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Question Number 140703 by rs4089 last updated on 11/May/21

Answered by Dwaipayan Shikari last updated on 11/May/21

$$x=t+1$$$${\int }_1^{\mathrm{\infty }}\frac{dt}{{(t+1)}^{p+1}{t}^q}t=\frac{1}{g}$$$$={\int }_0^1{g}^{q-2}{(1+\frac{1}{g})}^{-p-1}dg={\int }_0^1{g}^{p+q-1}{(g+1)}^{-p-1}dg$$$${}_2{F}_1(a,b;c;z)=\frac{\mathrm{\Gamma }\left(c\right)}{\mathrm{\Gamma }(c-b)\mathrm{\Gamma }\left(b\right)}{\int }_0^1{x}^{b-1}{(1-x)}^{c-b-1}{(1-zx)}^{-a}dx$$$${}_2{F}_1(p+1,p+q;p+q+1,-1)=\frac{\mathrm{\Gamma }(p+q+1)}{\mathrm{\Gamma }\left(1\right)\mathrm{\Gamma }(p+q)}{\int }_0^1{x}^{p+q-1}{(x+1)}^{-p-1}dx$$$$So$$$$\frac{1}{p+q}{}_2{F}_1(p+1,p+q;p+q+1;-1)={\int }_0^1{g}^{p+q-1}{(g+1)}^{-p-1}dg$$

Answered by Mathspace last updated on 12/May/21

$$B(p,q)={\int }_0^{\mathrm{\infty }}{x}^{p-1}{(1-x)}^{q-1}dx$$$$=\frac{\mathrm{\Gamma }\left(p\right).\mathrm{\Gamma }\left(q\right)}{\mathrm{\Gamma }(p+q)}\Rightarrow$$$${\int }_0^{\mathrm{\infty }}\frac{dx}{x^{p+1}{(x-1)}^q}={\int }_0^{\mathrm{\infty }}{x}^{-p-1}{(-1)}^{-q}{(1-x)}^{-q}$$$$={(-1)}^{-q}{\int }_0^{\mathrm{\infty }}{x}^{-p-1}{(1-x)}^{1-q-1}$$$$={(-1)}^{-q}B(-p,1-q)$$$$={(-1)}^{-p}\frac{\mathrm{\Gamma }(-p).\mathrm{\Gamma }(1-q)}{\mathrm{\Gamma }(1-p-q)}$$

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