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Question Number 127779 by Bird last updated on 02/Jan/21
find A_n = ∫_0 ^(+∞) (dx/((x^2 +1)^n ))
$${find}\:{A}_{{n}} =\:\int_{\mathrm{0}} ^{+\infty} \frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}} } \\ $$
Answered by Dwaipayan Shikari last updated on 02/Jan/21
∫_0 ^∞ (1/((1+x^2 )^n ))dx x^2 =u⇒2x=(du/dx) =(1/2)∫_0 ^∞ (u^(−(1/2)) /((1+u)^n ))du (u/(1+u))=t⇒(1/((1+u)^2 ))=(dt/du) =(1/2)∫_0 ^1 ((t/(1−t)))^(−(1/2)) (1+u)^(2−n) dt=(1/2)∫_0 ^1 t^(−(1/2)) (1−t)^(1/2) (1−t)^(n−2) dt =(1/2)β((1/2),n−(1/2))=(1/2).((Γ((1/2))Γ(n−(1/2)))/(Γ(n)))=((√π)/2).((Γ(n−(1/2)))/(Γ(n)))
$$\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{1}}{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}} }{dx}\:\:\:\:\:\:\:{x}^{\mathrm{2}} ={u}\Rightarrow\mathrm{2}{x}=\frac{{du}}{{dx}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\left(\mathrm{1}+{u}\right)^{{n}} }{du}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\frac{{u}}{\mathrm{1}+{u}}={t}\Rightarrow\frac{\mathrm{1}}{\left(\mathrm{1}+{u}\right)^{\mathrm{2}} }=\frac{{dt}}{{du}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \left(\frac{{t}}{\mathrm{1}−{t}}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}+{u}\right)^{\mathrm{2}−{n}} {dt}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} {t}^{−\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{t}\right)^{\frac{\mathrm{1}}{\mathrm{2}}} \left(\mathrm{1}−{t}\right)^{{n}−\mathrm{2}} {dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{2}},{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\mathrm{1}}{\mathrm{2}}.\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}\right)}=\frac{\sqrt{\pi}}{\mathrm{2}}.\frac{\Gamma\left({n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left({n}\right)} \\ $$
Answered by Ar Brandon last updated on 02/Jan/21
A_n =∫_0 ^∞ (dx/((x^2 +1)^n )) x^2 =u ⇒2xdx=du A_n =(1/2)∫_0 ^∞ (u^(−(1/2)) /( (1+u)^n ))du=(1/2)β((1/2),n−(1/2)) =(1/2)∙((Γ((1/2))Γ(n−(1/2)))/(Γ(n)))=((√π)/(2(n−1)!))Γ(n−(1/2)) Γ((n−1)+(1/2))=((√π)/2^(2n−3) )∙((Γ(2n−2))/(Γ(n−1)))=((√π)/2^(2n−3) )∙(((2n−3)!)/((n−2)!)) A_n =((√π)/(2(n−1)!))∙((√π)/2^(2n−3) )∙(((2n−3)!)/((n−2)!))=((π(2n−3)!)/(2^(2n−2) (n−1)!(n−2)!))
$$\mathrm{A}_{\mathrm{n}} =\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{dx}}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{n}} } \\ $$$$\mathrm{x}^{\mathrm{2}} =\mathrm{u}\:\Rightarrow\mathrm{2xdx}=\mathrm{du} \\ $$$$\mathrm{A}_{\mathrm{n}} =\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\infty} \frac{\mathrm{u}^{−\frac{\mathrm{1}}{\mathrm{2}}} }{\:\left(\mathrm{1}+\mathrm{u}\right)^{\mathrm{n}} }\mathrm{du}=\frac{\mathrm{1}}{\mathrm{2}}\beta\left(\frac{\mathrm{1}}{\mathrm{2}},\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\centerdot\frac{\Gamma\left(\frac{\mathrm{1}}{\mathrm{2}}\right)\Gamma\left(\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right)}{\Gamma\left(\mathrm{n}\right)}=\frac{\sqrt{\pi}}{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)!}\Gamma\left(\mathrm{n}−\frac{\mathrm{1}}{\mathrm{2}}\right) \\ $$$$\Gamma\left(\left(\mathrm{n}−\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}\right)=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2n}−\mathrm{3}} }\centerdot\frac{\Gamma\left(\mathrm{2n}−\mathrm{2}\right)}{\Gamma\left(\mathrm{n}−\mathrm{1}\right)}=\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2n}−\mathrm{3}} }\centerdot\frac{\left(\mathrm{2n}−\mathrm{3}\right)!}{\left(\mathrm{n}−\mathrm{2}\right)!} \\ $$$$\mathrm{A}_{\mathrm{n}} =\frac{\sqrt{\pi}}{\mathrm{2}\left(\mathrm{n}−\mathrm{1}\right)!}\centerdot\frac{\sqrt{\pi}}{\mathrm{2}^{\mathrm{2n}−\mathrm{3}} }\centerdot\frac{\left(\mathrm{2n}−\mathrm{3}\right)!}{\left(\mathrm{n}−\mathrm{2}\right)!}=\frac{\pi\left(\mathrm{2n}−\mathrm{3}\right)!}{\mathrm{2}^{\mathrm{2n}−\mathrm{2}} \left(\mathrm{n}−\mathrm{1}\right)!\left(\mathrm{n}−\mathrm{2}\right)!} \\ $$

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