Question Number 66467 by mathmax by abdo last updated on 15/Aug/19
$${calculate}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({n}+{x}^{{n}} \right)^{\mathrm{2}} }\:\:\:{with}\:{n}>\mathrm{1} \\ $$
Commented bymathmax by abdo last updated on 16/Aug/19
$${we}\:{have}\:{proved}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{2}} }\:=\frac{\pi}{{nsin}\left(\frac{\pi}{{n}}\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){a}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} \\ $$ $${a}={n}\:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({n}+{x}^{{n}} \right)^{{n}} }\:=\frac{\pi}{{nsin}\left(\frac{\pi}{{n}}\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){n}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} \\ $$ $$=\frac{\pi\left({n}−\mathrm{1}\right)}{{n}^{\mathrm{2}} {sin}\left(\frac{\pi}{{n}}\right)}\left(^{{n}} \sqrt{{n}}\right)×\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi\left({n}−\mathrm{1}\right)\left(^{{n}} \sqrt{{n}}\right)}{{n}^{\mathrm{4}} {sin}\left(\frac{\pi}{{n}}\right)}\:\:\:{with}\:\:{n}>\mathrm{1} \\ $$