If ${I_{m,n}} = \int\limits_0^1 {{x^{m - 1}}{{(1 - x)}^{n - 1}}dx}$, for m, $n \ge 1$, and
$$\int\limits_0^1 {{{{x^{m - 1}} + {x^{n - 1}}} \over {{{(1 + x)}^{m + 1}}}}} dx = \alpha {I_{m,n}}\alpha \in R$$, then $\alpha$ equals ___________.

${I_{m,n}} = \int\limits_0^1 {{x^{m - 1}}} .{(1 - x)^{n - 1}}dx$<br><br>Put $x = {1 \over {y + 1}} \Rightarrow dx = {{ - 1} \over {{{(y + 1)}^2}}}dy$<br><br>$1 - x = {y \over {y + 1}}$<br><br>$\therefore$ $${I_{m,n}} = \int\limits_\infty ^0 {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}( - 1)dy = } \int\limits_0^\infty {{{{y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $$ .... (i)<br><br>Similarly, ${I_{m,n}} = \int\limits_0^1 {{x^{n - 1}}.{{(1 - x)}^{m - 1}}dx}$<br><br>$$ \Rightarrow {I_{m,n}} = \int\limits_0^\infty {{{{y^{m - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $$ .... (ii)<br><br>From (i) &amp; (ii)<br><br>$$2{I_{m,n}} = \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} $$<br><br>$$ \Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_0^\infty {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} = {I_1} + {I_2}$$<br><br>Put $y = {1 \over z}$ in I₂<br><br>$dy = - {1 \over {{z^2}}}dz$<br><br>$$ \Rightarrow 2{I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} + \int\limits_1^0 {{{{z^{m + 1}} + {z^{n - 1}}} \over {{{(z + 1)}^{m + n}}}}( - dz)} $$<br><br>$$ \Rightarrow {I_{m,n}} = \int\limits_0^1 {{{{y^{m + 1}} + {y^{n - 1}}} \over {{{(y + 1)}^{m + n}}}}dy} \Rightarrow \alpha = 1$$