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Question Number 34277 by math khazana by abdo last updated on 03/May/18

calculate A_n =∫_0 ^∞      (dx/((x+1)(x+2)....(x+n)))  n integr≥2 .

$${calculate}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)....\left({x}+{n}\right)} \\ $$$${n}\:{integr}\geqslant\mathrm{2}\:. \\ $$

Commented by candre last updated on 03/May/18

(1/(Π_(i=1) ^n (x+i)))=Σ_(i=1) ^n (C_i /(x+i))  1=Σ_(j=1) ^n C_j Π_(i=1∧i≠j) ^n (x+i)  x=−j⇒1=C_j Π_(i=1∧i≠j) ^n (i−j)

$$\frac{\mathrm{1}}{\underset{{i}=\mathrm{1}} {\overset{{n}} {\prod}}\left({x}+{i}\right)}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{C}_{{i}} }{{x}+{i}} \\ $$$$\mathrm{1}=\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}{C}_{{j}} \underset{{i}=\mathrm{1}\wedge{i}\neq{j}} {\overset{{n}} {\prod}}\left({x}+{i}\right) \\ $$$${x}=−{j}\Rightarrow\mathrm{1}={C}_{{j}} \underset{{i}=\mathrm{1}\wedge{i}\neq{j}} {\overset{{n}} {\prod}}\left({i}−{j}\right) \\ $$

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