{\displaystyle (a_{1})}
(
a
2
)
{\displaystyle (a_{2})}
(
a
1
)
{\displaystyle (a_{1})}
z
≡
π
y
2
{\displaystyle z\equiv \pi y^{2}}
(
a
1
)
=
|
1
2
π
∫
π
D
y
2
∞
e
−
z
z
−
1
/
2
d
z
|
<
e
−
π
D
y
2
2
π
D
y
{\displaystyle (a_{1})=\left|{\frac {1}{2{\sqrt {\pi }}}}\int _{\pi D_{y}^{2}}^{\infty }e^{-z}z^{-1/2}dz\right|<{\frac {e^{-\pi D_{y}^{2}}}{2\pi D_{y}}}}
也就是說，只要
D
y
{\displaystyle D_{y}}
而
(
b
1
)
{\displaystyle (b_{1})}
(
b
2
)
{\displaystyle (b_{2})}
(
b
1
)
≤
|
∫
D
y
b
/
s
[
h
(
s
y
)
h
(
0
)
]
max
e
M
m
(
s
y
)
d
y
|
{\displaystyle (b_{1})\leq \left|\int _{D_{y}}^{b/s}\left[{\frac {h(sy)}{h(0)}}\right]_{\text{max}}e^{Mm(sy)}dy\right|}
其中
m
(
x
)
≥
1
M
ln
h
(
x
)
h
(
0
)
+
g
(
x
)
−
g
(
0
)
as
x
∈
[
s
D
y
,
b
]
{\displaystyle m(x)\geq {\frac {1}{M}}\ln {\frac {h(x)}{h(0)}}+g(x)-g(0)\,\,{\text{as}}\,\,x\in [sD_{y},b]}
並且
h
(
x
)
{\displaystyle h(x)}
h
(
0
)
{\displaystyle h(0)}
x
=
s
D
y
{\displaystyle x=sD_{y}}
m
(
x
)
{\displaystyle m(x)}
m
(
s
y
)
=
g
(
s
D
y
)
−
g
(
0
)
+
g
′
(
s
D
y
)
(
s
y
−
s
D
y
)
{\displaystyle m(sy)=g(sD_{y})-g(0)+g'(sD_{y})\left(sy-sD_{y}\right)}
m
(
x
)
{\displaystyle m(x)}
x
=
s
D
y
{\displaystyle x=sD_{y}}
由圖可以看出，當
s
{\displaystyle s}
D
y
{\displaystyle D_{y}}
D
y
{\displaystyle D_{y}}
e
−
α
x
{\displaystyle e^{-\alpha x}}
由泰勒展開我們得知，
M
[
g
(
s
D
y
)
−
g
(
0
)
]
=
M
[
g
″
(
0
)
2
s
2
D
y
2
+
g
‴
(
ξ
)
6
s
3
D
y
3
]
as
ξ
∈
[
0
,
s
D
y
]
=
−
π
D
y
2
+
(
2
π
)
3
/
2
g
‴
(
ξ
)
D
y
3
6
M
|
g
″
(
0
)
|
3
/
2
,
{\displaystyle {\begin{aligned}M\left[g(sD_{y})-g(0)\right]&=M\left[{\frac {g''(0)}{2}}s^{2}D_{y}^{2}+{\frac {g'''(\xi )}{6}}s^{3}D_{y}^{3}\right]\,\,{\text{as}}\,\,\xi \in [0,sD_{y}]\\&=-\pi D_{y}^{2}+{\frac {(2\pi )^{3/2}g'''(\xi )D_{y}^{3}}{6{\sqrt {M}}|g''(0)|^{3/2}}},\end{aligned}}}
且
M
s
g
′
(
s
D
y
)
=
M
s
(
g
″
(
0
)
s
D
y
+
g
‴
(
ζ
)
2
s
2
D
y
2
)
,
as
ζ
∈
[
0
,
s
D
y
]
=
−
2
π
D
y
+
2
M
(
π
|
g
″
(
0
)
|
)
3
/
2
g
‴
(
ζ
)
D
y
2
{\displaystyle {\begin{aligned}Msg'(sD_{y})&=Ms\left(g''(0)sD_{y}+{\frac {g'''(\zeta )}{2}}s^{2}D_{y}^{2}\right),\,\,{\text{as}}\,\,\zeta \in [0,sD_{y}]\\&=-2\pi D_{y}+{\sqrt {\frac {2}{M}}}\left({\frac {\pi }{|g''(0)|}}\right)^{3/2}g'''(\zeta )D_{y}^{2}\end{aligned}}}
然後將它們代回
(
b
1
)
{\displaystyle (b_{1})}
M
{\displaystyle M}
(
b
1
)
≤
|
[
h
(
s
y
)
h
(
0
)
]
max
e
−
π
D
y
2
∫
0
b
/
s
−
D
y
e
−
2
π
D
y
y
d
y
|
≤
|
[
h
(
s
y
)
h
(
0
)
]
max
e
−
π
D
y
2
1
2
π
D
y
|
.
