Question

If $I_{m,n}=\underset{0}{\overset{1}{\int }}{x}^{m-1}(1-x{)}^{n-1}dx,$ for $m,n\ge 1$ and $\underset{0}{\overset{1}{\int }}\frac{x^{m-1}+x^{n-1}}{(1+x{)}^{m+n}}dx=\alpha I_{m,n},\alpha \in R,$ then $\alpha$ equals ____

Answer 1

Answer: 1

$I_{m,n}=\underset{0}{\overset{1}{\int }}{x}^{m-1}(1-x{)}^{n-1}dx=I_{n,m}$
Now Let $x=\frac{1}{y+1}\Rightarrow dx=-\frac{1}{(y+1{)}^2}dy$
So
$I_{m,n}=-\underset{\infty }{\overset{0}{\int }}\frac{1}{(y+1{)}^{m-1}}\frac{y^{n-1}}{(y+1{)}^{n-1}}\frac{dy}{(y+1{)}^2}=\underset{0}{\overset{\infty }{\int }}\frac{y^{n-1}}{(1+y{)}^{m+n}}\mathrm{dy}$
similarly $I_{m,n}=\underset{0}{\overset{\infty }{\int }}\frac{y^{m-1}}{(1+y{)}^{m+n}}dy$
Now $2I_{m,n}=\underset{0}{\overset{\infty }{\int }}\frac{y^{m-1}+y^{n-1}}{(1+y{)}^{m+n}}dy$
$=\underset{0}{\overset{\infty }{\int }}\frac{y^{m-1}+y^{n-1}}{(1+y{)}^{m+n}}dy$
$=\underset{0}{\overset{1}{\int }}\frac{y^{m-1}+y^{n-1}}{(1+y{)}^{m+n}}dy+\underset{\text{substitute }y=\frac{1}{t}}{\underbrace{\underset{1}{\overset{\infty }{\int }}\frac{y^{m-1}+y^{n-1}}{(1+y{)}^{m+n}}dy}}$
$\Rightarrow 2I_{m,n}=\underset{0}{\overset{1}{\int }}\frac{y^{m-1}+y^{n-1}}{(1+y{)}^{m+n}}dy-\underset{1}{\overset{0}{\int }}\frac{t^{n-1}+t^{m-1}}{t^{m+n-2}}\frac{t^{m+n}}{(1+t{)}^{m+n}}\frac{dt}{t^2}$
$\Rightarrow$ Hence $2I_{m,n}=2\underset{0}{\overset{1}{\int }}\frac{y^{m-1}+y^{n-1}}{(1+y{)}^{m+n}}dy\Rightarrow \alpha =1$