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Question Number 62330 by maxmathsup by imad last updated on 19/Jun/19

$$findthevalueof{\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(1+t)}^2}dtwith0<a<1$$

Commented bymathmax by abdo last updated on 04/Jul/19

$$wehaveprovedthatB(x,y)={\int }_0^{\mathrm{\infty }}\frac{t^{x-1}}{{(1+t)}^{x+y}}dtlettakex=aandx+y=2\Rightarrow$$ $$x=aandy=2-a\Rightarrow$$ $${\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(1+t)}^2}dt=B(a,2-a)=\frac{\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(2-a)}{\mathrm{\Gamma }(a+2-a)}=\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(2-a)(\mathrm{\Gamma }\left(2\right)=1!=1)\Rightarrow$$ $${\int }_0^{\mathrm{\infty }}\frac{t^{a-1}}{{(1+t)}^2}dt=\mathrm{\Gamma }\left(a\right)\mathrm{\Gamma }(2-a)$$

Answered by tanmay last updated on 19/Jun/19

$$t={tan}^2\theta \to dt=2tan\theta {sec}^2\theta d\theta$$ $${\int }_0^{\frac{\pi }{2}}\frac{{\left({tan}^2\theta \right)}^{a-1}\times 2tan\theta {sec}^2\theta d\theta }{{sec}^4\theta }$$ $$2{\int }_0^{\frac{\pi }{2}}{\left(sin\theta \right)}^{2a-2+1}\times {cos}^2\theta \times \frac{1}{{\left(cos\theta \right)}^{2a-2+1}}d\theta$$ $$2{\int }_0^{\frac{\pi }{2}}{\left(sin\theta \right)}^{2a-1}\times {\left(cos\theta \right)}^{2-2a+1}d\theta$$ $$formula2{\int }_0^{\frac{\pi }{2}}{\left(sin\theta \right)}^{2p-1}{\left(cos\theta \right)}^{2q-1}d\theta$$ $$=\frac{\lceil \left(p\right)\lceil \left(q\right)}{\lceil (p+q)}here2p-1=2a-1p=a$$ $$2q-1=3-2a$$ $$2q=4-2a\to q=2-a\leftarrow lookhere$$ $$answer=\frac{\lceil \left(p\right)\lceil \left(q\right)}{\lceil (p+q)}=\frac{\lceil \left(a\right)\times \lceil (2-a)}{\lceil (a+2-a)}$$ $$=\frac{\lceil \left(a\right)\lceil (2-a)}{1}$$ $$$$ $$$$ $$$$ $$$$

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