朗伯
W
{\displaystyle W\,}
函数的积分形式为
W
(
x
)
=
x
π
∫
0
π
(
1
−
v
cot
v
)
2
+
v
2
x
+
v
csc
v
⋅
e
−
v
cot
v
d
v
,
|
arg
(
x
)
|
<
π
{\displaystyle W(x)={\frac {x}{\pi }}\int _{0}^{\pi }{\frac {\left(1-v\cot v\right)^{2}+v^{2}}{x+v\csc v\cdot e^{-v\cot v}}}{\rm {d}}v,|\arg \left(x\right)|<\pi \,}
W
(
x
)
=
∫
−
∞
−
1
e
−
1
π
ℑ
[
d
d
x
W
(
x
)
]
ln
(
1
−
z
x
)
d
x
{\displaystyle W(x)=\int _{-\infty }^{-{\frac {1}{e}}}{-{\frac {1}{\pi }}}\Im \left[{\frac {\rm {d}}{{\rm {d}}x}}W(x)\right]\ln \left(1-{\frac {z}{x}}\right){\rm {d}}x\,}
若
x
∉
[
−
1
e
,
0
]
,
k
∈
Z
{\displaystyle x\not \in \left[-{\frac {1}{e}},0\right],k\in {\mathbb {Z} }\,}
,若
x
∈
(
−
1
e
,
0
)
,
k
=
1
,
±
2
,
±
3
,
.
.
.
{\displaystyle x\in \left(-{\frac {1}{e}},0\right),k=1,\pm 2,\pm 3,...\,}
W
k
(
x
)
=
1
+
(
ln
x
−
1
+
2
k
π
i
)
e
i
2
π
∫
0
∞
ln
t
−
ln
t
+
ln
x
+
(
2
k
+
1
)
π
i
t
−
ln
t
+
ln
x
+
(
2
k
−
1
)
π
i
⋅
d
t
t
+
1
=
1
+
(
ln
x
−
1
+
2
k
π
i
)
e
i
2
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
4
k
2
−
1
)
π
2
+
2
π
(
t
−
ln
t
+
ln
x
)
i
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
{\displaystyle W_{k}(x)=1+\left(\ln x-1+2k\pi {\rm {i}}\right)e^{{\frac {\rm {i}}{2\pi }}\int _{0}^{\infty }\ln {\frac {t-\ln t+\ln x+(2k+1)\pi {\rm {i}}}{t-\ln t+\ln x+(2k-1)\pi {\rm {i}}}}\cdot {\frac {{\rm {d}}t}{t+1}}}=1+\left(\ln x-1+2k\pi {\rm {i}}\right)e^{{\frac {\rm {i}}{2\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}+2\pi \left(t-\ln t+\ln x\right){\rm {i}}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}}\,}
把被积函数的实部和虚部分离出来:
W
k
(
x
)
=
1
+
(
ln
x
−
1
+
2
k
π
i
)
e
i
2
π
∫
0
∞
[
1
2
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
+
i
arctan
2
π
(
t
−
ln
t
+
ln
x
)
(
t
−
ln
t
+
ln
x
)
2
+
(
4
k
2
−
1
)
π
2
]
⋅
d
t
t
+
1
{\displaystyle W_{k}(x)=1+\left(\ln x-1+2k\pi {\rm {i}}\right)e^{{\frac {\rm {i}}{2\pi }}\int _{0}^{\infty }\left[{\frac {1}{2}}\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}+{\rm {i}}\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\right]\cdot {\frac {{\rm {d}}t}{t+1}}}}
W
k
(
x
)
=
1
+
(
ln
x
−
1
)
cos
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
−
2
k
π
sin
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
+
i
[
(
ln
x
−
1
)
sin
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
+
2
k
π
cos
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
]
e
1
2
π
∫
0
∞
arctan
2
π
(
t
−
ln
t
+
ln
x
)
(
t
−
ln
t
+
ln
x
)
2
+
(
4
k
2
−
1
)
π
2
⋅
d
t
t
+
1
{\displaystyle {}_{W_{k}(x)=1+{\frac {\left(\ln x-1\right)\cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}-2k\pi \sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}+{\rm {i}}\left[\left(\ln x-1\right)\sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}+2k\pi \cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}\right]}{e^{{\frac {1}{2\pi }}\int _{0}^{\infty }\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\cdot {\frac {\rm {{d}t}}{t+1}}}}}}}
设
W
k
(
x
)
=
u
+
v
i
,
x
=
t
+
s
i
{\displaystyle W_{k}(x)=u+v{\rm {i}},x=t+s{\rm {i}}}
