{\displaystyle W\_{0}(x)}的图像，{\displaystyle -{\frac {1}{e}}\leq x\leq 4}
**朗伯W函数**（英語：，又称为**欧米加函数**或**乘积对数**），是{\displaystyle f(w)=we^{w}}的反函数，其中{\displaystyle e^{w}}是指数函数，{\displaystyle w}是任意复数。对于任何复数{\displaystyle z}，都有：

{\displaystyle z=W(z)e^{W(z)}.}
由于函数{\displaystyle f}不是单射，因此函数{\displaystyle W}是多值的（除了0以外）。如果我们把{\displaystyle x}限制为实数，并要求{\displaystyle w}是实数，那么函数仅对于{\displaystyle x\geq {\frac {1}{e}}}有定义，在{\displaystyle (-{\frac {1}{e}},0)}内是多值的；如果加上{\displaystyle w\geq -1}的限制，则定义了一个单值函数{\displaystyle W\_{0}(x)}（见图）。我们有{\displaystyle W\_{0}(0)=0}，{\displaystyle W\_{0}(-{\frac {1}{e}})=-1}。而在{\displaystyle [-{\frac {1}{e}},0)}内的{\displaystyle w\leq -1}分支，则记为{\displaystyle W\_{-1}(x)}，从{\displaystyle W\_{-1}(-{\frac {1}{e}})=-1}递减为{\displaystyle W\_{-1}(0^{-})=-\infty }。

朗伯{\displaystyle W}函数不能用初等函数来表示。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{\displaystyle y(t)=ay(t-1)}。

复平面上的朗伯W函数的函數圖形

微分和积分
-----

朗伯 {\displaystyle W\,}函数的积分形式为

{\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 \,}
{\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 {\displaystyle x\not \in \left[-{\frac {1}{e}},0\right],k\in {\mathbb {Z} }\,} ,若 {\displaystyle x\in \left(-{\frac {1}{e}},0\right),k=1,\pm 2,\pm 3,...\,}

{\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：

{\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}}}}
{\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 {\displaystyle W\_{k}(x)=u+v{\rm {i}},x=t+s{\rm {i}}} ，则有 {\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，

{\displaystyle e^{u}\left(u\cos v-v\sin v\right)=t,e^{u}\left(u\sin v+v\cos v\right)=s},当{\displaystyle s=0}时，易知 {\displaystyle u=-v\cot v}

  

{\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}},}}
{\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 {\displaystyle x>{\frac {1}{e}}} ，上式还可化为{\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{\displaystyle W\,}函数满足以下的微分方程：

{\displaystyle z\left[1+W(z)\right]{\frac {\rm {d}}{{\rm {d}}z}}W(z)=W(z)}，{\displaystyle z\neq -{\frac {1}{e}}\,,}
因此：

{\displaystyle {\frac {\rm {d}}{{\rm {d}}z}}W(z)={\frac {W(z)}{z\left[1+W(z)\right]}}}，{\displaystyle z\neq -{\frac {1}{e}}\,.}
函数{\displaystyle W(x)\,}，以及许多含有{\displaystyle W(x)\,}的表达式，都可以用{\displaystyle w=W(x)\,}的变量代换来积分，也就是说{\displaystyle x=we^{w}\,}

{\displaystyle \int W(x){\rm {d}}x=x\left[W(x)+{\frac {1}{W(x)}}-1\right]+C}
{\displaystyle \int \_{0}^{1}W(x){\rm {d}}x=\Omega +{\frac {1}{\Omega }}-2\approx 0.330366}
其中{\displaystyle \Omega }為欧米加常数。

性质
--

{\displaystyle 1\,}、{\displaystyle z^{z^{z^{z^{z^{.^{.^{.}}}}}}}=\lim \_{n\to \infty }(z\upuparrows n)=-{\frac {W(-\ln z)}{\ln z}}}，

其中{\displaystyle \upuparrows }是高德納箭號表示法。

{\displaystyle 2\,}、若{\displaystyle z>0\,}，则{\displaystyle \ln W(z)=\ln z-W(z)\,}

泰勒级数
----

{\displaystyle W\_{0}\,}在{\displaystyle x=0\,}的泰勒级数如下：

{\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 {\displaystyle {\frac {1}{e}}\,} 。

加法定理
----

{\displaystyle W(x)+W(y)=W\left[{\frac {xy}{W(x)}}+{\frac {xy}{W(y)}}\right]\,}
{\displaystyle x>0,y>0\,}

