Question Number 56932 by turbo msup by abdo last updated on 27/Mar/19
$${let}\:{f}_{{n}} \left({t}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+{t}^{\mathrm{2}} \right)^{{n}} } \\ $$$${with}\:{n}\:{from}\:{N}\:{and}\:{n}\geqslant\mathrm{1} \\ $$$$\mathrm{1}.\:{find}\:{a}\:{explicit}\:{form}\:{of}\:{f}_{{n}} \left({t}\right) \\ $$$$\mathrm{2}.\:{what}\:{is}\:{the}\:{value}\:{of} \\ $$$${g}_{{n}} \left({t}\right)=\int_{\mathrm{0}} ^{\infty} \:\frac{{t}\:{dx}}{\left({x}^{\mathrm{2}} +{t}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\:? \\ $$$$\mathrm{3}.\:{calculate}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{3}\right)^{\mathrm{4}} } \\ $$$${and}\:\int_{\mathrm{0}} ^{\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} \:+\mathrm{16}\right)^{\mathrm{3}} } \\ $$
Answered by Smail last updated on 27/Mar/19
![let x=t×tanθ⇒dx=((tdθ)/(cos^2 θ)) f_n (t)=∫_0 ^(π/2) (t/(t^(2n) cos^2 θ(1+tan^2 θ)^n ))dθ =t^(1−2n) ∫_0 ^(π/2) cos^(2(n−1)) θdθ let A_(n−1) =∫_0 ^(π/2) cos^(2(n−1)) θdθ =∫_0 ^(π/2) (cos^(2(n−2)) θ−sin^2 θcos^(2(n−2)) θ)dθ By parts u=sinθ⇒u′=cosθ v′=sinθcos^(2n−4) θ⇒v=−(1/(2n−3))cos^(2n−3) A_(n−1) =A_(n−2) −(1/(2n−3))A_(n−1) A_(n−1) =((2n−3)/(2n−2))A_(n−2) =(((2n−3)(2n−5))/((2n−2)(2n−4)))A_(n−3) =(((2(n−1)−1)(2(n−2)−1)(2(n−3)−1)...(2(n−(n−1))−1))/((2(n−1))(2(n−2))...(2(n−(n−1))))A_(n−(n−1)) =(((2n−3)(2n−5)(2n−7)...1)/(2^(n−1) (n−1)!))∫_0 ^(π/2) cos^2 θdθ =(((2n−2)(2n−3)(2n−4)(2n−5)(2n−6)...1)/(2^(n−1) (n−1)!×2^(n−1) (n−1)!))×(1/2)[θ+(1/2)sin2θ]_0 ^(π/2) =(((2n−2)!)/(2^(2(n−1)) ((n−1)!)^2 ))×(π/(2×2)) A_(n−1) =((π(2n−2)!)/(2^(2n) ((n−1)!)^2 )) f_n (t)=t^(1−2n) ×((π(2n−2)!)/(2^(2n) ((n−1)!)^2 )) f_n (t)=((π(2n−2)!)/(2^(2n) ((n−1)!)^2 ))t^(1−2n)](https://www.tinkutara.com/question/Q56990.png)
$${let}\:\:{x}={t}×{tan}\theta\Rightarrow{dx}=\frac{{td}\theta}{{cos}^{\mathrm{2}} \theta} \\ $$$${f}_{{n}} \left({t}\right)=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{t}}{{t}^{\mathrm{2}{n}} {cos}^{\mathrm{2}} \theta\left(\mathrm{1}+{tan}^{\mathrm{2}} \theta\right)^{{n}} }{d}\theta \\ $$$$={t}^{\mathrm{1}−\mathrm{2}{n}} \int_{\mathrm{0}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{2}\left({n}−\mathrm{1}\right)} \theta{d}\theta \\ $$$${let}\:\:{A}_{{n}−\mathrm{1}} =\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{2}\left({n}−\mathrm{1}\right)} \theta{d}\theta \\ $$$$=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \left({cos}^{\mathrm{2}\left({n}−\mathrm{2}\right)} \theta−{sin}^{\mathrm{2}} \theta{cos}^{\mathrm{2}\left({n}−\mathrm{2}\right)} \theta\right){d}\theta \\ $$$${By}\:{parts} \\ $$$${u}={sin}\theta\Rightarrow{u}'={cos}\theta \\ $$$${v}'={sin}\theta{cos}^{\mathrm{2}{n}−\mathrm{4}} \theta\Rightarrow{v}=−\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{3}}{cos}^{\mathrm{2}{n}−\mathrm{3}} \\ $$$${A}_{{n}−\mathrm{1}} ={A}_{{n}−\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{2}{n}−\mathrm{3}}{A}_{{n}−\mathrm{1}} \\ $$$${A}_{{n}−\mathrm{1}} =\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{2}{n}−\mathrm{2}}{A}_{{n}−\mathrm{2}} =\frac{\left(\mathrm{2}{n}−\mathrm{3}\right)\left(\mathrm{2}{n}−\mathrm{5}\right)}{\left(\mathrm{2}{n}−\mathrm{2}\right)\left(\mathrm{2}{n}−\mathrm{4}\right)}{A}_{{n}−\mathrm{3}} \\ $$$$=\frac{\left(\mathrm{2}\left({n}−\mathrm{1}\right)−\mathrm{1}\right)\left(\mathrm{2}\left({n}−\mathrm{2}\right)−\mathrm{1}\right)\left(\mathrm{2}\left({n}−\mathrm{3}\right)−\mathrm{1}\right)…\left(\mathrm{2}\left({n}−\left({n}−\mathrm{1}\right)\right)−\mathrm{1}\right)}{\left(\mathrm{2}\left({n}−\mathrm{1}\right)\right)\left(\mathrm{2}\left({n}−\mathrm{2}\right)\right)…\left(\mathrm{2}\left({n}−\left({n}−\mathrm{1}\right)\right)\right.}{A}_{{n}−\left({n}−\mathrm{1}\right)} \\ $$$$=\frac{\left(\mathrm{2}{n}−\mathrm{3}\right)\left(\mathrm{2}{n}−\mathrm{5}\right)\left(\mathrm{2}{n}−\mathrm{7}\right)…\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} {cos}^{\mathrm{2}} \theta{d}\theta \\ $$$$=\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)\left(\mathrm{2}{n}−\mathrm{3}\right)\left(\mathrm{2}{n}−\mathrm{4}\right)\left(\mathrm{2}{n}−\mathrm{5}\right)\left(\mathrm{2}{n}−\mathrm{6}\right)…\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!×\mathrm{2}^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)!}×\frac{\mathrm{1}}{\mathrm{2}}\left[\theta+\frac{\mathrm{1}}{\mathrm{2}}{sin}\mathrm{2}\theta\right]_{\mathrm{0}} ^{\pi/\mathrm{2}} \\ $$$$=\frac{\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}\left({n}−\mathrm{1}\right)} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }×\frac{\pi}{\mathrm{2}×\mathrm{2}} \\ $$$${A}_{{n}−\mathrm{1}} =\frac{\pi\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} } \\ $$$${f}_{{n}} \left({t}\right)={t}^{\mathrm{1}−\mathrm{2}{n}} ×\frac{\pi\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} } \\ $$$${f}_{{n}} \left({t}\right)=\frac{\pi\left(\mathrm{2}{n}−\mathrm{2}\right)!}{\mathrm{2}^{\mathrm{2}{n}} \left(\left({n}−\mathrm{1}\right)!\right)^{\mathrm{2}} }{t}^{\mathrm{1}−\mathrm{2}{n}} \\ $$