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Question Number 66468 by mathmax by abdo last updated on 15/Aug/19

calculate I_n = ∫_0 ^∞ (dx/((x^n +3)^2 )) with n>1

$${calculate}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\:\frac{{dx}}{\left({x}^{{n}} \:+\mathrm{3}\right)^{\mathrm{2}} }\:\:{with}\:{n}>\mathrm{1} \\ $$

Commented bymathmax by abdo last updated on 16/Aug/19

let f(a) =∫_0 ^∞ (dx/(a +x^n )) with a>0 ⇒f(a) =(1/a)∫_0 ^∞ (dx/((1+(x^n /a)))) let use the changement (x^n /a) =u^n ⇒x^n =a u^n ⇒x=a^(1/n) u ⇒ f(a) =(1/a)∫_0 ^∞ (1/(1+u^n ))a^(1/n) du =a^((1/n)−1) ∫_0 ^∞ (du/(1+u^n )) changement u=α^(1/n) give ∫_0 ^∞ (1/(1+α))(1/n)α^((1/n)−1) dα =(1/n) ∫_0 ^∞ (α^((1/n)−1) /(1+α))dα =(1/n) (π/(sin((π/n)))) ⇒ f(a) =a^((1/n)−1) ×(π/(nsin((π/n)))) =((π a^((1/n)−1) )/(nsin((π/n)))) and we have f^′ (a) =−∫_0 ^∞ (dx/((a+x^n )^2 )) ⇒∫_0 ^∞ (dx/((a+x^n )^2 )) =−f^′ (a) f^′ (a) =((π((1/n)−1)a^((1/n)−2) )/(nsin((π/n)))) ⇒∫_0 ^∞ (dx/((a+x^n )^2 )) =((π(1−(1/n))a^((1/n)−2) )/(nsin((π/n)))) a=3 ⇒ ∫_0 ^∞ (dx/((3+x^n )^2 )) =((π(1−(1/n))3^((1/n)−2) )/(nsin((π/n)))) =I_n

$${let}\:{f}\left({a}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{{a}\:+{x}^{{n}} }\:\:\:{with}\:{a}>\mathrm{0}\:\Rightarrow{f}\left({a}\right)\:=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left(\mathrm{1}+\frac{{x}^{{n}} }{{a}}\right)} \\ $$ $${let}\:{use}\:{the}\:{changement}\:\:\frac{{x}^{{n}} }{{a}}\:={u}^{{n}} \:\Rightarrow{x}^{{n}} \:={a}\:{u}^{{n}} \:\Rightarrow{x}={a}^{\frac{\mathrm{1}}{{n}}} \:{u}\:\Rightarrow \\ $$ $${f}\left({a}\right)\:=\frac{\mathrm{1}}{{a}}\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{\mathrm{1}+{u}^{{n}} }{a}^{\frac{\mathrm{1}}{{n}}} \:{du}\:={a}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{du}}{\mathrm{1}+{u}^{{n}} }\:{changement}\:{u}=\alpha^{\frac{\mathrm{1}}{{n}}} \\ $$ $${give}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\mathrm{1}}{\mathrm{1}+\alpha}\frac{\mathrm{1}}{{n}}\alpha^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} \:{d}\alpha\:=\frac{\mathrm{1}}{{n}}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{\alpha^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{\mathrm{1}+\alpha}{d}\alpha\:=\frac{\mathrm{1}}{{n}}\:\frac{\pi}{{sin}\left(\frac{\pi}{{n}}\right)}\:\Rightarrow \\ $$ $${f}\left({a}\right)\:={a}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} ×\frac{\pi}{{nsin}\left(\frac{\pi}{{n}}\right)}\:=\frac{\pi\:{a}^{\frac{\mathrm{1}}{{n}}−\mathrm{1}} }{{nsin}\left(\frac{\pi}{{n}}\right)}\:\:{and}\:{we}\:{have} \\ $$ $${f}^{'} \left({a}\right)\:=−\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{2}} }\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{2}} }\:=−{f}^{'} \left({a}\right) \\ $$ $${f}^{'} \left({a}\right)\:=\frac{\pi\left(\frac{\mathrm{1}}{{n}}−\mathrm{1}\right){a}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} }{{nsin}\left(\frac{\pi}{{n}}\right)}\:\Rightarrow\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({a}+{x}^{{n}} \right)^{\mathrm{2}} }\:=\frac{\pi\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right){a}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} }{{nsin}\left(\frac{\pi}{{n}}\right)} \\ $$ $${a}=\mathrm{3}\:\Rightarrow\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left(\mathrm{3}+{x}^{{n}} \right)^{\mathrm{2}} }\:=\frac{\pi\left(\mathrm{1}−\frac{\mathrm{1}}{{n}}\right)\mathrm{3}^{\frac{\mathrm{1}}{{n}}−\mathrm{2}} }{{nsin}\left(\frac{\pi}{{n}}\right)}\:={I}_{{n}} \\ $$ $$ \\ $$

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