Question

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

• A. $I(m,n)=\underset{0}{\overset{\infty }{\int }}\frac{x^{m-1}}{(1+x{)}^{m+n-1}}dx$
• B. $I(m,n)=\underset{0}{\overset{\infty }{\int }}\frac{x^m}{(1+x{)}^{m+n}}dx$
• C. $I(m,n)=\underset{0}{\overset{\infty }{\int }}\frac{x^{n-1}}{(1+x{)}^{m+n}}dx$
• D. $I(m,n)=\underset{0}{\overset{\infty }{\int }}\frac{x^n}{(1+x{)}^{m+n}}dx$

Answer 1

Answer: (C)

Putting $x=\frac{1}{1+y},dx$
$=-\frac{1}{(1+y{)}^2}dy$, we get
$I(m,n)=\underset{0}{\overset{1}{\int }}{x}^{m-1}(1-x{)}^{n-1}dx$
$=\underset{\infty }{\overset{0}{\int }}\frac{1}{(1+y{)}^{m-1}}{\left(1-\frac{1}{1+y}\right)}^{n-1}\frac{(-1)}{(1+y{)}^2}dy$
$=\underset{0}{\overset{\infty }{\int }}\frac{y^{n-1}}{(1+y{)}^{m+n}}dy={\int }_0^{\infty }\frac{x^{n-1}}{(1+x{)}^{m+n}}dx$
Since $I(m,n)=I(n,m),I(m,n)$
$={\int }_0^{\infty }\frac{x^{m-1}}{(1+x{)}^{m+n}}dx=\underset{0}{\overset{\infty }{\int }}\frac{x^{n-1}}{(1+x{)}^{m+n}}dx$