If I m , n = ∫ 0 1 x m - 1 1 - x n - 1 d x , for m , n ⩾ 1 , and ∫ 0 1 x m - 1 + x n - 1…

$I_{m,n}={\int }_0^1{x}^{m-1}{\left(1-x\right)}^{n-1}dx=I_{n,m}$

Now Let $x=\frac{1}{y+1}\Rightarrow dx=-\frac{1}{{\left(y+1\right)}^2}dy$

So, $I_{m,n}=-{\int }_{\infty }^0\frac{1}{{\left(y+1\right)}^{m-1}}\frac{y^{n-1}}{{\left(y+1\right)}^{n-1}}\frac{dy}{{\left(y+1\right)}^2}={\int }_0^{\infty }\frac{y^{n-1}}{{\left(1+y\right)}^{m+n}}dy$

similarly $I_{m,n}={\int }_0^{\infty }\frac{y^{m-1}}{{\left(1+y\right)}^{m+n}}dy$

Now $2I_{m,n}={\int }_0^{\infty }\frac{y^{m-1}+y^{n-1}}{{\left(1+y\right)}^{m+n}}dy$

$={\int }_0^{\infty }\frac{y^{m-1}+y^{n-1}}{{\left(1+y\right)}^{m+n}}dy$

$={\int }_0^1\frac{y^{m-1}+y^{n-1}}{{\left(1+y\right)}^{m+n}}dy+\underset{\mathrm{substitute} y=\frac{1}{t}}{\underbrace{{\int }_1^{\infty }\frac{y^{m-1}+y^{n-1}}{{\left(1+y\right)}^{m+n}}dy}}$

$\Rightarrow 2I_{m,n}={\int }_0^1\frac{y^{m-1}+y^{n-1}}{{\left(1+y\right)}^{m+n}}dy-{\int }_1^0\frac{t^{n-1}+t^{m-1}}{t^{m+n-2}}\frac{t^{m+n}}{{\left(1+t\right)}^{m+n}}\frac{dt}{t^2}$

$\Rightarrow$ Hence $2I_{m,n}=2{\int }_0^1\frac{y^{m-1}+y^{n-1}}{{\left(1+y\right)}^{m+n}}dy\Rightarrow \alpha =1$