# 朗伯W函数

${\displaystyle z=W(z)e^{W(z)}.}$

## 微分和积分

${\displaystyle W(x)={\frac {x}{\pi }}\int _{0}^{\pi }{\frac {\left(1-v\cot v\right)^{2}+v^{2}}{x+v\csc v\cdot e^{-v\cot v}}}{\rm {d}}v,|\arg \left(x\right)|<\pi \,}$
${\displaystyle W(x)=\int _{-\infty }^{-{\frac {1}{e}}}{-{\frac {1}{\pi }}}\Im \left[{\frac {\rm {d}}{{\rm {d}}x}}W(x)\right]\ln \left(1-{\frac {z}{x}}\right){\rm {d}}x\,}$

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

${\displaystyle W_{k}(x)=1+\left(\ln x-1+2k\pi {\rm {i}}\right)e^{{\frac {\rm {i}}{2\pi }}\int _{0}^{\infty }\ln {\frac {t-\ln t+\ln x+(2k+1)\pi {\rm {i}}}{t-\ln t+\ln x+(2k-1)\pi {\rm {i}}}}\cdot {\frac {{\rm {d}}t}{t+1}}}=1+\left(\ln x-1+2k\pi {\rm {i}}\right)e^{{\frac {\rm {i}}{2\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}+2\pi \left(t-\ln t+\ln x\right){\rm {i}}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}}\,}$

${\displaystyle W_{k}(x)=1+\left(\ln x-1+2k\pi {\rm {i}}\right)e^{{\frac {\rm {i}}{2\pi }}\int _{0}^{\infty }\left[{\frac {1}{2}}\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}+{\rm {i}}\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\right]\cdot {\frac {{\rm {d}}t}{t+1}}}}$
${\displaystyle {}_{W_{k}(x)=1+{\frac {\left(\ln x-1\right)\cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}-2k\pi \sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}+{\rm {i}}\left[\left(\ln x-1\right)\sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}+2k\pi \cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}\right]}{e^{{\frac {1}{2\pi }}\int _{0}^{\infty }\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\cdot {\frac {\rm {{d}t}}{t+1}}}}}}}$

${\displaystyle W_{k}(x)=u+v{\rm {i}},x=t+s{\rm {i}}}$ ，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 ${\displaystyle u=-v\cot v}$

${\displaystyle {}_{W_{k}(x)={\frac {\left(1-\ln x\right)\sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}-2k\pi \cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}}{e^{{\frac {1}{2\pi }}\int _{0}^{\infty }\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\cdot {\frac {\rm {{d}t}}{t+1}}}}}\cot {\frac {\left(\ln x-1\right)\sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}+2k\pi \cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}}{e^{{\frac {1}{2\pi }}\int _{0}^{\infty }\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\cdot {\frac {\rm {{d}t}}{t+1}}}}}+{\frac {\left(\ln x-1\right)\sin {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}+2k\pi \cos {\frac {1}{4\pi }}\int _{0}^{\infty }\ln {\frac {\left(t-\ln t+\ln x\right)^{2}+\left(2k+1\right)^{2}\pi ^{2}}{\left(t-\ln t+\ln x\right)^{2}+\left(2k-1\right)^{2}\pi ^{2}}}\cdot {\frac {{\rm {d}}t}{t+1}}}{e^{{\frac {1}{2\pi }}\int _{0}^{\infty }\arctan {\frac {2\pi \left(t-\ln t+\ln x\right)}{\left(t-\ln t+\ln x\right)^{2}+\left(4k^{2}-1\right)\pi ^{2}}}\cdot {\frac {\rm {{d}t}}{t+1}}}}}{\rm {i}},}}$
${\displaystyle W_{0}(x)=1+\left(\ln x-1\right)e^{-{\frac {1}{\pi }}\int _{0}^{\infty }\arg \left(t-\ln t+\ln x+\pi {\rm {i}}\right)\cdot {\frac {\rm {{d}t}}{t+1}}},x>0}$

${\displaystyle x>{\frac {1}{e}}}$ ，上式还可化为${\displaystyle W_{0}(x)=1+\left(\ln x-1\right)e^{-{\frac {1}{\pi }}\int _{0}^{\infty }\arctan {\frac {\pi }{t-\ln t+\ln x}}\cdot {\frac {\rm {{d}t}}{t+1}}}}$

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

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

${\displaystyle \int W(x){\rm {d}}x=x\left[W(x)+{\frac {1}{W(x)}}-1\right]+C}$
${\displaystyle \int _{0}^{1}W(x){\rm {d}}x=\Omega +{\frac {1}{\Omega }}-2\approx 0.330366}$

## 性质

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

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

## 泰勒级数

${\displaystyle W_{0}\,}$${\displaystyle x=0\,}$的泰勒级数如下：

${\displaystyle W_{0}(x)=\sum _{n=1}^{\infty }{\frac {(-n)^{n-1}}{n!}}\ x^{n}=x-x^{2}+{\frac {3}{2}}x^{3}-{\frac {8}{3}}x^{4}+{\frac {125}{24}}x^{5}-\cdots }$

