W
0
(
x
)
{\displaystyle W_{0}(x)}
−
1
e
≤
x
≤
4
{\displaystyle -{\frac {1}{e}\leq x\leq 4}
朗伯W函数 （英語：Lambert W function ，又称为欧米加函数 或乘积对数 ），是
f
(
w
)
=
w
e
w
{\displaystyle f(w)=we^{w}
反函数 ，其中
e
w
{\displaystyle e^{w}
指数函数 ，
w
{\displaystyle w}
复数 。对于任何复数
z
{\displaystyle z}
z
=
W
(
z
)
e
W
(
z
)
.
{\displaystyle z=W(z)e^{W(z)}.}
由于函数
f
{\displaystyle f}
单射 ，因此函数
W
{\displaystyle W}
多值 的（除了0以外）。如果我们把
x
{\displaystyle x}
w
{\displaystyle w}
x
≥
1
e
{\displaystyle x\geq {\frac {1}{e}
(
−
1
e
,
0
)
{\displaystyle (-{\frac {1}{e},0)}
w
≥
−
1
{\displaystyle w\geq -1}
W
0
(
x
)
{\displaystyle W_{0}(x)}
W
0
(
0
)
=
0
{\displaystyle W_{0}(0)=0}
W
0
(
−
1
e
)
=
−
1
{\displaystyle W_{0}(-{\frac {1}{e})=-1}
[
−
1
e
,
0
)
{\displaystyle [-{\frac {1}{e},0)}
w
≤
−
1
{\displaystyle w\leq -1}
W
−
1
(
x
)
{\displaystyle W_{-1}(x)}
W
−
1
(
−
1
e
)
=
−
1
{\displaystyle W_{-1}(-{\frac {1}{e})=-1}
W
−
1
(
0
−
)
=
−
∞
{\displaystyle W_{-1}(0^{-})=-\infty }
朗伯
W
{\displaystyle W}
初等函数 来表示。它在组合数学 中有许多用途，例如树 的计算。它可以用来解许多含有指数的方程，也出现在某些微分方程 的解中，例如
y
(
t
)
=
a
y
(
t
−
1
)
{\displaystyle y(t)=ay(t-1)}
复平面上的朗伯W函数的函數圖形
朗伯
W
{\displaystyle W\,}
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} }\,}
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}
W
k
(
x
)
=
u
+
v
i
,
x
=
t
+
s
i
{\displaystyle W_{k}(x)=u+v{\rm {i},x=t+s{\rm {i}
(
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}
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}
s
=
0
{\displaystyle s=0}
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}
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\,}
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)\,}
W
(
x
)
{\displaystyle W(x)\,}
w
=
W
(
x
)
{\displaystyle w=W(x)\,}
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 }
欧米加常数 。
1
{\displaystyle 1\,}
z
z
z
z
z
.
.
.
=
lim
n
→
∞
(
z
⇈
n
)
=
−
W
(
−
ln
z
)
ln
z
{\displaystyle z^{z^{z^{z^{z^{.^{.^{.}=\lim _{n\to \infty }(z\upuparrows n)=-{\frac {W(-\ln z)}{\ln z}
其中
⇈
{\displaystyle \upuparrows }
高德納箭號表示法 。
2
{\displaystyle 2\,}
z
>
0
{\displaystyle z>0\,}
ln
W
(
z
)
=
ln
z
−
W
(
z
)
{\displaystyle \ln W(z)=\ln z-W(z)\,}
W
0
{\displaystyle W_{0}\,}
x
=
0
{\displaystyle x=0\,}
W
0
(
x
)
=
∑
n
=
1
∞
(
−
n
)
n
−
1
n
!
x
n
=
x
−
x
2
+
3
2
x
3
−
8
3
x
4
+
125
24
x
5
−
⋯
{\displaystyle W_{0}(x)=\sum _{n=1}^{\infty }{\frac {(-n)^{n-1}{n!}\ x^{n}=x-x^{2}+{\frac {3}{2}x^{3}-{\frac {8}{3}x^{4}+{\frac {125}{24}x^{5}-\cdots }
收敛半径 为
1
e
{\displaystyle {\frac {1}{e}\,}
W
(
x
)
+
W
(
y
)
=
W
[
x
y
W
(
x
)
+
x
y
W
(
y
)
]
{\displaystyle W(x)+W(y)=W\left[{\frac {xy}{W(x)}+{\frac {xy}{W(y)}\right]\,}
x
>
0
,
y
>
0
{\displaystyle x>0,y>0\,}
實部
ℜ
[
W
(
x
+
y
i
)
]
=
∑
k
=
1
∞
(
−
k
)
k
−
1
k
!