{\displaystyle {\begin{aligned}(b_{1})&\leq \left|\left[{\frac {h(sy)}{h(0)}}\right]_{\text{max}}e^{-\pi D_{y}^{2}}\int _{0}^{b/s-D_{y}}e^{-2\pi D_{y}y}dy\right|\\&\leq \left|\left[{\frac {h(sy)}{h(0)}}\right]_{\text{max}}e^{-\pi D_{y}^{2}}{\frac {1}{2\pi D_{y}}}\right|.\end{aligned}}}
所以，原則上也是
D
y
{\displaystyle D_{y}}
m
(
x
)
{\displaystyle m(x)}
D
y
{\displaystyle D_{y}}
至於靠近極大值的點的積分，我們一樣可以利用泰勒展開來求，當
h
(
x
)
{\displaystyle h(x)}
(
c
)
≤
∫
−
D
y
D
y
e
−
π
y
2
|
s
h
′
(
ξ
)
h
(
0
)
y
|
d
y
<
2
π
M
|
g
″
(
0
)
|
|
h
′
(
ξ
)
h
(
0
)
y
|
max
(
1
−
e
−
π
D
y
2
)
{\displaystyle {\begin{aligned}(c)&\leq \int _{-D_{y}}^{D_{y}}e^{-\pi y^{2}}\left|{\frac {sh'(\xi )}{h(0)}}y\right|\,dy\\&<{\sqrt {\frac {2}{\pi M|g''(0)|}}}\left|{\frac {h'(\xi )}{h(0)}}y\right|_{\text{max}}\left(1-e^{-\pi D_{y}^{2}}\right)\end{aligned}}}
會看到它原則上與
M
{\displaystyle M}
h
(
x
)
{\displaystyle h(x)}
所以，概括的講，靠近極大點的附近的積分原則上會隨着
M
{\displaystyle {\sqrt {M}}}
D
y
{\displaystyle D_{y}}
D
y
{\displaystyle D_{y}}
m
(
x
)
{\displaystyle m(x)}
g
(
x
)
−
g
(
0
)
{\displaystyle g(x)-g(0)}
m
(
x
)
{\displaystyle m(x)}
x
=
s
D
y
{\displaystyle x=sD_{y}}
D
y
{\displaystyle D_{y}}
M
{\displaystyle {\sqrt {M}}}
M
{\displaystyle M}
D
y
{\displaystyle D_{y}}
至於多維的情形，讓我們令
x
{\displaystyle \mathbf {x} }
n
{\displaystyle n}
f
(
x
)
{\displaystyle f(\mathbf {x} )}
∫
e
M
f
(
x
)
d
n
x
≈
(
2
π
M
)
n
/
2
|
−
H
(
f
)
(
x
0
)
|
−
1
/
2
e
M
f
(
x
0
)
as
M
→
∞
{\displaystyle \int e^{Mf(\mathbf {x} )}\,d^{n}x\approx \left({\frac {2\pi }{M}}\right)^{n/2}|-H(f)(\mathbf {x} _{0})|^{-1/2}e^{Mf(\mathbf {x} _{0})}{\text{ as }}M\to \infty \,}
其中的
H
(
f
)
(
x
0
)
{\displaystyle H(f)(\mathbf {x} _{0})}
x
0
{\displaystyle \mathbf {x} _{0}}
海森矩陣 ，而
|
⋅
|
{\displaystyle |\cdot |}
行列式 ；此外，與單變數的拉普拉斯方法類似，這裏的 海森矩陣 必須為 負定矩陣 ，即該矩陣的所有本徵值皆小於0，這樣才會是極大值所在。.[ 2]
此拉普拉斯方法可以被推廣到 複分析 裏頭使用，搭配 留數定理 ，以找出一個過最速下降點的 contour (翻路徑的話，會與path integral 相衝，所以，這裏還是以英文原字稱呼) 的 曲線積分 ，用來取代原有的複數空間的 contour積分。因為有時
f
(
z
)
{\displaystyle f(z)}
x
{\displaystyle x}
steepest descents 的章節。
以下就是該方法在z 平面下的形式：
∫
a
b
e
M
f
(
z
)
d
z
≈
2
π
−
M
f
″
(
z
0
)
e
M
f
(
z
0
)
as
M
→
∞
.
