Question Number 40159 by maxmathsup by imad last updated on 16/Jul/18
$${let}\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{{n}} } \\ $$$${find}\:{a}\:{relation}\:{etween}\:{I}_{{n}} \:{and}\:{I}_{{n}+\mathrm{1}} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{I}_{\mathrm{1}\:} \:{and}\:{I}_{\mathrm{2}} \\ $$
Commented by prof Abdo imad last updated on 17/Jul/18
![1) we have I_n = ∫_0 ^∞ ((1+x^3 )/((1+x^3 )^(n+1) ))dx =I_(n+1) + ∫_0 ^∞ (x^3 /((1+x^3 )^(n+1) ))dx by parts ∫_0 ^∞ (x^3 /((1+x^3 )^3 ))dx =(1/3)∫_0 ^∞ x(((3x^2 )/((1+x^3 )^(n+1) )))dx =(1/3)∫_0 ^∞ x ( 3x^2 (1+x^3 )^(−n−1) )dx =(1/3)[−(x/n)(1+x^3 )^(−n) ]_0 ^(+∞) −(1/3)∫_0 ^∞ 1 (−(1/n))(1/((1+x^3 )^n ))dx = (1/(3n)) I_n ⇒ I_n = I_(n+1) +(1/(3n)) I_n ⇒ (1−(1/(3n))) I_n = I_(n+1) ⇒ I_(n+1) =((3n−1)/(3n)) I_n 2) we have I_1 = ∫_0 ^∞ (dx/(1+x^3 )) =_(x^ =t^(1/3) ) ∫_0 ^∞ (1/(1+t)) (1/3)t^((1/3)−1) dt =(1/3) ∫_0 ^∞ (t^((1/3)−1) /(1+t))dt =(1/3) (π/(sin((π/3)))) =(π/(3.((√3)/2))) = ((2π)/(3(√3))) I_2 = (2/3) I_1 = (2/3) ((2π)/(3(√3))) = ((4π)/(9(√3))) .](Q40219.png)
$$\left.\mathrm{1}\right)\:{we}\:{have}\:\:{I}_{{n}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}+{x}^{\mathrm{3}} }{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} }{dx} \\ $$$$={I}_{{n}+\mathrm{1}} \:\:+\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} }{dx}\:\:{by}\:{parts} \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{x}^{\mathrm{3}} }{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{\mathrm{3}} }{dx}\:=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\:{x}\left(\frac{\mathrm{3}{x}^{\mathrm{2}} }{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{{n}+\mathrm{1}} }\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\:{x}\:\left(\:\mathrm{3}{x}^{\mathrm{2}} \:\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{−{n}−\mathrm{1}} \right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\left[−\frac{{x}}{{n}}\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{−{n}} \right]_{\mathrm{0}} ^{+\infty} \:−\frac{\mathrm{1}}{\mathrm{3}}\int_{\mathrm{0}} ^{\infty} \:\:\mathrm{1}\:\left(−\frac{\mathrm{1}}{{n}}\right)\frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{3}} \right)^{{n}} }{dx} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{3}{n}}\:\:{I}_{{n}} \:\Rightarrow\:{I}_{{n}} =\:{I}_{{n}+\mathrm{1}} \:\:\:+\frac{\mathrm{1}}{\mathrm{3}{n}}\:{I}_{{n}} \:\Rightarrow \\ $$$$\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{3}{n}}\right)\:{I}_{{n}} \:=\:{I}_{{n}+\mathrm{1}} \:\:\Rightarrow\:{I}_{{n}+\mathrm{1}} =\frac{\mathrm{3}{n}−\mathrm{1}}{\mathrm{3}{n}}\:{I}_{{n}} \\ $$$$\left.\mathrm{2}\right)\:{we}\:{have}\:{I}_{\mathrm{1}} =\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\mathrm{1}+{x}^{\mathrm{3}} } \\ $$$$=_{{x}^{} ={t}^{\frac{\mathrm{1}}{\mathrm{3}}} } \:\:\:\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{\mathrm{1}}{\mathrm{1}+{t}}\:\frac{\mathrm{1}}{\mathrm{3}}{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} {dt}\:=\frac{\mathrm{1}}{\mathrm{3}}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{t}^{\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}} }{\mathrm{1}+{t}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\:\frac{\pi}{{sin}\left(\frac{\pi}{\mathrm{3}}\right)}\:=\frac{\pi}{\mathrm{3}.\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\:=\:\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}} \\ $$$${I}_{\mathrm{2}} =\:\frac{\mathrm{2}}{\mathrm{3}}\:{I}_{\mathrm{1}} =\:\frac{\mathrm{2}}{\mathrm{3}}\:\frac{\mathrm{2}\pi}{\mathrm{3}\sqrt{\mathrm{3}}}\:=\:\frac{\mathrm{4}\pi}{\mathrm{9}\sqrt{\mathrm{3}}}\:\:. \\ $$