,则有
(
u
+
v
i
)
e
u
+
v
i
=
t
+
s
i
{\displaystyle \left(u+v{\rm {i}}\right)e^{u+v{\rm {i}}}=t+s{\rm {i}}}
,展开分离出实部和虚部,
e
u
(
u
cos
v
−
v
sin
v
)
=
t
,
e
u
(
u
sin
v
+
v
cos
v
)
=
s
{\displaystyle e^{u}\left(u\cos v-v\sin v\right)=t,e^{u}\left(u\sin v+v\cos v\right)=s}
,当
s
=
0
{\displaystyle s=0}
时,易知
u
=
−
v
cot
v
{\displaystyle u=-v\cot v}
W
k
(
x
)
=
(
1
−
ln
x
)
sin
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
−
2
k
π
cos
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
e
1
2
π
∫
0
∞
arctan
2
π
(
t
−
ln
t
+
ln
x
)
(
t
−
ln
t
+
ln
x
)
2
+
(
4
k
2
−
1
)
π
2
⋅
d
t
t
+
1
cot
(
ln
x
−
1
)
sin
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
+
2
k
π
cos
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
e
1
2
π
∫
0
∞
arctan
2
π
(
t
−
ln
t
+
ln
x
)
(
t
−
ln
t
+
ln
x
)
2
+
(
4
k
2
−
1
)
π
2
⋅
d
t
t
+
1
+
(
ln
x
−
1
)
sin
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
+
2
k
π
cos
1
4
π
∫
0
∞
ln
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
+
1
)
2
π
2
(
t
−
ln
t
+
ln
x
)
2
+
(
2
k
−
1
)
2
π
2
⋅
d
t
t
+
1
e
1
2
π
∫
0
∞
arctan
2
π
(
t
−
ln
t
+
ln
x
)
(
t
−
ln
t
+
ln
x
)
2
+
(
4
k
2
−
1
)
π
2
⋅
d
t
t
+
1
i
,
{\displaystyle {}_{W_{k}(x)={\frac {\left(1-\ln x\right)\sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}-2k\pi \cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}}{e^{{\frac {1}{2\pi }}\int _{0}^{\infty }\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\cdot {\frac {\rm {{d}t}}{t+1}}}}}\cot {\frac {\left(\ln x-1\right)\sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}+2k\pi \cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}}{e^{{\frac {1}{2\pi }}\int _{0}^{\infty }\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\cdot {\frac {\rm {{d}t}}{t+1}}}}}+{\frac {\left(\ln x-1\right)\sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}+2k\pi \cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}}{e^{{\frac {1}{2\pi }}\int _{0}^{\infty }\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\cdot {\frac {\rm {{d}t}}{t+1}}}}}{\rm {i}},}}
W
0
(
x
)
=
1
+
(
ln
x
−
1
)
e
−
1
π
∫
0
∞
arg
(
t
−
ln
t
+
ln
x
+
π
i
)
⋅
d
t
t
+
1
,
x
>
0
{\displaystyle W_{0}(x)=1+\left(\ln x-1\right)e^{-{\frac {1}{\pi }}\int _{0}^{\infty }\arg \left(t-\ln t+\ln x+\pi {\rm {i}}\right)\cdot {\frac {\rm {{d}t}}{t+1}}},x>0}
若
x
>
1
e
{\displaystyle x>{\frac {1}{e}}}
,上式还可化为
W
0
(
x
)
=
1
+
(
ln
x
−
1
)
e
−
1
π
∫
0
∞
arctan
π
t
−
ln
t
+
ln
x
⋅
d
t
t
+
1
{\displaystyle W_{0}(x)=1+\left(\ln x-1\right)e^{-{\frac {1}{\pi }}\int _{0}^{\infty }\arctan {\frac {\pi }{t-\ln t+\ln x}}\cdot {\frac {\rm {{d}t}}{t+1}}}}
由隐函数 的求导法则,朗伯
W
{\displaystyle W\,}
函数满足以下的微分方程 :
z
[
1
+
W
(
z
)
]
d
d
z
W
(
z
)
=
W
(
z
)
{\displaystyle z\left[1+W(z)\right]{\frac {\rm {d}}{{\rm {d}}z}}W(z)=W(z)}
,
z
≠
−
1
e
,
{\displaystyle z\neq -{\frac {1}{e}}\,,}
因此:
d
d
z
W
(
z
)
=
W
(
z
)
z
[
1
+
W
(
z
)
]
{\displaystyle {\frac {\rm {d}}{{\rm {d}}z}}W(z)={\frac {W(z)}{z\left[1+W(z)\right]}}}
,
z
≠
−
1
e
.
