Editing Lambert W function (section)

=== 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.