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Question Number 106745 by qwertyu last updated on 06/Aug/20

Commented by Her_Majesty last updated on 06/Aug/20

$$alwaysthesamequestions...$$$$howdoyoudefine‘‘x{!}^{″}?$$

Commented by qwertyu last updated on 06/Aug/20

$${\int }_0^1\mathit{x}!=?$$

Commented by Her_Majesty last updated on 06/Aug/20

$$again,howdoyoudefinex!$$$$x!=1\times 2\times 3\times ...\times (x-1)\times xforx⩾1with$$$$0!:=1$$$$thennointegralexists$$

Commented by floor(10²Eta[1]) last updated on 06/Aug/20

$$\mathrm{you}\mathrm{should}\mathrm{search}\mathrm{more}\mathrm{before}\mathrm{assuming}$$$$\mathrm{things}\mathrm{like}\mathrm{that}!$$

Commented by floor(10²Eta[1]) last updated on 06/Aug/20

$$\mathrm{the}\mathrm{gamma}\mathrm{function}{\mathrm{it}}^{\prime }\mathrm{s}\mathrm{an}\mathrm{extension}$$$$\mathrm{of}\mathrm{the}\mathrm{factorial}\mathrm{function}\mathrm{and}\mathrm{is}\mathrm{defined}$$$$\mathrm{for}\mathrm{all}\mathrm{complex}\mathrm{numbers}\mathrm{except}\mathrm{the}$$$$\mathrm{non}-\mathrm{positive}\mathrm{integers}.$$$$\mathrm{\Gamma }\left(\mathrm{z}\right)=(\mathrm{z}-1)!={\int }_0^{\mathrm{\infty }}{\mathrm{x}}^{\mathrm{z}-1}{\mathrm{e}}^{-\mathrm{x}}\mathrm{dx}$$

Commented by Her_Majesty last updated on 06/Aug/20

$$Iknowthesethings,{don}^{\prime }thavetosearch$$$$ButIoftenwonderifthepostersofthese$$$$questionsknowwhattheyareposting$$$$$$$$asyouwrote\mathrm{\Gamma }(x+1)istheextensionofx!$$$$butnonetheless\int x!dx{doesn}^{\prime }texist;only$$$$theextension\int \mathrm{\Gamma }(x+1)dxexists$$$$$$$$youguys{don}^{\prime }tdopropermathssometimes$$$$itseems$$

Commented by Sarah85 last updated on 07/Aug/20

$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{another}\mathrm{pandemic}\mathrm{it}\mathrm{seems}...$$

Commented by 1549442205PVT last updated on 07/Aug/20

$$\mathrm{Since}\mathrm{the}\mathrm{function}\mathrm{x}!{\mathrm{isn}}^{\prime }\mathrm{t}\mathrm{a}\mathrm{continue}$$$$\mathrm{function},\mathrm{it}\mathrm{is}\mathrm{interrupt}\mathrm{everywhere}$$$$,\mathrm{so}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{exist}\mathrm{defined}\mathrm{integral}{\int }_{\mathrm{a}}^{\mathrm{b}}\mathrm{x}!\mathrm{dx}$$

Answered by mathmax by abdo last updated on 06/Aug/20

$${\int }_0^1\mathrm{x}!\mathrm{dx}={\int }_0^1\mathrm{\Gamma }(\mathrm{x}+1)\mathrm{dx}{=}_{\mathrm{x}+1=\mathrm{t}}{\int }_1^2\mathrm{\Gamma }\left(\mathrm{t}\right)\mathrm{dt}$$$$={\int }_1^2{\lim }_{\mathrm{n}\to +\mathrm{\infty }}\frac{\mathrm{n}!{\mathrm{n}}^{\mathrm{t}}}{\mathrm{t}(\mathrm{t}+1)(\mathrm{t}+2)....(\mathrm{t}+\mathrm{n})}\mathrm{dt}\mathrm{if}\mathrm{we}\mathrm{have}\mathrm{the}\mathrm{convergencedominee}$$$$\mathrm{we}\mathrm{can}\mathrm{do}$$$$\mathrm{I}={\lim }_{\mathrm{n}\to +\mathrm{\infty }}\mathrm{n}!{\int }_1^2\frac{{\mathrm{n}}^{\mathrm{t}}}{\mathrm{t}(\mathrm{t}+1)(\mathrm{t}+2)...(\mathrm{t}+\mathrm{n})}\mathrm{dt}\mathrm{after}\mathrm{we}\mathrm{decompose}$$$$\mathrm{F}\left(\mathrm{t}\right)=\frac{{\mathrm{n}}^{\mathrm{t}}}{\mathrm{t}(\mathrm{t}+1)(\mathrm{t}+2)....(\mathrm{t}+\mathrm{n})}=\sum _{\mathrm{k}=0}^{\mathrm{n}}\frac{{\mathrm{a}}_{\mathrm{k}}}{\mathrm{t}+\mathrm{k}}\mathrm{be}\mathrm{continued}...$$$$$$

Answered by Sarah85 last updated on 07/Aug/20

$$\underset{0}{\stackrel{1}{\int }}\mathrm{\Gamma }(x+1)dx\approx .922745950259$$

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