{\displaystyle z\neq -{\frac {1}{e}}\,.}
函数
W
(
x
)
{\displaystyle W(x)\,}
,以及许多含有
W
(
x
)
{\displaystyle W(x)\,}
的表达式,都可以用
w
=
W
(
x
)
{\displaystyle w=W(x)\,}
的变量代换 来积分,也就是说
x
=
w
e
w
{\displaystyle x=we^{w}\,}
∫
W
(
x
)
d
x
=
x
[
W
(
x
)
+
1
W
(
x
)
−
1
]
+
C
{\displaystyle \int W(x){\rm {d}}x=x\left[W(x)+{\frac {1}{W(x)}}-1\right]+C}
∫
0
1
W
(
x
)
d
x
=
Ω
+
1
Ω
−
2
≈
0.330366
{\displaystyle \int _{0}^{1}W(x){\rm {d}}x=\Omega +{\frac {1}{\Omega }}-2\approx 0.330366}
其中
Ω
{\displaystyle \Omega }
为欧米加常数 。
许多含有指数的方程都可以用
W
{\displaystyle W\,}
函数来解出。一般的方法是把未知数都移到方程的一侧,并设法化为
Y
=
X
e
X
{\displaystyle Y=Xe^{X}\,}
的形式。
例子
例子1
2
t
=
5
t
{\displaystyle 2^{t}=5t\,}
⇒
1
=
5
t
2
t
{\displaystyle \Rightarrow 1={\frac {5t}{2^{t}}}\,}
⇒
1
=
5
t
e
−
t
ln
2
{\displaystyle \Rightarrow 1=5t\,e^{-t\ln 2}\,}
⇒
1
5
=
t
e
−
t
ln
2
{\displaystyle \Rightarrow {\frac {1}{5}}=t\,e^{-t\ln 2}\,}
⇒
−
ln
2
5
=
(
−
t
ln
2
)
e
−
t
ln
2
{\displaystyle \Rightarrow -{\frac {\ln 2}{5}}=(-\,t\,\ln 2)\,e^{-t\ln 2}\,}
⇒
−
t
ln
2
=
W
k
(
−
ln
2
5
)
{\displaystyle \Rightarrow -t\ln 2=W_{k}\left(-{\frac {\ln 2}{5}}\right)\,}
⇒
t
=
−
W
k
(
−
ln
2
5
)
ln
2
{\displaystyle \Rightarrow t=-{\frac {W_{k}\left(-{\frac {\ln 2}{5}}\right)}{\ln 2}}\,}
更一般地,以下的方程
Q
a
x
+
b
=
c
x
+
d
{\displaystyle Q^{ax+b}=cx+d\,}
其中
Q
>
0
∧
Q
≠
1
∧
c
≠
0
{\displaystyle Q>0\land Q\neq 1\land c\neq 0}
两边同乘:
a
c
{\displaystyle {\frac {a}{c}}}
,
得到:
a
c
Q
a
x
+
b
=
a
x
+
a
d
c
{\displaystyle {\frac {a}{c}}Q^{ax+b}=ax+{\frac {ad}{c}}\,}
同除以:
Q
a
x
{\displaystyle Q^{ax}\,}
,
得到:
a
c
Q
b
=
(
a
x
+
a
d
c
)
Q
−
a
x
{\displaystyle {\frac {a}{c}}Q^{b}=\left(ax+{\frac {ad}{c}}\right)Q^{-ax}\,}
同除:
Q
a
d
c
{\displaystyle Q^{\frac {ad}{c}}\,}
,
a
c
Q
b
−
a
d
c
=
(
a
x
+
a
d
c
)
Q
−
(
a
x
+
a
d
c
)
{\displaystyle {\frac {a}{c}}Q^{b-{\frac {ad}{c}}}=\left(ax+{\frac {ad}{c}}\right)Q^{-\left(ax+{\frac {ad}{c}}\right)}\,}
可以用变量代换
令
t
=
a
x
+
a
d
c
{\displaystyle t=ax+{\frac {ad}{c}}}
化为
t
Q
−
t
=
a
c
Q
b
−
a
d
c
{\displaystyle tQ^{-t}={\frac {a}{c}}Q^{b-{\frac {ad}{c}}}}
即:
t
(
e
ln
Q
)
−
t
=
a
c
Q
b
−
a
d
c
{\displaystyle t\left(e^{\ln Q}\right)^{-t}={\frac {a}{c}}Q^{b-{\frac {ad}{c}}}}
同乘:
ln
Q
{\displaystyle {\ln Q}\,}
得出
t
ln
Q
⋅
e
−
t
ln
Q
=
ln
