Editing Lambert W function (section)

== Other formulas ==

=== Definite integrals ===
There are several useful definite integral formulas involving the principal branch of the {{mvar|W}} function, including the following:
: <math>\begin{align}
& \int_0^\pi W_0\left( 2\cot^2x \right)\sec^2 x\,dx = 4\sqrt{\pi}, \\[5pt]
& \int_0^\infty \frac{W_0(x)}{x\sqrt{x}}\,dx = 2\sqrt{2\pi}, \\[5pt]
& \int_0^\infty W_0\left(\frac{1}{x^2}\right)\,dx = \sqrt{2\pi}, \text{ and more generally}\\[5pt]
& \int_0^\infty W_0\left(\frac{1}{x^N}\right)\,dx = N^{1-\frac1N} \Gamma\left(1-\frac1N\right)\qquad \text{for }N > 0
\end{align}</math>
where <math>\Gamma</math> denotes the [[gamma function]].

The first identity can be found by writing the [[Gaussian integral]] in [[polar coordinates]].

The second identity can be derived by making the substitution {{math|1=''u'' = ''W''<sub>0</sub>(''x'')}}, which gives
: <math>\begin{align}
x & =ue^u, \\[5pt]
\frac{dx}{du} & =(u+1)e^u.
\end{align}</math>

Thus
: <math>\begin{align}
\int_0^\infty \frac{W_0(x)}{x\sqrt{x}}\,dx &=\int_0^\infty \frac{u}{ue^{u}\sqrt{ue^{u}}}(u+1)e^u \, du \\[5pt]
&=\int_0^\infty \frac{u+1}{\sqrt{ue^u}}du \\[5pt]
&=\int_0^\infty \frac{u+1}{\sqrt{u}}\frac{1}{\sqrt{e^u}}du\\[5pt]
&=\int_0^\infty u^\tfrac12 e^{-\frac{u}{2}}du+\int_0^\infty u^{-\tfrac12} e^{-\frac{u}{2}}du\\[5pt]
&=2\int_0^\infty (2w)^\tfrac12 e^{-w} \, dw+2\int_0^\infty (2w)^{-\tfrac12} e^{-w} \, dw && \quad (u =2w) \\[5pt]
&=2\sqrt{2}\int_0^\infty w^\tfrac12 e^{-w} \, dw + \sqrt{2} \int_0^\infty w^{-\tfrac12} e^{-w} \, dw \\[5pt]
&=2\sqrt{2} \cdot \Gamma \left (\tfrac32 \right )+\sqrt{2} \cdot \Gamma \left (\tfrac12 \right ) \\[5pt]
&=2\sqrt{2} \left (\tfrac12\sqrt{\pi} \right )+\sqrt{2}\left(\sqrt{\pi}\right) \\[5pt]
&=2\sqrt{2\pi}.
\end{align}</math>

The third identity may be derived from the second by making the substitution {{math|1=''u'' = ''x''<sup>−2</sup>}} and the first can also be derived from the third by the substitution {{math|1=''z'' = {{sfrac|1|{{sqrt|2}}}} tan ''x''}}. Deriving its generalization, the fourth identity, is only slightly more involved and can be done by substituting, in turn, <math>u = x^{\frac1N}</math>, <math>t = W_0(u)</math>, and <math>z = \frac tN</math>, observing that one obtains two integrals matching the definition of the gamma function, and finally using the properties of the gamma function to collect terms and simplify.

Except for {{mvar|z}} along the branch cut {{open-closed|−∞, −{{sfrac|1|''e''}}}} (where the integral does not converge), the principal branch of the Lambert {{mvar|W}} function can be computed by the following integral:<ref>{{cite web|title=The Lambert ''W'' Function|url=http://www.orcca.on.ca/LambertW/|publisher=Ontario Research Centre for Computer Algebra}}</ref>
: <math>\begin{align}
W_0(z)&=\frac{z}{2\pi}\int_{-\pi}^\pi\frac{\left(1-\nu\cot\nu\right)^2+\nu^2}{z+\nu\csc\left(\nu\right) e^{-\nu\cot\nu}} \, d\nu \\[5pt]
&= \frac{z}{\pi} \int_0^\pi \frac{\left(1-\nu\cot\nu\right)^2+\nu^2}{z + \nu \csc\left(\nu\right) e^{-\nu\cot\nu}} \, d\nu,
\end{align}</math>
where the two integral expressions are equivalent due to the symmetry of the integrand.

=== Indefinite integrals ===
<math display="block">\int \frac{ W(x) }{x} \, dx \; = \; \frac{ W(x)^2}{2} + W(x) + C </math>
{{math proof|title=1st proof|proof=
Introduce substitution variable <math> u= W(x) \rightarrow ue^u=x \;\;\;\; \frac{ d }{ du } ue^u = (u+1)e^u </math>
: <math>\int \frac{ W(x) }{x} \, dx \; = \; \int \frac{u}{ue^u}(u+1)e^u \, du </math>
: <math>\int \frac{ W(x) }{x} \, dx \; = \; \int \frac{ \cancel{\color{OliveGreen}{u}} }{ \cancel{\color{OliveGreen}{u}} \cancel{\color{BrickRed}{e^u}} }\left(u+1\right)\cancel{\color{BrickRed}{e^u}} \, du </math>
: <math>\int \frac{ W(x) }{x} \, dx \; = \; \int (u+1) \, du </math>
: <math>\int \frac{ W(x) }{x} \, dx \; = \; \frac{u^2}{2} + u + C </math>
:: <math> u= W(x) </math>
: <math>\int \frac{ W(x) }{x} \, dx \; = \; \frac{ W(x) ^2}{2} + W(x) + C </math>
}}
{{math proof|title=2nd proof|proof=
<math> W(x) e^{ W(x) } = x \rightarrow \frac{ W(x) }{ x } = e^{ - W(x) } </math>