(
x
2
+
y
2
)
k
cos
(
k
arctan
x
y
)
{\displaystyle \Re \left[W(x+y{\rm {i})\right]=\sum _{k=1}^{\infty }{\frac {(-k)^{k-1}{k!}{\sqrt {(x^{2}+y^{2})^{k}\cos \left(k\arctan {\frac {x}{y}\right)\,}
x
2
+
y
2
<
1
e
2
{\displaystyle x^{2}+y^{2}<{\frac {1}{e^{2}\,}
虛部
ℑ
[
W
(
x
+
y
i
)
]
=
∑
k
=
1
∞
(
−
k
)
k
−
1
k
!
(
x
2
+
y
2
)
k
sin
(
k
arctan
x
y
)
{\displaystyle \Im \left[W(x+y{\rm {i})\right]=\sum _{k=1}^{\infty }{\frac {(-k)^{k-1}{k!}{\sqrt {(x^{2}+y^{2})^{k}\sin \left(k\arctan {\frac {x}{y}\right)\,}
x
2
+
y
2
<
1
e
2
{\displaystyle x^{2}+y^{2}<{\frac {1}{e^{2}\,}
模長
|
W
(
x
+
y
i
)
|
=
W
(
x
+
y
)
{\displaystyle |W(x+y{\rm {i})|=W({\sqrt {x+y})\,}
模角
arg
[
W
(
x
+
y
i
)
]
=
∑
k
=
1
∞
(
−
k
)
k
−
1
k
!
arctan
[
cot
(
k
arctan
x
y
)
]
{\displaystyle \arg \left[W(x+y{\rm {i})\right]=\sum _{k=1}^{\infty }{\frac {(-k)^{k-1}{k!}\arctan \left[\cot(k\arctan {\frac {x}{y})\right]\,}
x
2
+
y
2
<
1
e
2
{\displaystyle x^{2}+y^{2}<{\frac {1}{e^{2}\,}
共軛值
W
(
x
+
y
i
)
¯
=
∑
k
=
1
∞
(
−
k
)
k
−
1
k
!
(
x
2
+
y
2
)
k
[
cos
(
k
arctan
x
y
)
−
i
sin
(
k
arctan
x
y
)
]
{\displaystyle {\overline {W(x+y{\rm {i})}=\sum _{k=1}^{\infty }{\frac {(-k)^{k-1}{k!}{\sqrt {(x^{2}+y^{2})^{k}\left[\cos \left(k\arctan {\frac {x}{y}\right)-{\rm {i}\sin \left(k\arctan {\frac {x}{y}\right)\right]\,}
x
2
+
y
2
<
1
e
2
{\displaystyle x^{2}+y^{2}<{\frac {1}{e^{2}\,}
W
(
−
π
2
)
=
π
2
i
{\displaystyle W\left(-{\frac {\pi }{2}\right)={\frac {\pi }{2}i}
W
(
−
ln
2
2
)
=
−
ln
2
{\displaystyle W\left(-{\frac {\ln 2}{2}\right)=-\ln 2}
W
(
−
1
e
)
=
−
1
{\displaystyle W\left(-{1 \over e}\right)=-1}
W
(
1
)
=
Ω
=
1
∫
−
∞
∞
d
x
(
e
x
−
x
)
2
+
π
2
−
1
≈
0.56714329
…
{\displaystyle W\left(1\right)=\Omega ={\frac {1}{\int _{-\infty }^{\infty }{\frac {\rm {d}x}{(e^{x}-x)^{2}+\pi ^{2}-1\approx 0.56714329\dots \,}
欧米加常数 ）
W
(
e
)
=
1
{\displaystyle W(e)=1\,}
W
(
e
e
+
1
)
=
e
{\displaystyle W(e^{e+1})=e\,}
W
(
1
e
1
−
1
e
)
=
1
e
{\displaystyle W\left({\frac {1}{e^{1-{\frac {1}{e}\right)={\frac {1}{e}
W
(
π
e
π
)
=
π
{\displaystyle W({\pi }e^{\pi })=\pi }
W
(
k
ln
k
)
=
ln
k
{\displaystyle W(k{\ln k})={\ln k}
(
k
>
0
)
{\displaystyle (k>0)}
W
(
i
π
)
=
−
i
π
{\displaystyle W({\rm {i}\pi )=-{\rm {i}\pi }
W
(
−
i
π
)
=
i
π
{\displaystyle W(-{\rm {i}\pi )={\rm {i}\pi }
W
(
i
cos
1
−
sin
1
)
=
i
{\displaystyle W({\rm {i}\cos 1-\sin 1)={\rm {i}
W
(
−
3
2
π
)
=
−
3
2
π
i
{\displaystyle W(-{\frac {3}{2}{\pi })=-{\frac {3}{2}{\pi }{\rm {i}
W
(
−
8
7
7
ln
2
)
=
−
32
7
ln
2
{\displaystyle W(-{\frac {\sqrt[{7}]{8}{7}{\ln 2})=-{\frac {32}{7}{\ln 2}
W
(
−
3
54
ln
3
)
=
−
9
2
ln
3
{\displaystyle W(-{\frac {\sqrt {3}{54}{\ln 3})=-{\frac {9}{2}{\ln 3}
W
(
−
ln
2
4
)
=
−
4
ln
2
{\displaystyle W(-{\frac {\ln 2}{4})=-4{\ln 2}
W
(
−
1
)
=
e
1
2
π
∫
0
∞
1
t
+
1
arctan
2
π
t
−
ln
t
d
t
−
cos
[
1
4
π
∫
0
∞
1
t
+
1
ln
(
t
−
ln
t
)
2
4
