微分和積分
編輯
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}}}\,} 把被積函數的實部和虛部分離出來：
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}}}}
由隱函數 的求導法則，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}}\,.} 函數W ( x ) {\displaystyle W(x)\,} W ( x ) {\displaystyle W(x)\,} w = W ( x ) {\displaystyle w=W(x)\,} 變量代換 來積分，也就是說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 } 歐米加常數 。
泰勒級數
編輯
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 } 收斂半徑 為 1 e {\displaystyle {\frac {1}{e}}\,}
加法定理
編輯
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 \,} 歐米加常數 ）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}
許多含有指數的方程都可以用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} 兩邊同乘: 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)}\,}
可以用變量代換
令t = a x + a d c {\displaystyle t=ax+{\frac {ad}{c}}}
化為
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)}
帶入t = a x + a d c {\displaystyle t=ax+{\frac {ad}{c}}}
為
( 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)} 因此最終的解為
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}}} 若輔助方程：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)} 輔助方程無實數解，原方程亦無實解；
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 )}
輔助方程有一實數解，原方程有一實解：
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)}
輔助方程有二實解，設為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)}
為
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 用類似的方法，可知以下方程的解
x x = t , {\displaystyle x^{x}={\mathrm {t} }\,,} 為
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\,} 取對數，
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}}\,} 取倒數，
( 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)}}\,} 最終解為 : 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}\,} 兩邊開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}\,}
化為
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)\,}
兩邊同乘
− 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)\,}
最終得
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}}\,}
k ∈ Z {\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：
e − c x = a o ( x − r ) ( 1 ) {\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x = r + W ( c e − c r a o ) c {\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] e − c x = a o ( x − r 1 ) ( x − r 2 ) ( 2 ) {\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」（超幾何分布）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）氫分子離子 。[5] x ： e − c x = a o ∏ i = 1 ∞ ( x − r i ) ∏ i = 1 ∞ ( x − s i ) ( 3 ) {\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 。由於哈代 的「虛假導數」概念，多根的特殊情況得以解決[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] 黎曼假設 的 Keiper-Li 準則 [8]
朗伯W函數在複平面上的圖像
z = Re(W0 (x + i y ))
z = Im(W0 (x + i y ))
W 函數可以用以下的遞推關係 算出：
w j + 1 = w j − w j e w j − z e w j ( w j + 1 ) − ( w j + 2 ) ( w j e w j − z ) 2 w j + 2 {\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}}}}} 參考來源
編輯
^ 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] （頁面存檔備份 ，存於網際網路檔案館 ）
^ T.C. Scott, G. Fee and J. Grotendorst, "Asymptotic series of Generalized Lambert W Function" （頁面存檔備份 ，存於網際網路檔案館 ）, 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" （頁面存檔備份 ，存於網際網路檔案館 ）, 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] （頁面存檔備份 ，存於網際網路檔案館 ）
^ 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] （頁面存檔備份 ，存於網際網路檔案館 ）; Arxiv [6] （頁面存檔備份 ，存於網際網路檔案館 ）
^ 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] （頁面存檔備份 ，存於網際網路檔案館 ）
^ 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
外部連結
編輯
![{\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))
