Question-168710

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

Commented by bekzodjumayev last updated on 16/Apr/22

$${Please}\:{help} \\ $$

Commented by mather last updated on 16/Apr/22

$${x}!=\mathrm{1}\: \\ $$over the interval of integration. So answer is 1.
!

Commented by alephzero last updated on 16/Apr/22

$${you}\:{are}\:{wrong}!\:{x}!\:{is}\:{can}\:{be} \\ $$$${defined}\:{as}\:\Gamma\left({x}+\mathrm{1}\right),\:{which}\:{is}\:{not} \\ $$$${equals}\:\mathrm{1}\:{over}\:{the}\:{interval}\:\left[\mathrm{1},\:\mathrm{2}\right]! \\ $$

Commented by MJS_new last updated on 16/Apr/22

$$\mathrm{I}\:\mathrm{learned}\:\mathrm{it}\:\mathrm{makes}\:\mathrm{no}\:\mathrm{sense}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{uphold} \\ $$$$\mathrm{precise}\:\mathrm{mathematical}\:\mathrm{language}. \\ $$$$\mathrm{BUT} \\ $$$${x}!\:\mathrm{is}\:\mathrm{only}\:\mathrm{defined}\:\mathrm{for}\:{x}\in\mathbb{N} \\ $$$$\Gamma\left({x}+\mathrm{1}\right)\:\mathrm{is}\:\mathrm{its}\:\mathrm{expansion}\:\mathrm{for}\:{x}\in\mathbb{R}\backslash\mathbb{Z}^{−} \:\left(\mathrm{and}\right. \\ $$$$\left.\mathrm{beyond}\right) \\ $$$$\Rightarrow \\ $$$$\int{x}!{dx}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{exist} \\ $$$$\mathrm{but}\:\mathrm{of}\:\mathrm{course}\:\mathrm{everybody}\:\mathrm{is}\:\mathrm{free}\:\mathrm{to}\:\mathrm{use} \\ $$$$\mathrm{wishy}−\mathrm{washy}\:\mathrm{language}\:\mathrm{instead} \\ $$

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