π
2
+
(
t
−
ln
t
)
2
d
t
]
+
π
sin
[
1
4
π
∫
0
∞
1
t
+
1
ln
(
t
−
ln
t
)
2
4
π
2
+
(
t
−
ln
t
)
2
d
t
]
−
i
{
π
cos
[
1
4
π
∫
0
∞
1
t
+
1
ln
(
t
−
ln
t
)
2
4
π
2
+
(
t
−
ln
t
)
2
d
t
]
+
sin
[
1
4
π
∫
0
∞
1
t
+
1
ln
(
t
−
ln
t
)
2
4
π
2
+
(
t
−
ln
t
)
2
d
t
]
}
e
1
2
π
∫
0
∞
1
t
+
1
arctan
2
π
t
−
ln
t
d
t
≈
−
0.31813
−
1.33723
i
{\displaystyle W\left(-1\right)={\frac {e^{\frac {1}{2\pi }\int _{0}^{\infty }{1 \over t+1}\arctan {2\pi \over t-\ln t}{\rm {d}t}-\cos \left[{\frac {1}{4\pi }\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}{\rm {d}t\right]+\pi \sin \left[{\frac {1}{4\pi }\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}{\rm {d}t\right]-{\rm {i}\left\{\pi \cos \left[{\frac {1}{4\pi }\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}{\rm {d}t\right]+\sin \left[{\frac {1}{4\pi }\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}{\rm {d}t\right]\right\}{e^{\frac {1}{2\pi }\int _{0}^{\infty }{1 \over t+1}\arctan {2\pi \over t-\ln t}{\rm {d}t}\approx -0.31813-1.33723{\rm {i}
W
(
−
ln
k
k
)
=
−
ln
k
{\displaystyle W(-{\frac {\ln k}{k})=-\ln k}
W
[
−
ln
(
x
+
1
)
x
(
x
+
1
)
1
x
]
=
−
x
+
1
x
ln
(
x
+
1
)
>
,
−
1
<
x
<
0
{\displaystyle W\left[-{\frac {\ln(x+1)}{x(x+1)^{\frac {1}{x}\right]=-{\frac {x+1}{x}\ln(x+1)>,-1<x<0}
许多含有指数的方程都可以用
W
{\displaystyle W\,}
Y
=
X
e
X
{\displaystyle Y=Xe^{X}\,}
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}
−
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\,}
a
{\displaystyle a\,}
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}\,}
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{\displaystyle k\in {\mathbb {Z} }\,}
標準的 Lambert W 函數可用來表示以下超越代數方程式的解：
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{\displaystyle e^{-cx}=a_{o}(x-r)~~\quad \qquad \qquad \qquad \qquad (1)}
其中 a 0 , c 與 r 為實常數。
其解為
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{\textstyle x=r+{\tfrac {W\left({\frac {ce^{-cr}{a_{o}\right)}{c}
Lambert W 函數之一般化[ 1] [ 2] [ 3]
一項在低維空間內廣義相對論 與量子力學 的應用（量子引力 ），實際上一種以前未知的 連結 於此二區域中，如 “Journal of Classical and Quantum Gravity”[ 4]
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{\displaystyle e^{-cx}=a_{o}(x-r_{1})(x-r_{2})~~\qquad \qquad (2)}
其中 r 1 和 r 2 是不同實常數，為二維多項式的根。於此函數解有單一引數 x 但 r i 和 a o 為函數的參數。如此一來，此一般式類似於 “hypergeometric”（超几何分布）函數與 “Meijer G“，但屬於不同類函數。當 r 1 = r 2 ，(2)的兩方可分解為 (1) 因此其解簡化為標準 W 函數。(2)式代表著 “dilaton”（軸子 ）場的方程，可據此推導線性，雙體重力問題 1+1 維（一空間維與一時間維）當兩不等（靜止）質量，以及，量子力學的特徵能Delta位勢阱 給不等電位於一維空間。 量子力學的一特例特徵能的分析解三體問題，亦即（三維）氢分子離子 。[ 5] x ：
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{\displaystyle e^{-cx}=a_{o}{\frac {\prod _{i=1}^{\infty }(x-r_{i})}{\prod _{i=1}^{\infty }(x-s_{i})}\qquad \qquad \qquad (3)}
其中 r i 與 s i 是相異實常數而 x 是特徵能和內核距離R之函數。式 (3) 與其特例表示於 (1) 和 (2) 是與一更大類型延遲微分方程 。由于哈代 的“虚假导数”概念，多根的特殊情况得以解决[ 6] Lambert "W" 函數於基礎物理問題之應用並未完全即使標準情況如 (1) 最近在原子，分子，與光學物理領域可見[ 7] 黎曼假设 的 Keiper-Li 准则 [ 8]
