朗伯W函数

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。

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

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

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

${\displaystyle W_{0}\,}$ 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 ${\displaystyle {\frac {1}{e}}\,}$ 。

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

${\displaystyle x>0,y>0\,}$

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

${\displaystyle {\overline {W(x+y{\rm {i}})}}=\sum _{k=1}^{\infty }{\frac {(-k)^{k-1}}{k!}}{\sqrt {(x^{2}+y^{2})^{k}}}\left[\cos \left(k\arctan {\frac {x}{y}}\right)-{\rm {i}}\sin \left(k\arctan {\frac {x}{y}}\right)\right]\,}$ , ${\displaystyle x^{2}+y^{2}<{\frac {1}{e^{2}}}\,}$

${\displaystyle W\left(-{\frac {\pi }{2}}\right)={\frac {\pi }{2}}i}$

${\displaystyle W\left(-{\frac {\ln 2}{2}}\right)=-\ln 2}$

${\displaystyle W\left(-{1 \over e}\right)=-1}$

${\displaystyle W\left(1\right)=\Omega ={\frac {1}{\int _{-\infty }^{\infty }{\frac {{\rm {d}}x}{(e^{x}-x)^{2}+\pi ^{2}}}}}-1\approx 0.56714329\dots \,}$ （欧米加常数）

${\displaystyle W(e)=1\,}$

${\displaystyle W(e^{e+1})=e\,}$

${\displaystyle W\left({\frac {1}{e^{1-{\frac {1}{e}}}}}\right)={\frac {1}{e}}}$

${\displaystyle W({\pi }e^{\pi })=\pi }$

${\displaystyle W(k{\ln k})={\ln k}}$ ${\displaystyle (k>0)}$

${\displaystyle W({\rm {i}}\pi )=-{\rm {i}}\pi }$

${\displaystyle W(-{\rm {i}}\pi )={\rm {i}}\pi }$

${\displaystyle W({\rm {i}}\cos 1-\sin 1)={\rm {i}}}$

${\displaystyle W(-{\frac {3}{2}}{\pi })=-{\frac {3}{2}}{\pi }{\rm {i}}}$

${\displaystyle W(-{\frac {\sqrt[{7}]{8}}{7}}{\ln 2})=-{\frac {32}{7}}{\ln 2}}$

${\displaystyle W(-{\frac {\sqrt {3}}{54}}{\ln 3})=-{\frac {9}{2}}{\ln 3}}$

${\displaystyle W(-{\frac {\ln 2}{4}})=-4{\ln 2}}$

${\displaystyle W\left(-1\right)={\frac {e^{{\frac {1}{2\pi }}\int _{0}^{\infty }{1 \over t+1}\arctan {2\pi \over t-\ln t}{\rm {d}}t}-\cos \left[{\frac {1}{4\pi }}\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}}{\rm {d}}t\right]+\pi \sin \left[{\frac {1}{4\pi }}\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}}{\rm {d}}t\right]-{\rm {i}}\left\{\pi \cos \left[{\frac {1}{4\pi }}\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}}{\rm {d}}t\right]+\sin \left[{\frac {1}{4\pi }}\int _{0}^{\infty }{1 \over t+1}\ln {\left(t-\ln t\right)^{2} \over 4\pi ^{2}+\left(t-\ln t\right)^{2}}{\rm {d}}t\right]\right\}}{e^{{\frac {1}{2\pi }}\int _{0}^{\infty }{1 \over t+1}\arctan {2\pi \over t-\ln t}{\rm {d}}t}}}\approx -0.31813-1.33723{\rm {i}}}$

${\displaystyle W(-{\frac {\ln k}{k}})=-\ln k}$

${\displaystyle W\left[-{\frac {\ln(x+1)}{x(x+1)^{\frac {1}{x}}}}\right]=-{\frac {x+1}{x}}\ln(x+1)>,-1<x<0}$

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

${\displaystyle x_{k}=-{\frac {n}{c\ln u}}W_{k}\left[-{\frac {c\ln u}{na}}u^{{\frac {d}{n}}-{\frac {cb}{na}}}\left(\cos {\frac {2k\pi }{n}}+{\rm {i}}\sin {\frac {2k\pi }{n}}\right)\right]-{\frac {b}{a}}\,}$

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

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 a₀, 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^[1][2][3] 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”^[4] 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 r₁ 和 r₂ 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 r₁ = r₂，(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]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^[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] 以及黎曼假设的 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：

${\displaystyle w_{j+1}=w_{j}-{\frac {w_{j}e^{w_{j}}-z}{e^{w_{j}}(w_{j}+1)-{\frac {(w_{j}+2)(w_{j}e^{w_{j}}-z)}{2w_{j}+2}}}}}$

• MathWorld （页面存档备份，存于互联网档案馆）