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Question Number 165757 by Zaynal last updated on 07/Feb/22

Answered by Mathspace last updated on 08/Feb/22

p(x)=Π_(n=1) ^(100) (x+n) ⇒((p^′ (x))/(p(x))) =Σ_(n=1) ^(100) (1/(x+n)) ⇒ ∫_0 ^1 {Π_(n=1) ^(100) (x+n)(Σ_(n=1) ^(100 ) (1/(x+n)))dx =∫_0 ^1 p(x).((p^′ (x))/(p(x)))dx=∫_0 ^1 p^′ (x)dx =p(1)−p(0) =Π_(n=1) ^(100) (n+1)−Π_(n=1) ^(100) n =Π_(n=2) ^(101) n−100! =101!−100! =101.100!−100! =100×(100)!

$${p}\left({x}\right)=\prod_{{n}=\mathrm{1}} ^{\mathrm{100}} \left({x}+{n}\right)\:\Rightarrow\frac{{p}^{'} \left({x}\right)}{{p}\left({x}\right)} \\ $$$$=\sum_{{n}=\mathrm{1}} ^{\mathrm{100}} \frac{\mathrm{1}}{{x}+{n}}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \left\{\prod_{{n}=\mathrm{1}} ^{\mathrm{100}} \left({x}+{n}\right)\left(\sum_{{n}=\mathrm{1}} ^{\mathrm{100}\:} \frac{\mathrm{1}}{{x}+{n}}\right){dx}\right. \\ $$$$=\int_{\mathrm{0}} ^{\mathrm{1}} {p}\left({x}\right).\frac{{p}^{'} \left({x}\right)}{{p}\left({x}\right)}{dx}=\int_{\mathrm{0}} ^{\mathrm{1}} {p}^{'} \left({x}\right){dx} \\ $$$$={p}\left(\mathrm{1}\right)−{p}\left(\mathrm{0}\right) \\ $$$$=\prod_{{n}=\mathrm{1}} ^{\mathrm{100}} \left({n}+\mathrm{1}\right)−\prod_{{n}=\mathrm{1}} ^{\mathrm{100}} {n} \\ $$$$=\prod_{{n}=\mathrm{2}} ^{\mathrm{101}} {n}−\mathrm{100}! \\ $$$$=\mathrm{101}!−\mathrm{100}! \\ $$$$=\mathrm{101}.\mathrm{100}!−\mathrm{100}! \\ $$$$=\mathrm{100}×\left(\mathrm{100}\right)! \\ $$

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