朗伯W函数在复平面上的图像
z = Re(W0 (x + i y ))
z = Im(W0 (x + i y ))
W 函数可以用以下的递推关系 算出：
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{\displaystyle w_{j+1}=w_{j}-{\frac {w_{j}e^{w_{j}-z}{e^{w_{j}(w_{j}+1)-{\frac {(w_{j}+2)(w_{j}e^{w_{j}-z)}{2w_{j}+2}
^ T.C. Scott and R.B. Mann, General Relativity and Quantum Mechanics: Towards a Generalization of the Lambert W Function , AAECC (Applicable Algebra in Engineering, Communication and Computing), vol. 17, no. 1, (April 2006), pp.41-47, [1] ; Arxiv [2] （页面存档备份 ，存于互联网档案馆 ）
^ T.C. Scott, G. Fee and J. Grotendorst, "Asymptotic series of Generalized Lambert W Function" （页面存档备份 ，存于互联网档案馆 ）, SIGSAM, vol. 47, no. 3, (September 2013), pp. 75-83
^ T.C. Scott, G. Fee, J. Grotendorst and W.Z. Zhang, "Numerics of the Generalized Lambert W Function" （页面存档备份 ，存于互联网档案馆 ）, SIGSAM, vol. 48, no. 2, (June 2014), pp. 42-56
^ P.S. Farrugia, R.B. Mann, and T.C. Scott, N-body Gravity and the Schrödinger Equation , Class. Quantum Grav. vol. 24, (2007), pp. 4647-4659, [3] ; Arxiv [4] （页面存档备份 ，存于互联网档案馆 ）
^ T.C. Scott, M. Aubert-Frécon and J. Grotendorst, New Approach for the Electronic Energies of the Hydrogen Molecular Ion , Chem. Phys. vol. 324, (2006), pp. 323-338, [5] （页面存档备份 ，存于互联网档案馆 ）; Arxiv [6] （页面存档备份 ，存于互联网档案馆 ）
^ Aude Maignan, T.C. Scott, "Fleshing out the Generalized Lambert W Function", SIGSAM, vol. 50, no. 2, (June 2016), pp. 45-60
^ T.C. Scott, A. Lüchow, D. Bressanini and J.D. Morgan III, The Nodal Surfaces of Helium Atom Eigenfunctions , Phys. Rev. A 75, (2007), p. 060101, [7] （页面存档备份 ，存于互联网档案馆 ）
^ R.C. McPhedran, T.C Scott and Aude Maignan, "The Keiper-Li Criterion for the Riemann Hypothesis and Generalized Lambert Functions", ACM Commun. Comput. Algebra, vol. 57, no. 3, (December 2023), pp. 85-110
![{\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\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3bf1ea9825edb9dc37486b29ef3629f54f52c52c)
![{\displaystyle x\not \in \left[-{\frac {1}{e},0\right],k\in {\mathbb {Z} }\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/53fd6c87a7b01214078d06687892fd33a9691232)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/babe26f83ddb9f0591fcea00bae08e480e322f84)
![{\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}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1bbe97f2e0426384d7058ef799b21c75d99ae587)
![{\displaystyle z\left[1+W(z)\right]{\frac {\rm {d}{\rm {d}z}W(z)=W(z)}](https://wikimedia.org/api/rest_v1/media/math/render/svg/335f854707acfd8a0fcc7c61e0964924f072f15a)
![{\displaystyle {\frac {\rm {d}{\rm {d}z}W(z)={\frac {W(z)}{z\left[1+W(z)\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/bd4f89b7cbdee99ee248268cf3ebda1d4171b3f5)
![{\displaystyle \int W(x){\rm {d}x=x\left[W(x)+{\frac {1}{W(x)}-1\right]+C}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9bc6b4621f45014de0d670d2f3c67fc3ecad49e6)
