Question Number 37600 by prof Abdo imad last updated on 15/Jun/18
$${n}\:{integr}\:{natural}\:{calculate} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)……\left({x}+{n}\right)} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 16/Jun/18
$$\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)…\left({x}+{n}\right)}={e}^{{t}} \\ $$$$−\left\{{ln}\left({x}+\mathrm{1}\right)+{ln}\left({x}+\mathrm{2}\right)+{ln}\left({x}+\mathrm{3}\right)..+{ln}\left({x}+{n}\right)=\right. \\ $$$$\:\:\:{t} \\ $$$${dt}=−\left\{\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{{x}+\mathrm{2}}+…+\frac{\mathrm{1}}{{x}+{n}}\right\}{dx} \\ $$$$=−\left\{\frac{\frac{{e}^{{t}} }{{x}+\mathrm{1}}+\frac{{e}^{{t}} }{{x}+\mathrm{2}}+…+\frac{{e}^{{t}} }{{x}+{n}}}{\left({x}+\mathrm{1}\right)\left({x}+\mathrm{2}\right)\left({x}+\mathrm{3}\right)..\left({x}+{n}\right)}\right\}{contd} \\ $$
Commented by math khazana by abdo last updated on 17/Jun/18
$${you}\:{are}\:{really}\:{a}\:{rail}\:{ways}… \\ $$
Answered by ajfour last updated on 17/Jun/18
$${I}=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dx}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\left({n}−\mathrm{2}\right)!}\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dx}}{{x}+\mathrm{2}} \\ $$$$−\frac{\mathrm{1}}{\mathrm{2}!\left({n}−\mathrm{3}\right)!}\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dx}}{{x}+\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}!\left({n}−\mathrm{4}\right)!}\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dx}}{{x}+\mathrm{4}} \\ $$$$…..−\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}−\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\:\:\infty} \frac{{dx}}{{x}+{n}} \\ $$$$\:\:\:=\underset{{x}\rightarrow\infty} {\mathrm{lim}}\Sigma\frac{\left(−\mathrm{1}\right)^{{r}} }{\left({r}−\mathrm{1}\right)!\left({n}−{r}\right)!}\mathrm{ln}\:\left(\mathrm{1}+\frac{{x}}{{r}}\right) \\ $$$$\:\:\:=\frac{\mathrm{1}}{\left({n}−\mathrm{1}\right)!}\Sigma\left(−\mathrm{1}\right)^{{r}} \underset{{x}\rightarrow\infty} {\mathrm{lim}^{{n}−\mathrm{1}} }{C}_{{r}−\mathrm{1}} \mathrm{ln}\:\left(\mathrm{1}+\frac{{x}}{{r}}\right) \\ $$$$….. \\ $$