Q
⋅
a
c
Q
b
−
a
d
c
{\displaystyle t{\ln Q}\cdot e^{-t\ln Q}={\ln Q}\cdot {\frac {a}{c}}Q^{b-{\frac {ad}{c}}}}
故
t
ln
Q
=
−
W
k
(
−
a
ln
Q
c
Q
b
−
a
d
c
)
{\displaystyle t{\ln Q}=-W_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}
带入
t
=
a
x
+
a
d
c
{\displaystyle t=ax+{\frac {ad}{c}}}
为
(
a
x
+
a
d
c
)
ln
Q
=
−
W
k
(
−
a
ln
Q
c
Q
b
−
a
d
c
)
{\displaystyle \left(ax+{\frac {ad}{c}}\right){\ln Q}=-W_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}
因此最终的解为
x
=
−
W
k
(
−
a
ln
Q
c
Q
b
−
a
d
c
)
a
ln
Q
−
d
c
{\displaystyle x=-{\frac {W_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}{a\ln Q}}-{\frac {d}{c}}}
若辅助方程:
x
e
x
=
−
a
ln
Q
c
Q
b
−
a
d
c
{\displaystyle xe^{x}=-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}}
中,
−
a
ln
Q
c
Q
b
−
a
d
c
∈
(
−
∞
,
−
1
e
)
{\displaystyle -{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\in \left(-\infty ,-{\frac {1}{e}}\right)}
,辅助方程无实数解,原方程亦无实解;
若:
−
a
ln
Q
c
Q
b
−
a
d
c
∈
{
−
1
e
}
∪
[
0
,
+
∞
)
{\displaystyle -{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\in \left\{-{\frac {1}{e}}\right\}\cup \mathbf {[} 0,+\infty )}
,
辅助方程有一实数解,原方程有一实解:
x
=
−
W
k
(
−
a
ln
Q
c
Q
b
−
a
d
c
)
a
ln
Q
−
d
c
{\displaystyle x=-{\frac {W_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}{a\ln Q}}-{\frac {d}{c}}}
若:
−
a
ln
Q
c
Q
b
−
a
d
c
∈
(
−
1
e
,
0
)
{\displaystyle -{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\in \left(-{\frac {1}{e}},0\right)}
,
辅助方程有二实解,设为
W
(
−
a
ln
Q
c
Q
b
−
a
d
c
)
{\displaystyle W\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}
,
W
−
1
(
−
a
ln
Q
c
Q
b
−
a
d
c
)
{\displaystyle {\rm {W}}_{-1}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}
,
为
x
1
=
−
W
(
−
a
ln
Q
c
Q
b
−
a
d
c
)
a
ln
Q
−
d
c
{\displaystyle x_{1}=-{\frac {W\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}{a\ln Q}}-{\frac {d}{c}}}
x
2
=
−
W
−
1
(
−
a
ln
Q
c
Q
b
−
a
d
c
)
a
ln
Q
−
d
c
{\displaystyle x_{2}=-{\frac {{\rm {W}}_{-1}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}{a\ln Q}}-{\frac {d}{c}}}
例子2 用类似的方法,可知以下方程的解
x
x
=
t
,
{\displaystyle x^{x}={\mathrm {t} }\,,}
为
x
=
ln
t
W
(
ln
t
)
{\displaystyle x={\frac {\ln {\rm {t}}}{W(\ln {\rm {t}})}}\,}
或
x
=
exp
(
W
k
[
ln
(
t
)
]
)
.
