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Question Number 62330 by maxmathsup by imad last updated on 19/Jun/19
find the value of ∫_0 ^∞ (t^(a−1) /((1+t)^2 ))dt with 0<a<1
$${find}\:{the}\:{value}\:{of}\:\int_{\mathrm{0}} ^{\infty} \:\:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}\:\:\:{with}\:\:\:\mathrm{0}<{a}<\mathrm{1} \\ $$
Commented by mathmax by abdo last updated on 04/Jul/19
we have proved that B(x,y) =∫_0 ^∞ (t^(x−1) /((1+t)^(x+y) )) dt let take x=a and x+y=2 ⇒ x=a and y =2−a ⇒ ∫_0 ^∞ (t^(a−1) /((1+t)^2 ))dt =B(a,2−a) =((Γ(a)Γ(2−a))/(Γ(a+2−a))) =Γ(a)Γ(2−a) (Γ(2)=1! =1) ⇒ ∫_0 ^∞ (t^(a−1) /((1+t)^2 ))dt =Γ(a)Γ(2−a)
$${we}\:{have}\:{proved}\:{that}\:{B}\left({x},{y}\right)\:=\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{x}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{{x}+{y}} }\:{dt}\:\:{let}\:{take}\:{x}={a}\:{and}\:{x}+{y}=\mathrm{2}\:\Rightarrow \\ $$$${x}={a}\:{and}\:{y}\:=\mathrm{2}−{a}\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}\:={B}\left({a},\mathrm{2}−{a}\right)\:\:\:\:=\frac{\Gamma\left({a}\right)\Gamma\left(\mathrm{2}−{a}\right)}{\Gamma\left({a}+\mathrm{2}−{a}\right)}\:=\Gamma\left({a}\right)\Gamma\left(\mathrm{2}−{a}\right)\:\:\:\:\left(\Gamma\left(\mathrm{2}\right)=\mathrm{1}!\:=\mathrm{1}\right)\:\Rightarrow \\ $$$$\int_{\mathrm{0}} ^{\infty} \:\:\:\frac{{t}^{{a}−\mathrm{1}} }{\left(\mathrm{1}+{t}\right)^{\mathrm{2}} }{dt}\:=\Gamma\left({a}\right)\Gamma\left(\mathrm{2}−{a}\right) \\ $$
Answered by tanmay last updated on 19/Jun/19
t=tan^2 θ→dt=2tanθsec^2 θdθ ∫_0 ^(π/2) (((tan^2 θ)^(a−1) ×2tanθsec^2 θdθ)/(sec^4 θ)) 2∫_0 ^(π/2) (sinθ)^(2a−2+1) ×cos^2 θ×(1/((cosθ)^(2a−2+1) ))dθ 2∫_0 ^(π/2) (sinθ)^(2a−1) ×(cosθ)^(2−2a+1) dθ formula 2∫_0 ^(π/2) (sinθ)^(2p−1) (cosθ)^(2q−1) dθ =((⌈(p)⌈(q))/(⌈(p+q))) here 2p−1=2a−1 p=a 2q−1=3−2a 2q=4−2a→q=2−a←look here answer =((⌈(p)⌈(q))/(⌈(p+q)))=((⌈(a)×⌈(2−a))/(⌈(a+2−a))) =((⌈(a)⌈(2−a))/1)
$${t}={tan}^{\mathrm{2}} \theta\rightarrow{dt}=\mathrm{2}{tan}\theta{sec}^{\mathrm{2}} \theta{d}\theta\:\: \\ $$$$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left({tan}^{\mathrm{2}} \theta\right)^{{a}−\mathrm{1}} ×\mathrm{2}{tan}\theta{sec}^{\mathrm{2}} \theta{d}\theta}{{sec}^{\mathrm{4}} \theta} \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\theta\right)^{\mathrm{2}{a}−\mathrm{2}+\mathrm{1}} ×{cos}^{\mathrm{2}} \theta×\frac{\mathrm{1}}{\left({cos}\theta\right)^{\mathrm{2}{a}−\mathrm{2}+\mathrm{1}} }{d}\theta \\ $$$$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\theta\right)^{\mathrm{2}{a}−\mathrm{1}} ×\left({cos}\theta\right)^{\mathrm{2}−\mathrm{2}{a}+\mathrm{1}} {d}\theta \\ $$$${formula}\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\theta\right)^{\mathrm{2}{p}−\mathrm{1}} \left({cos}\theta\right)^{\mathrm{2}{q}−\mathrm{1}} {d}\theta \\ $$$$=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)}\:{here}\:\mathrm{2}{p}−\mathrm{1}=\mathrm{2}{a}−\mathrm{1}\:\:{p}={a} \\ $$$$\mathrm{2}{q}−\mathrm{1}=\mathrm{3}−\mathrm{2}{a} \\ $$$$\mathrm{2}{q}=\mathrm{4}−\mathrm{2}{a}\rightarrow{q}=\mathrm{2}−{a}\leftarrow{look}\:{here} \\ $$$${answer}\:=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)}=\frac{\lceil\left({a}\right)×\lceil\left(\mathrm{2}−{a}\right)}{\lceil\left({a}+\mathrm{2}−{a}\right)} \\ $$$$=\frac{\lceil\left({a}\right)\lceil\left(\mathrm{2}−{a}\right)}{\mathrm{1}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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