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calculate-A-n-0-dx-n-x-n-2-with-n-gt-1-




Question Number 66467 by mathmax by abdo last updated on 15/Aug/19
calculate A_n =∫_0 ^∞     (dx/((n+x^n )^2 ))   with n>1
$${calculate}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({n}+{x}^{{n}} \right)^{\mathrm{2}} }\:\:\:{with}\:{n}>\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 16/Aug/19
we have proved  ∫_0 ^∞  (dx/((a+x^n )^2 )) =(π/(nsin((π/n))))(1−(1/n))a^((1/n)−2)   a=n ⇒ ∫_0 ^∞    (dx/((n+x^n )^n )) =(π/(nsin((π/n))))(1−(1/n))n^((1/n)−2)   =((π(n−1))/(n^2 sin((π/n))))(^n (√n))×(1/n^2 ) =((π(n−1)(^n (√n)))/(n^4 sin((π/n))))   with  n>1
$${we}\:{have}\:{proved}\:\:\int_{\mathrm{0}} ^{\infty} \:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{2}} }\:=\frac{\pi}{{nsin}\left(\frac{\pi}{{n}}\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){a}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} \\ $$$${a}={n}\:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({n}+{x}^{{n}} \right)^{{n}} }\:=\frac{\pi}{{nsin}\left(\frac{\pi}{{n}}\right)}\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){n}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} \\ $$$$=\frac{\pi\left({n}−\mathrm{1}\right)}{{n}^{\mathrm{2}} {sin}\left(\frac{\pi}{{n}}\right)}\left(^{{n}} \sqrt{{n}}\right)×\frac{\mathrm{1}}{{n}^{\mathrm{2}} }\:=\frac{\pi\left({n}−\mathrm{1}\right)\left(^{{n}} \sqrt{{n}}\right)}{{n}^{\mathrm{4}} {sin}\left(\frac{\pi}{{n}}\right)}\:\:\:{with}\:\:{n}>\mathrm{1} \\ $$

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