{\displaystyle x=\exp \left(W_{k}\left[\ln({\rm {t}})\right]\right).}
例子3 以下方程的解
x
log
b
x
=
a
{\displaystyle x\log _{b}{x}=a\,}
具有形式
x
=
a
ln
b
W
k
(
a
ln
b
)
{\displaystyle x={\frac {a{\ln b}}{W_{k}\left(a{\ln b}\right)}}}
例子4
x
a
−
b
x
=
0
{\displaystyle x^{a}-b^{x}=0\,}
a
>
0
{\displaystyle a>0\,}
:
b
>
0
{\displaystyle b>0\,}
:
x
>
0
{\displaystyle x>0\,}
取对数,
a
ln
x
=
x
ln
b
{\displaystyle a\ln x=x\ln b\,}
ln
x
x
=
ln
b
a
{\displaystyle {\frac {\ln x}{x}}={\frac {\ln b}{a}}\,}
e
ln
x
x
=
e
ln
b
a
{\displaystyle e^{\frac {\ln x}{x}}=e^{\frac {\ln b}{a}}\,}
x
1
x
=
b
1
a
{\displaystyle x^{\frac {1}{x}}=b^{\frac {1}{a}}\,}
取倒数,
(
1
x
)
1
x
=
b
−
1
a
{\displaystyle \left({\frac {1}{x}}\right)^{\frac {1}{x}}=b^{-{\frac {1}{a}}}\,}
1
x
=
−
ln
b
a
W
(
−
1
a
ln
b
)
{\displaystyle {\frac {1}{x}}=-{\frac {\ln b}{aW\left(-{\frac {1}{a}}\ln b\right)}}\,}
最终解为 :
x
=
−
a
ln
b
W
k
(
−
ln
b
a
)
{\displaystyle x=-{\frac {a}{\ln b}}W_{k}\left(-{\frac {\ln b}{a}}\right)\,}
例子5
(
a
x
+
b
)
n
=
u
c
x
+
d
{\displaystyle (ax+b)^{n}=u^{cx+d}\,}
两边开
n
{\displaystyle n\,}
次方并除以
a
{\displaystyle a\,}
得
x
+
b
a
=
u
c
n
x
+
d
n
a
(
cos
2
k
π
n
+
i
sin
2
k
π
n
)
{\displaystyle x+{\frac {b}{a}}={\frac {u^{{\frac {c}{n}}x+{\frac {d}{n}}}}{a}}\left(\cos {\frac {2k\pi }{n}}+{\rm {i}}\sin {\frac {2k\pi }{n}}\right)\,}
令
u
=
e
ln
u
{\displaystyle u=e^{\ln u}\,}
,
化为
x
+
b
a
=
e
c
ln
u
n
x
+
d
ln
u
n
a
(
cos
2
k
π
n
+
i
sin
2
k
π
n
)
{\displaystyle x+{\frac {b}{a}}={\frac {e^{{\frac {c\ln u}{n}}x+{\frac {d\ln u}{n}}}}{a}}\left(\cos {\frac {2k\pi }{n}}+{\rm {i}}\sin {\frac {2k\pi }{n}}\right)\,}
两边同乘
−
c
ln
u
n
u
−
c
n
x
−
c
b
n
a
{\displaystyle -{\frac {c\ln u}{n}}u^{-{\frac {c}{n}}x-{\frac {cb}{na}}}\,}
,
(
−
c
ln
u
n
x
−
c
b
ln
u
n
a
)
e
−
c
ln
u
n
x
−
c
b
ln
u
n
a
=
−
c
ln
u
n
a
u
d
n
−
c
b
n
a
(
cos
2
k
π
n
+
i
sin
2
k
π
n
)
{\displaystyle \left(-{\frac {c\ln u}{n}}x-{\frac {cb\ln u}{na}}\right)e^{-{\frac {c\ln u}{n}}x-{\frac {cb\ln u}{na}}}=-{\frac {c\ln u}{na}}u^{{\frac {d}{n}}-{\frac {cb}{na}}}\left(\cos {\frac {2k\pi }{n}}+{\rm {i}}\sin {\frac {2k\pi }{n}}\right)\,}
最终得
x
k
=
−
n
c
ln
u
W
k
[
−
c
ln
u
n
a
u
d
n
−
c
b
n
a
(
cos
2
k
π
n
+
i
sin
2
k
π
n
)
]
−
b
a
{\displaystyle x_{k}=-{\frac {n}{c\ln u}}W_{k}\left[-{\frac {c\ln u}{na}}u^{{\frac {d}{n}}-{\frac {cb}{na}}}\left(\cos {\frac {2k\pi }{n}}+{\rm {i}}\sin {\frac {2k\pi }{n}}\right)\right]-{\frac {b}{a}}\,}
k
∈
Z
{\displaystyle k\in {\mathbb {Z} }\,}