Question and Answers Forum

All Questions      Topic List

Integration Questions

Previous in All Question      Next in All Question      

Previous in Integration      Next in Integration      

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 bymathmax 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)

$${t}={tan}^{\mathrm{2}} \theta\rightarrow{dt}=\mathrm{2}{tan}\theta{sec}^{\mathrm{2}} \theta{d}\theta\:\: \\ $$ $$\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \frac{\left({tan}^{\mathrm{2}} \theta\right)^{{a}−\mathrm{1}} ×\mathrm{2}{tan}\theta{sec}^{\mathrm{2}} \theta{d}\theta}{{sec}^{\mathrm{4}} \theta} \\ $$ $$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\theta\right)^{\mathrm{2}{a}−\mathrm{2}+\mathrm{1}} ×{cos}^{\mathrm{2}} \theta×\frac{\mathrm{1}}{\left({cos}\theta\right)^{\mathrm{2}{a}−\mathrm{2}+\mathrm{1}} }{d}\theta \\ $$ $$\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\theta\right)^{\mathrm{2}{a}−\mathrm{1}} ×\left({cos}\theta\right)^{\mathrm{2}−\mathrm{2}{a}+\mathrm{1}} {d}\theta \\ $$ $${formula}\:\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} \left({sin}\theta\right)^{\mathrm{2}{p}−\mathrm{1}} \left({cos}\theta\right)^{\mathrm{2}{q}−\mathrm{1}} {d}\theta \\ $$ $$=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)}\:{here}\:\mathrm{2}{p}−\mathrm{1}=\mathrm{2}{a}−\mathrm{1}\:\:{p}={a} \\ $$ $$\mathrm{2}{q}−\mathrm{1}=\mathrm{3}−\mathrm{2}{a} \\ $$ $$\mathrm{2}{q}=\mathrm{4}−\mathrm{2}{a}\rightarrow{q}=\mathrm{2}−{a}\leftarrow{look}\:{here} \\ $$ $${answer}\:=\frac{\lceil\left({p}\right)\lceil\left({q}\right)}{\lceil\left({p}+{q}\right)}=\frac{\lceil\left({a}\right)×\lceil\left(\mathrm{2}−{a}\right)}{\lceil\left({a}+\mathrm{2}−{a}\right)} \\ $$ $$=\frac{\lceil\left({a}\right)\lceil\left(\mathrm{2}−{a}\right)}{\mathrm{1}} \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$ $$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com