$Factorials$

$Gaussians$

{\mathcal {G}}(n)\ =\ \int _{0}^{\infty }{e^{-x^{n}}\ dx}

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{\mathcal {G}}(0)\ =\ \infty \ \qquad \qquad \qquad \quad \ {\mathcal {G}}(\infty )\ =\ 1\ \qquad \qquad \qquad \quad \ {\mathcal {G}}'\left(2{\tfrac {1}{6}}\right)\ =\ 0^{^{+}}

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The factorial of a positive number is the Gaussian of its reciprocal or multiplicative inverse:

n!\ =\ {\mathcal {G}}{\Big (}{\tfrac {1}{n}}{\Big )}\ =\ \int _{0}^{\infty }{e^{-{\sqrt[{n}]{x}}}\ dx}\qquad \qquad \qquad \qquad \qquad \qquad \Gamma ^{'}{\Big (}1{\tfrac {6}{13}}{\Big )}\ =\ 0^{^{-}}

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n!\ =\ \int _{0}^{\infty }{e^{-{\sqrt[{n}]{x}}}\ dx}\ =\ \int _{0}^{\infty }{\frac {\ x^{n}}{e^{x}}}\ dx\ =\ \int _{0}^{1}{{\Big |}\ln x{\Big |}^{n}\ dx}\ =\ \int _{0}^{1}{{\Big (}\!\!-\ln x{\Big )}^{n}\ dx}

n!\ =\ \int _{0}^{1}{\ln ^{n}{\Big (}{\tfrac {1}{x}}{\Big )}\ dx}\ =\ \int _{1}^{\infty }{{\frac {\ln ^{n}x}{x^{2}}}\ dx}\ \simeq \ \sum _{k=1}^{\infty }{\frac {\ln ^{n}k}{k^{2}}}\ =\ {\bigg |}\zeta ^{(n)}(2){\bigg |}

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{\underset {_{{\text{k}}\,=\,0}}{\overset {n}{\mathbf {\Xi } }}}\ {\Bigg (}a_{_{\text{k}}}\ ,\ b_{_{\text{k}}},\ {\frac {1}{N_{_{\text{k}}}}}{\Bigg )}\ =\ {\sqrt[{^{N_{_{\text{0}}}}}]{a_{_{\text{0}}}\ +\ b_{_{\text{0}}}{\sqrt[{^{N_{_{\text{1}}}}}]{a_{_{\text{1}}}\ +\ b_{_{\text{1}}}{\sqrt[{^{N_{_{\text{2}}}}}]{\ldots \ {\sqrt[{^{N_{_{n}}}}]{a_{_{n}}}}}}}}}}

{\underset {_{{\text{k}}\,=\,0}}{\overset {n}{\mathbf {\Xi } }}}\ {\Big (}a_{_{\text{k}}}\ ,\ b_{_{\text{k}}},\ -1{\Big )}\ =\ {\cfrac {1}{a_{_{0}}\ +\ {\cfrac {b_{_{0}}}{a_{_{1}}\ +\ {\cfrac {b_{_{1}}}{\ddots \ {a_{_{n}}}}}}}}}

{\underset {_{{\text{k}}\,=\,0}}{\overset {n}{\mathbf {\Xi } }}}\ {\Big (}a_{_{\text{k}}}\ ,\ 1,\ 1{\Big )}\ =\ \sum _{k\,=\,0}^{n}a_{_{\text{k}}}\qquad \qquad ;\qquad \qquad {\underset {_{{\text{k}}\,=\,0}}{\overset {n}{\mathbf {\Xi } }}}\ {\Big (}0,\ a_{_{\text{k}}}\ ,\ 1{\Big )}\ =\ \prod _{k\,=\,0}^{n}a_{_{\text{k}}}

{\underset {_{{\text{k}}\,=\,0}}{\overset {n}{\mathbf {\Xi } }}}\ {\Big (}a_{_{\text{k}}}\ ,\ x,\ 1{\Big )}\ =\ \sum _{k\,=\,0}^{n}a_{_{\text{k}}}\ x^{k}\ =\ P_{n}(x)

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