![{\displaystyle W(x)+W(y)=W\left[{\frac {xy}{W(x)}+{\frac {xy}{W(y)}\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3d1c5c1b56c0679636684dbde2bf7e21ed8ac788)
![{\displaystyle \Re \left[W(x+y{\rm {i})\right]=\sum _{k=1}^{\infty }{\frac {(-k)^{k-1}{k!}{\sqrt {(x^{2}+y^{2})^{k}\cos \left(k\arctan {\frac {x}{y}\right)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/0d051f4afb9dbd7a1cfeca6d0db44afc98e3327a)
![{\displaystyle \Im \left[W(x+y{\rm {i})\right]=\sum _{k=1}^{\infty }{\frac {(-k)^{k-1}{k!}{\sqrt {(x^{2}+y^{2})^{k}\sin \left(k\arctan {\frac {x}{y}\right)\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/80cce84264c6e3322d91899fd86de05534bfd65a)
![{\displaystyle \arg \left[W(x+y{\rm {i})\right]=\sum _{k=1}^{\infty }{\frac {(-k)^{k-1}{k!}\arctan \left[\cot(k\arctan {\frac {x}{y})\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/fd8d40bbf60b7b82f166a0f81eb76d2b6dfbe23f)
![{\displaystyle {\overline {W(x+y{\rm {i})}=\sum _{k=1}^{\infty }{\frac {(-k)^{k-1}{k!}{\sqrt {(x^{2}+y^{2})^{k}\left[\cos \left(k\arctan {\frac {x}{y}\right)-{\rm {i}\sin \left(k\arctan {\frac {x}{y}\right)\right]\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2c7d6b51b9d52b3188295b5fe32a07bc8c18f0d6)
![{\displaystyle W(-{\frac {\sqrt[{7}]{8}{7}{\ln 2})=-{\frac {32}{7}{\ln 2}](https://wikimedia.org/api/rest_v1/media/math/render/svg/88c2c31ae34096cc653a48ab1459d7e3595fda3f)
![{\displaystyle W\left(-1\right)={\frac {e^{\frac {1}{2\pi }\int _{0}^{\infty }{1 \over t+1}\arctan {2\pi \over t-\ln t}{\rm {d}t}-\cos \left[{\frac {1}{4\pi }\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}{\rm {d}t\right]+\pi \sin \left[{\frac {1}{4\pi }\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}{\rm {d}t\right]-{\rm {i}\left\{\pi \cos \left[{\frac {1}{4\pi }\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}{\rm {d}t\right]+\sin \left[{\frac {1}{4\pi }\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}{\rm {d}t\right]\right\}{e^{\frac {1}{2\pi }\int _{0}^{\infty }{1 \over t+1}\arctan {2\pi \over t-\ln t}{\rm {d}t}\approx -0.31813-1.33723{\rm {i}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c240aacbf3ac467f466b19f28417ce10f6f7d6b9)
![{\displaystyle W\left[-{\frac {\ln(x+1)}{x(x+1)^{\frac {1}{x}\right]=-{\frac {x+1}{x}\ln(x+1)>,-1<x<0}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cbfbc0f57e483d72daadd901434f3654ec5974e4)
![{\displaystyle x=\exp \left(W_{k}\left[\ln({\rm {t})\right]\right).}](https://wikimedia.org/api/rest_v1/media/math/render/svg/04f2ee3ce598c11cbb8a14c20aedfbc1b2de9b99)
![{\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}\,}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a45d01eb537904ece465cdbec27d8b26b104ab99)
z = Re(W0(x + i y))
z = Im(W0(x + i y))