複數值
---

實部

{\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)\,} , {\displaystyle x^{2}+y^{2}<{\frac {1}{e^{2}}}\,}
虛部

{\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)\,}, {\displaystyle x^{2}+y^{2}<{\frac {1}{e^{2}}}\,}
模長

{\displaystyle |W(x+y{\rm {i}})|=W({\sqrt {x+y}})\,}
模角

{\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]\,}, {\displaystyle x^{2}+y^{2}<{\frac {1}{e^{2}}}\,}
共軛值

{\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]\,}, {\displaystyle x^{2}+y^{2}<{\frac {1}{e^{2}}}\,}

特殊值
---

{\displaystyle W\left(-{\frac {\pi }{2}}\right)={\frac {\pi }{2}}i}
{\displaystyle W\left(-{\frac {\ln 2}{2}}\right)=-\ln 2}
{\displaystyle W\left(-{1 \over e}\right)=-1}
{\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）
{\displaystyle W(e)=1\,}
{\displaystyle W(e^{e+1})=e\,}
{\displaystyle W\left({\frac {1}{e^{1-{\frac {1}{e}}}}}\right)={\frac {1}{e}}}
{\displaystyle W({\pi }e^{\pi })=\pi }
{\displaystyle W(k{\ln k})={\ln k}} {\displaystyle (k>0)}
{\displaystyle W({\rm {i}}\pi )=-{\rm {i}}\pi }
{\displaystyle W(-{\rm {i}}\pi )={\rm {i}}\pi }
{\displaystyle W({\rm {i}}\cos 1-\sin 1)={\rm {i}}}
{\displaystyle W(-{\frac {3}{2}}{\pi })=-{\frac {3}{2}}{\pi }{\rm {i}}}
{\displaystyle W(-{\frac {\sqrt[{7}]{8}}{7}}{\ln 2})=-{\frac {32}{7}}{\ln 2}}
{\displaystyle W(-{\frac {\sqrt {3}}{54}}{\ln 3})=-{\frac {9}{2}}{\ln 3}}
{\displaystyle W(-{\frac {\ln 2}{4}})=-4{\ln 2}}
{\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}}}
{\displaystyle W(-{\frac {\ln k}{k}})=-\ln k}
{\displaystyle W\left[-{\frac {\ln(x+1)}{x(x+1)^{\frac {1}{x}}}}\right]=-{\frac {x+1}{x}}\ln(x+1)>,-1<x<0}

应用
--

许多含有指数的方程都可以用{\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{\displaystyle Y=Xe^{X}\,}的形式。

### 例子

例子1
{\displaystyle 2^{t}=5t\,}
{\displaystyle \Rightarrow 1={\frac {5t}{2^{t}}}\,}
{\displaystyle \Rightarrow 1=5t\,e^{-t\ln 2}\,}
{\displaystyle \Rightarrow {\frac {1}{5}}=t\,e^{-t\ln 2}\,}
{\displaystyle \Rightarrow -{\frac {\ln 2}{5}}=(-\,t\,\ln 2)\,e^{-t\ln 2}\,}
{\displaystyle \Rightarrow -t\ln 2=W\_{k}\left(-{\frac {\ln 2}{5}}\right)\,}
{\displaystyle \Rightarrow t=-{\frac {W\_{k}\left(-{\frac {\ln 2}{5}}\right)}{\ln 2}}\,}
更一般地，以下的方程

{\displaystyle Q^{ax+b}=cx+d\,}
其中

{\displaystyle Q>0\land Q\neq 1\land c\neq 0}
两边同乘: {\displaystyle {\frac {a}{c}}}，

得到：{\displaystyle {\frac {a}{c}}Q^{ax+b}=ax+{\frac {ad}{c}}\,}

同除以：{\displaystyle Q^{ax}\,}，

得到：{\displaystyle {\frac {a}{c}}Q^{b}=\left(ax+{\frac {ad}{c}}\right)Q^{-ax}\,}

同除：{\displaystyle Q^{\frac {ad}{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{\displaystyle t=ax+{\frac {ad}{c}}}

化为

{\displaystyle tQ^{-t}={\frac {a}{c}}Q^{b-{\frac {ad}{c}}}}
即：{\displaystyle t\left(e^{\ln Q}\right)^{-t}={\frac {a}{c}}Q^{b-{\frac {ad}{c}}}}