<math>\int \frac{ W(x) }{x} \, dx \; = \; \int e^{ - W(x) } \, dx </math>
:: <math> u= W(x) \rightarrow ue^u=x \;\;\;\; \frac{ d }{ \, du } ue^u = \left(u+1\right)e^u </math>
<math>\int \frac{ W(x) }{x} \, dx \; = \; \int e^{-u} (u+1) e^u \, du </math>

<math>\int \frac{ W(x) }{x} \, dx \; = \; \int \cancel{\color{OliveGreen}{e^{ -u } }} \left( u+1 \right) \cancel{\color{OliveGreen}{ e^u }} \, du </math>

<math>\int \frac{ W(x) }{x} \, dx \; = \; \int (u+1) \, du </math>

<math>\int \frac{ W(x) }{x} \, dx \; = \; \frac{u^2}{2} + u + C </math>
:: <math> u= W(x) </math>
<math>\int \frac{ W(x) }{x} \, dx \; = \; \frac{ W(x) ^2}{2} + W(x) + C </math>
}}

<math display="block">\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(A e^{Bx}\right) ^2}{2B} + \frac{ W\left(A e^{Bx}\right) }{B} + C </math>
{{math proof|proof=

<math>\int W\left(A e^{Bx}\right) \, dx \; = \; \int W\left(A e^{Bx}\right) \, dx </math>
:: <math> u = Bx \rightarrow \frac{u}{B} = x \;\;\;\; \frac{ d }{ du } \frac{u}{B} = \frac{1}{B} </math>
<math>\int W\left(A e^{Bx}\right) \, dx \; = \; \int W\left(A e^{u}\right) \frac{1}{B} du </math>
:: <math> v = e^u \rightarrow \ln\left(v\right) = u \;\;\;\; \frac{ d }{ dv } \ln\left(v\right) = \frac{1}{v} </math>
<math>\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int \frac{W\left(A v\right)}{v} dv </math>
:: <math> w = Av \rightarrow \frac{w}{A} = v \;\;\;\; \frac{ d }{ dw } \frac{w}{A} = \frac{1}{A} </math>
<math>\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int \frac{\cancel{\color{OliveGreen}{A}} W(w)}{w} \cancel{\color{OliveGreen}{ \frac{1}{A} }} dw </math>
:: <math> t = W\left(w\right) \rightarrow te^t=w \;\;\;\; \frac{ d }{ dt } te^t = \left(t+1\right)e^t </math>
<math>\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int \frac{t}{te^t}\left(t+1\right)e^t dt </math>

<math>\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int \frac{ \cancel{\color{OliveGreen}{t}} }{ \cancel{\color{OliveGreen}{t}} \cancel{\color{BrickRed}{e^t}} }\left(t+1\right) \cancel{\color{BrickRed}{e^t}} dt </math>

<math>\int W\left(A e^{Bx}\right) \, dx \; = \; \frac{1}{B} \int (t+1) dt </math>

<math> \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{t^2}{2B} + \frac{t}{B} + C </math>
:: <math> t = W\left(w\right) </math>
<math> \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(w\right) ^2}{2B} + \frac{ W\left(w\right) }{B} + C </math>
:: <math> w = Av </math>
<math> \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(Av\right) ^2}{2B} + \frac{ W\left(Av\right) }{B} + C </math>
:: <math> v = e^u </math>
<math> \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(Ae^u\right) ^2}{2B} + \frac{ W\left(Ae^u\right) }{B} + C </math>
:: <math> u = Bx </math>
<math> \int W\left(A e^{Bx}\right) \, dx \; = \; \frac{ W\left(Ae^{Bx}\right) ^2}{2B} + \frac{ W\left(Ae^{Bx}\right) }{B} + C </math>
}}

<math display="block"> \int \frac{ W(x) }{x^2} \, dx \; = \; \operatorname{Ei}\left(- W(x) \right) - e^{ - W(x) } + C </math>
{{math proof|proof=
Introduce substitution variable <math> u = W(x)</math>, which gives us <math> ue^u = x </math> and <math>\frac{ d }{ du } ue^u = \left(u+1\right)e^u </math>

<math>\begin{align}
\int \frac{ W(x) }{x^2} \, dx \;
& = \; \int \frac{ u }{ \left(ue^u\right)^2 } \left(u+1\right)e^u du \\
& = \; \int \frac{ u+1 }{ ue^u } du \\
& = \; \int \frac{ u }{ ue^u } du \; + \; \int \frac{ 1 }{ ue^u } du \\
& = \; \int e^{-u} du \; + \; \int \frac{ e^{-u} }{ u } du
\end{align}</math>
:: <math> v = -u \rightarrow -v = u \;\;\;\; \frac{ d }{ dv } -v = -1 </math>
<math> \int \frac{ W(x) }{x^2} \, dx \; = \; \int e^{v} \left(-1\right) dv \; + \; \int \frac{ e^{-u} }{ u } du </math>

<math> \int \frac{ W(x) }{x^2} \, dx \; = \; - e^v + \operatorname{Ei}\left(-u\right) + C </math>
:: <math> v = -u </math>
<math> \int \frac{ W(x) }{x^2} \, dx \; = \; - e^{-u} + \operatorname{Ei}\left(-u\right) + C </math>
:: <math> u = W(x) </math>
<math>\begin{align}
 \int \frac{ W(x) }{x^2} \, dx \; &= \; - e^{- W(x) } + \operatorname{Ei}\left(- W(x) \right) + C \\
 &= \; \operatorname{Ei}\left(- W(x) \right) - e^{- W(x) } + C
\end{align}</math>
}}