Question and Answers Forum

All Questions      Topic List

None Questions

Previous in All Question      Next in All Question      

Previous in None      Next in None      

Question Number 168710 by bekzodjumayev last updated on 16/Apr/22

Commented by bekzodjumayev last updated on 16/Apr/22

$$Pleasehelp$$

Commented by mather last updated on 16/Apr/22

$$x!=1$$over the interval of integration. So answer is 1. !

Commented by alephzero last updated on 16/Apr/22

$$youarewrong!x!iscanbe$$$$definedas\mathrm{\Gamma }(x+1),whichisnot$$$$equals1overtheinterval[1,2]!$$

Commented by MJS_new last updated on 16/Apr/22

$$\mathrm{I}\mathrm{learned}\mathrm{it}\mathrm{makes}\mathrm{no}\mathrm{sense}\mathrm{trying}\mathrm{to}\mathrm{uphold}$$$$\mathrm{precise}\mathrm{mathematical}\mathrm{language}.$$$$\mathrm{BUT}$$$$x!\mathrm{is}\mathrm{only}\mathrm{defined}\mathrm{for}x\in \mathbb{N}$$$$\mathrm{\Gamma }(x+1)\mathrm{is}\mathrm{its}\mathrm{expansion}\mathrm{for}x\in \mathbb{R}\mathrm{\setminus }{\mathbb{Z}}^{-}(\mathrm{and}$$$$\mathrm{beyond})$$$$\Rightarrow$$$$\int x!dx{\mathrm{doesn}}^{\prime }\mathrm{t}\mathrm{exist}$$$$\mathrm{but}\mathrm{of}\mathrm{course}\mathrm{everybody}\mathrm{is}\mathrm{free}\mathrm{to}\mathrm{use}$$$$\mathrm{wishy}-\mathrm{washy}\mathrm{language}\mathrm{instead}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com