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。

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：

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

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

• MathWorld （页面存档备份，存于）