{\displaystyle \int _{a}^{b}\!e^{Mf(z)}\,dz\approx {\sqrt {\frac {2\pi }{-Mf''(z_{0})}}}e^{Mf(z_{0})}{\text{ as }}M\to \infty .\,}
其中 z 0 就是新的路徑通過的鞍點。
注意，開根號裏的負號是用來指定最速下降的方向，千萬別認為取
f
″
(
z
0
)
{\displaystyle f''(z_{0})}
亞純函數 ，就有必要將被新舊 contour 包到的極點所貢獻的留數給加入，範例請參考 Okounkov 的文章 arXiv:math/0309074 的第三章。
拉普拉斯方法可以用在推導 斯特靈公式 上；當 N 很大時，
N
!
≈
2
π
N
N
N
e
−
N
{\displaystyle N!\approx {\sqrt {2\pi N}}N^{N}e^{-N}\,}
證明：
由 Γ函數 的積分定義，我們可以得到
N
!
=
Γ
(
N
+
1
)
=
∫
0
∞
e
−
x
x
N
d
x
.
{\displaystyle N!=\Gamma (N+1)=\int _{0}^{\infty }e^{-x}x^{N}\,dx.}
接着讓我們做變數變換，
x
=
N
z
{\displaystyle x=Nz\,}
因此
d
x
=
N
d
z
.
{\displaystyle dx=N\,dz.}
將這些代回 Γ函數 的積分定義裏，我們可以得到
N
!
=
∫
0
∞
e
−
N
z
(
N
z
)
N
N
d
z
=
N
N
+
1
∫
0
∞
e
−
N
z
z
N
d
z
=
N
N
+
1
∫
0
∞
e
−
N
z
e
N
ln
z
d
z
=
N
N
+
1
∫
0
∞
e
N
(
ln
z
−
z
)
d
z
.
{\displaystyle {\begin{aligned}N!&=\int _{0}^{\infty }e^{-Nz}\left(Nz\right)^{N}N\,dz\\&=N^{N+1}\int _{0}^{\infty }e^{-Nz}z^{N}\,dz\\&=N^{N+1}\int _{0}^{\infty }e^{-Nz}e^{N\ln z}\,dz\\&=N^{N+1}\int _{0}^{\infty }e^{N(\ln z-z)}\,dz.\end{aligned}}}
經由此變數變換後，我們有了拉普拉斯方法所需要的
f
(
x
)
{\displaystyle f(x)}
f
(
z
)
=
ln
z
−
z
{\displaystyle f\left(z\right)=\ln {z}-z}
而它乃為二次可微函數，且
f
′
(
z
)
=
1
z
−
1
,
{\displaystyle f'(z)={\frac {1}{z}}-1,\,}
f
″
(
z
)
=
−
1
z
2
.
{\displaystyle f''(z)=-{\frac {1}{z^{2}}}.\,}
因此， ƒ (z ) 的極大值出現在 z 0 = 1 而且在該點的二次微分為
−
1
{\displaystyle -1}
N
!
≈
N
N
+
1
2
π
N
e
−
N
=
2
π
N
N
N
e
−
N
.
