Question

If $I(m,n)={\int }_0^1{x}^{(m-1)}(1-x{)}^{n-1}dx,(m,nϵI,m,n\ge 0)$, then

• A. $I(m,n)={\int }_0^{\infty }\frac{x^{m-1}}{(1+x{)}^{m-n}}dx$
• B. $I(m,n)={\int }_0^{\infty }\frac{x^m}{(1+x{)}^{m+n}}dx$
• C. $I(m,n)={\int }_0^{\infty }\frac{x^{n-1}}{(1+x{)}^{m+n}}dx$
• D. $I(m,n)={\int }_0^{\infty }\frac{x^n}{(1+x{)}^{m+n}}dx$

Answer 1

The correct option is C $I(m,n)={\int }_0^{\infty }\frac{x^{n-1}}{(1+x{)}^{m+n}}dx$

Given , $I_{m,n}={\int }_0^1{x}^{m-1}(1-x{)}^{n-1}dx$
Substitute $x=\frac{1}{1+y}$
$\Rightarrow dx=-\frac{1}{(1+y{)}^2}dy$
$I_{m,n}$
$={\int }_{\infty }^0\frac{1}{(1+y{)}^{m-1}}{(1-\frac{1}{1+y})}^{n-1}\frac{(-1)}{(1+y{)}^2}dy$
$={\int }_0^{\infty }\frac{y^{n-1}}{(1+y{)}^{m+n}}dy$
$={\int }_0^{\infty }\frac{x^{n-1}}{(1+x{)}^{m+n}}dx$
Hence, option C is correct