同乘：{\displaystyle {\ln Q}\,}

得出

{\displaystyle t{\ln Q}\cdot e^{-t\ln Q}={\ln Q}\cdot {\frac {a}{c}}Q^{b-{\frac {ad}{c}}}}
故{\displaystyle t{\ln Q}=-W\_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}

带入{\displaystyle t=ax+{\frac {ad}{c}}}

为

{\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

{\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：{\displaystyle xe^{x}=-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}}中，

{\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：{\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：

{\displaystyle x=-{\frac {W\_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}{a\ln Q}}-{\frac {d}{c}}}
若: {\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{\displaystyle W\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}，

{\displaystyle {\rm {W}}\_{-1}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}，

为

{\displaystyle x\_{1}=-{\frac {W\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}{a\ln Q}}-{\frac {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

{\displaystyle x^{x}={\mathrm {t} }\,,}
为

{\displaystyle x={\frac {\ln {\rm {t}}}{W(\ln {\rm {t}})}}\,}
或

{\displaystyle x=\exp \left(W\_{k}\left[\ln({\rm {t}})\right]\right).}
例子3
以下方程的解

{\displaystyle x\log \_{b}{x}=a\,}
具有形式

{\displaystyle x={\frac {a{\ln b}}{W\_{k}\left(a{\ln b}\right)}}}

  

例子4
{\displaystyle x^{a}-b^{x}=0\,}
{\displaystyle a>0\,} : {\displaystyle b>0\,} : {\displaystyle x>0\,}
取对数，

{\displaystyle a\ln x=x\ln b\,}
{\displaystyle {\frac {\ln x}{x}}={\frac {\ln b}{a}}\,}
{\displaystyle e^{\frac {\ln x}{x}}=e^{\frac {\ln b}{a}}\,}
{\displaystyle x^{\frac {1}{x}}=b^{\frac {1}{a}}\,}
取倒数，

{\displaystyle \left({\frac {1}{x}}\right)^{\frac {1}{x}}=b^{-{\frac {1}{a}}}\,}
{\displaystyle {\frac {1}{x}}=-{\frac {\ln b}{aW\left(-{\frac {1}{a}}\ln b\right)}}\,}
最终解为 : {\displaystyle x=-{\frac {a}{\ln b}}W\_{k}\left(-{\frac {\ln b}{a}}\right)\,}

例子5
{\displaystyle (ax+b)^{n}=u^{cx+d}\,}
两边开{\displaystyle n\,}次方并除以{\displaystyle a\,}得

{\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{\displaystyle u=e^{\ln u}\,} ，

化为

{\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

{\displaystyle -{\frac {c\ln u}{n}}u^{-{\frac {c}{n}}x-{\frac {cb}{na}}}\,} ，

{\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

{\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}}\,}

{\displaystyle k\in {\mathbb {Z} }\,}

一般化
---

標準的 Lambert W 函數可用來表示以下超越代數方程式的解：

{\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{\textstyle x=r+{\tfrac {W\left({\frac {ce^{-cr}}{a\_{o}}}\right)}{c}}}

Lambert W 函數之一般化 包括: 

* 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" 所示其 (1) 的右邊式現為二維多項式 x：

{\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。於此 (1)（或 (2)）的右手邊現為無限級數多項式之比於 *x*：

{\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。
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 以及黎曼假设的 Keiper-Li 准则

图象
--

* 朗伯W函数在复平面上的图像
* z = Re(W0(x + i y)) *z* = Re(W0(*x* + *i* *y*))
* z = Im(W0(x + i y)) *z* = Im(W0(*x* + *i* *y*))
* 
*

计算
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*W*函数可以用以下的递推关系算出：

{\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}}}}}

参考来源
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1. ↑  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, ; Arxiv  （页面存档备份，存于）
2. ↑  T.C. Scott, G. Fee and J. Grotendorst, "Asymptotic series of Generalized Lambert W Function" （页面存档备份，存于）, SIGSAM, vol. 47, no. 3, (September 2013), pp. 75-83
3. ↑  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）, SIGSAM, vol. 48, no. 2, (June 2014), pp. 42-56
4. ↑  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, ; Arxiv  （页面存档备份，存于）
5. ↑  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,  （页面存档备份，存于）; Arxiv  （页面存档备份，存于）
6. ↑  Aude Maignan, T.C. Scott, "Fleshing out the Generalized Lambert W Function", SIGSAM, vol. 50, no. 2, (June 2016), pp. 45-60
7. ↑  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,  （页面存档备份，存于）
8. ↑  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

外部链接
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* MathWorld （页面存档备份，存于）