{\displaystyle N!\approx N^{N+1}{\sqrt {\frac {2\pi }{N}}}e^{-N}={\sqrt {2\pi N}}N^{N}e^{-N}.\,}
Deift, P.; Zhou, X., A steepest descent method for oscillatory Riemann–Hilbert problems. Asymptotics for the MKdV equation, Ann. of Math. 137 (2), 1993, 137 (2): 295–368, doi:10.2307/2946540 Azevedo-Filho, A.; Shachter, R., Laplace's Method Approximations for Probabilistic Inference in Belief Networks with Continuous Variables, Mantaras, R.; Poole, D. (編), Uncertainty in Artificial Intelligence, San Francisco, CA: Morgan Kauffman, 1994, CiteSeerX : 10.1.1.91.2064 Erdelyi, A., Asymptotic Expansions , Dover Publications, 2012, ISBN 9780486155050 Arfken, G.B.; Weber, H.J., Mathematical Methods For Physicists International Student Edition , Elsevier Science, 2005, ISBN 9780080470696 
![{\displaystyle y\in [-D_{y},D_{y}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/8aa8dfe7fc811c2ef66966ca426f4b6ccb2c03f4)
![{\displaystyle e^{M[f(sy+x_{0})-f(x_{0})]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/99cbc90a19b0c254dda9ead41f20d7039d37ec83)
![{\displaystyle x\in [-sD_{y},sD_{y}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/547e961cdd1cd28b795ba67bc1f023e0b73932c3)
.gif)
![{\displaystyle f(x_{0})=\max _{[a,b]}f(x)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/42ed16a88fdef34ad75b360378e1fd7966076e08)
![{\displaystyle [a,b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9c4b788fc5c637e26ee98b45f89a5c08c85f7935)
![{\displaystyle \int _{a}^{b}\!h(x)e^{Mg(x)}\,dx=h(0)e^{Mg(0)}s\underbrace {\int _{a/s}^{b/s}{\frac {h(x)}{h(0)}}e^{M\left[g(sy)-g(0)\right]}dy} _{1+R}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d9d8c5f1a9260cdf3331a58e39a5f08d9c5e18f)
![{\displaystyle A\equiv {\frac {h(sy)}{h(0)}}e^{M\left[g(sy)-g(0)\right]}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6f87c80e60835a4456a81e52431528624040de69)
![{\displaystyle (b_{1})\leq \left|\int _{D_{y}}^{b/s}\left[{\frac {h(sy)}{h(0)}}\right]_{\text{max}}e^{Mm(sy)}dy\right|}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2c106d9c211f67ed5df83394527fa638f8ecc5d)
![{\displaystyle m(x)\geq {\frac {1}{M}}\ln {\frac {h(x)}{h(0)}}+g(x)-g(0)\,\,{\text{as}}\,\,x\in [sD_{y},b]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c2b1764e4a4886c377dafa566cfac99a2c33c7fe)
![{\displaystyle {\begin{aligned}M\left[g(sD_{y})-g(0)\right]&=M\left[{\frac {g''(0)}{2}}s^{2}D_{y}^{2}+{\frac {g'''(\xi )}{6}}s^{3}D_{y}^{3}\right]\,\,{\text{as}}\,\,\xi \in [0,sD_{y}]\\&=-\pi D_{y}^{2}+{\frac {(2\pi )^{3/2}g'''(\xi )D_{y}^{3}}{6{\sqrt {M}}|g''(0)|^{3/2}}},\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/10fecd2aa81f3bd1fb0de34b4c94f6761d4b80cf)
![{\displaystyle {\begin{aligned}Msg'(sD_{y})&=Ms\left(g''(0)sD_{y}+{\frac {g'''(\zeta )}{2}}s^{2}D_{y}^{2}\right),\,\,{\text{as}}\,\,\zeta \in [0,sD_{y}]\\&=-2\pi D_{y}+{\sqrt {\frac {2}{M}}}\left({\frac {\pi }{|g''(0)|}}\right)^{3/2}g'''(\zeta )D_{y}^{2}\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/71bbef45e68a3a8d2e5663ccb5639981c6d0365e)
![{\displaystyle {\begin{aligned}(b_{1})&\leq \left|\left[{\frac {h(sy)}{h(0)}}\right]_{\text{max}}e^{-\pi D_{y}^{2}}\int _{0}^{b/s-D_{y}}e^{-2\pi D_{y}y}dy\right|\\&\leq \left|\left[{\frac {h(sy)}{h(0)}}\right]_{\text{max}}e^{-\pi D_{y}^{2}}{\frac {1}{2\pi D_{y}}}\right|.\end{aligned}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/05bf30637cd4d06c8c8082e99d8f44c412637c09)
.