If I_{m, n} = int_{0}^{1} x^{m-1}(1-x)^{n-1} | vedclass

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(C) We are given $I_{m, n} = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx$. By substituting $x = \frac{1}{1+y}$,we have $dx = -\frac{1}{(1+y)^2} dy$. When $x=0,

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(C) We are given $I_{m, n} = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx$. By substituting $x = \frac{1}{1+y}$,we have $dx = -\frac{1}{(1+y)^2} dy$. When $x=0, y 	o \infty$ and when $x=1, y=0$. Thus,$I_{m, n} = \int_{\infty}^{0} (\frac{1}{1+y})^{m-1} (1 - \frac{1}{1+y})^{n-1} (-\frac{1}{(1+y)^2}) dy = \int_{0}^{\infty} \frac{y^{n-1}}{(1+y)^{m+n}} dy$. Similarly,$I_{m, n} = \int_{0}^{\infty} \frac{y^{m-1}}{(1+y)^{m+n}} dy$. Adding these,$2I_{m, n} = \int_{0}^{\infty} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy$. Splitting the integral at $y=1$: $2I_{m, n} = \int_{0}^{1} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy + \int_{1}^{\infty} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy$. In the second integral,substitute $y = \frac{1}{t}$,so $dy = -\frac{1}{t^2} dt$. $\int_{1}^{\infty} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy = \int_{1}^{0} \frac{(1/t)^{m-1} + (1/t)^{n-1}}{(1 + 1/t)^{m+n}} (-1/t^2) dt = \int_{0}^{1} \frac{t^{n-1} + t^{m-1}}{(t+1)^{m+n}} dt$. Thus,$2I_{m, n} = \int_{0}^{1} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy + \int_{0}^{1} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy = 2 \int_{0}^{1} \frac{y^{m-1} + y^{n-1}}{(1+y)^{m+n}} dy$. Therefore,$\int_{0}^{1} \frac{x^{m-1} + x^{n-1}}{(1+x)^{m+n}} dx = I_{m, n}$,which implies $\alpha = 1$.

If $I_{m, n} = \int_{0}^{1} x^{m-1}(1-x)^{n-1} dx$ for $m, n \geq 1$ and $\int_{0}^{1} \frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}} dx = \alpha I_{m, n}$,where $\alpha \in R$,then $\alpha$ equals .... .

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