Question Number 67006 by mathmax by abdo last updated on 21/Aug/19
$${calculae}\:{A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:\:\:{with}\:{n}\:{integr}\:{natural}\:{and}\:{n}>\mathrm{0} \\ $$
Commented by mathmax by abdo last updated on 22/Aug/19
![A_n =∫_0 ^∞ (dx/((x^2 +1)^n )) changement x=tanθ give A_n =∫_0 ^(π/2) (((1+tan^2 θ))/((1+tan^2 θ)^n ))dθ =∫_0 ^(π/2) (cos^2 θ)^(n−1) dθ =∫_0 ^(π/2) cos^(2n−2) θ dθ (its a wallis integral) we have A_(n+1) =∫_0 ^(π/2) cos^(2n) θdθ =∫_0 ^(π/2) (1−sin^2 θ)cos^(2n−2) θ dθ = A_(n−1) −∫_0 ^(π/2) sin^2 θ cos^(2n−2) θ dθ by parts u=sinθ and v^′ =sinθ cos^(2n−2) θ ⇒ ∫_0 ^(π/2) sin^2 θ cos^(2n−2) θ dθ =[−(1/(2n−1))sinθ cos^(2n−1) θ]_0 ^(π/2) +(1/(2n−1))∫_0 ^(π/2) cosθ cos^(2n−1) θ dθ =(1/(2n−1)) A_(n+1) ⇒A_(n+1) =A_(n−1) −(1/(2n−1)) A_(n+1) ⇒ (1+(1/(2n−1)))A_(n+1) =A_n ⇒((2n)/(2n−1)) A_(n+1) =A_n ⇒A_(n+1) =((2n−1)/(2n)) A_n ⇒ Π_(k=1) ^(n−1) A_(k+1) =Π_(k=1) ^(n−1) ((2k−1)/(2k)) ×Π_(k=1) ^(n−1) A_k ⇒ A_2 .A_3 .....A_n =((Π_(k=1) ^(n−1) (2k−1))/(Π_(k=1) ^(n−1) (2k)))×A_1 .A_2 .....A_(n−1 ) ⇒ A_n =((1.3.5......(2n−3))/(2^(n−1) (n−1)!)) A_1 =((1.2.3.4.5....(2n−3)(2n−2))/(2^(n−1) (n−1)!2.4....2(n−1)))A_1 =(((2n−2)!)/((2^(n−1) )^2 ((n−1)!)^2 )) (π/2) =(((2n−2)!)/(2^(2n−1) {(n−1)!}^2 ))×π](https://www.tinkutara.com/question/Q67107.png)
$${A}_{{n}} =\int_{\mathrm{0}} ^{\infty} \:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} }\:\:{changement}\:{x}={tan}\theta\:{give}\: \\ $$$${A}_{{n}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\:\:\:\frac{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)}{\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{{n}} }{d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left({cos}^{\mathrm{2}} \theta\right)^{{n}−\mathrm{1}} {d}\theta\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{n}−\mathrm{2}} \theta\:{d}\theta \\ $$$$\left({its}\:{a}\:{wallis}\:{integral}\right)\:{we}\:{have}\:{A}_{{n}+\mathrm{1}} =\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}^{\mathrm{2}{n}} \theta{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:\left(\mathrm{1}−{sin}^{\mathrm{2}} \theta\right){cos}^{\mathrm{2}{n}−\mathrm{2}} \theta\:{d}\theta\:=\:{A}_{{n}−\mathrm{1}} −\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{sin}^{\mathrm{2}} \theta\:{cos}^{\mathrm{2}{n}−\mathrm{2}} \theta\:{d}\theta \\ $$$${by}\:{parts}\:{u}={sin}\theta\:\:{and}\:{v}^{'} \:={sin}\theta\:{cos}^{\mathrm{2}{n}−\mathrm{2}} \theta\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}} \theta\:{cos}^{\mathrm{2}{n}−\mathrm{2}} \theta\:{d}\theta\:=\left[−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}{sin}\theta\:{cos}^{\mathrm{2}{n}−\mathrm{1}} \theta\right]_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \:{cos}\theta\:{cos}^{\mathrm{2}{n}−\mathrm{1}} \theta\:{d}\theta \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:{A}_{{n}+\mathrm{1}} \:\Rightarrow{A}_{{n}+\mathrm{1}} ={A}_{{n}−\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\:{A}_{{n}+\mathrm{1}} \:\Rightarrow \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{1}}\right){A}_{{n}+\mathrm{1}} ={A}_{{n}} \:\Rightarrow\frac{\mathrm{2}{n}}{\mathrm{2}{n}−\mathrm{1}}\:{A}_{{n}+\mathrm{1}} ={A}_{{n}} \:\Rightarrow{A}_{{n}+\mathrm{1}} =\frac{\mathrm{2}{n}−\mathrm{1}}{\mathrm{2}{n}}\:{A}_{{n}} \\ $$$$\Rightarrow\:\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{A}_{{k}+\mathrm{1}} =\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\mathrm{2}{k}−\mathrm{1}}{\mathrm{2}{k}}\:×\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:{A}_{{k}} \:\Rightarrow \\ $$$${A}_{\mathrm{2}} .{A}_{\mathrm{3}} …..{A}_{{n}} =\frac{\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}−\mathrm{1}\right)}{\prod_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \left(\mathrm{2}{k}\right)}×{A}_{\mathrm{1}} .{A}_{\mathrm{2}} …..{A}_{{n}−\mathrm{1}\:\:} \Rightarrow \\ $$$${A}_{{n}} =\frac{\mathrm{1}.\mathrm{3}.\mathrm{5}……\left(\mathrm{2}{n}−\mathrm{3}\right)}{\mathrm{2}^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}\:{A}_{\mathrm{1}} \:=\frac{\mathrm{1}.\mathrm{2}.\mathrm{3}.\mathrm{4}.\mathrm{5}….\left(\mathrm{2}{n}−\mathrm{3}\right)\left(\mathrm{2}{n}−\mathrm{2}\right)}{\mathrm{2}^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!\mathrm{2}.\mathrm{4}….\mathrm{2}\left({n}−\mathrm{1}\right)}{A}_{\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\left(\mathrm{2}^{{n}−\mathrm{1}} \right)^{\mathrm{2}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }\:\frac{\pi}{\mathrm{2}}\:=\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \left\{\left({n}−\mathrm{1}\right)!\right\}^{\mathrm{2}} }×\pi \\ $$$$ \\ $$$$ \\ $$