## 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

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

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

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

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

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

## 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

## 应用

### 例子

${\displaystyle 2^{t}=5t\,}$
${\displaystyle \Rightarrow 1={\frac {5t}{2^{t}}}\,}$
${\displaystyle \Rightarrow 1=5t\,e^{-t\ln 2}\,}$
${\displaystyle \Rightarrow {\frac {1}{5}}=t\,e^{-t\ln 2}\,}$
${\displaystyle \Rightarrow -{\frac {\ln 2}{5}}=(-\,t\,\ln 2)\,e^{-t\ln 2}\,}$
${\displaystyle \Rightarrow -t\ln 2=W_{k}\left(-{\frac {\ln 2}{5}}\right)\,}$
${\displaystyle \Rightarrow t=-{\frac {W_{k}\left(-{\frac {\ln 2}{5}}\right)}{\ln 2}}\,}$

${\displaystyle Q^{ax+b}=cx+d\,}$

${\displaystyle Q>0\land Q\neq 1\land c\neq 0}$

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

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

${\displaystyle tQ^{-t}={\frac {a}{c}}Q^{b-{\frac {ad}{c}}}}$

${\displaystyle t{\ln Q}\cdot e^{-t\ln Q}={\ln Q}\cdot {\frac {a}{c}}Q^{b-{\frac {ad}{c}}}}$

${\displaystyle t{\ln Q}=-W_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}$

${\displaystyle \left(ax+{\frac {ad}{c}}\right){\ln Q}=-W_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}$

${\displaystyle x=-{\frac {W_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}{a\ln Q}}-{\frac {d}{c}}}$

${\displaystyle -{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\in \left(-\infty ,-{\frac {1}{e}}\right)}$,

${\displaystyle x=-{\frac {W_{k}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}{a\ln Q}}-{\frac {d}{c}}}$

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

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

${\displaystyle x_{2}=-{\frac {{\rm {W}}_{-1}\left(-{\frac {a\ln Q}{c}}\,Q^{b-{\frac {ad}{c}}}\right)}{a\ln Q}}-{\frac {d}{c}}}$

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

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

${\displaystyle x=\exp \left(W_{k}\left[\ln({\rm {t}})\right]\right).}$

${\displaystyle x\log _{b}{x}=a\,}$

${\displaystyle x={\frac {a{\ln b}}{W_{k}\left(a{\ln b}\right)}}}$

${\displaystyle x^{a}-b^{x}=0\,}$
${\displaystyle a>0\,}$ : ${\displaystyle b>0\,}$ : ${\displaystyle x>0\,}$

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

${\displaystyle \left({\frac {1}{x}}\right)^{\frac {1}{x}}=b^{-{\frac {1}{a}}}\,}$
${\displaystyle {\frac {1}{x}}=-{\frac {\ln b}{aW\left(-{\frac {1}{a}}\ln b\right)}}\,}$

${\displaystyle (ax+b)^{n}=u^{cx+d}\,}$

${\displaystyle x+{\frac {b}{a}}={\frac {u^{{\frac {c}{n}}x+{\frac {d}{n}}}}{a}}\left(\cos {\frac {2k\pi }{n}}+{\rm {i}}\sin {\frac {2k\pi }{n}}\right)\,}$

${\displaystyle u=e^{\ln u}\,}$

${\displaystyle x+{\frac {b}{a}}={\frac {e^{{\frac {c\ln u}{n}}x+{\frac {d\ln u}{n}}}}{a}}\left(\cos {\frac {2k\pi }{n}}+{\rm {i}}\sin {\frac {2k\pi }{n}}\right)\,}$

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

${\displaystyle \left(-{\frac {c\ln u}{n}}x-{\frac {cb\ln u}{na}}\right)e^{-{\frac {c\ln u}{n}}x-{\frac {cb\ln u}{na}}}=-{\frac {c\ln u}{na}}u^{{\frac {d}{n}}-{\frac {cb}{na}}}\left(\cos {\frac {2k\pi }{n}}+{\rm {i}}\sin {\frac {2k\pi }{n}}\right)\,}$

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

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

## 一般化

${\displaystyle e^{-cx}=a_{o}(x-r)~~\quad \qquad \qquad \qquad \qquad (1)}$

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：
${\displaystyle e^{-cx}=a_{o}(x-r_{1})(x-r_{2})~~\qquad \qquad (2)}$

• 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
${\displaystyle e^{-cx}=a_{o}{\frac {\prod _{i=1}^{\infty }(x-r_{i})}{\prod _{i=1}^{\infty }(x-s_{i})}}\qquad \qquad \qquad (3)}$

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]

## 计算

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

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, [1]; Arxiv [2]页面存档备份，存于互联网档案馆
2. ^ 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
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，存于互联网档案馆）, 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, [3]; Arxiv [4]页面存档备份，存于互联网档案馆
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, [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
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, [7]页面存档备份，存于互联网档案馆
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