# in

Question:

In $I_{m, n}=\int_{0}^{1} x^{m-1}(1-x)^{n-1} d x$, for $\mathrm{m}, \mathrm{n} \geq 1$ and $\int_{0}^{1} \frac{\mathrm{x}^{\prime \prime-1}+\mathrm{x}^{n-1}}{(1+\mathrm{x})^{m+n}} \mathrm{dx}=\alpha \mathrm{I}_{\mathrm{m}, \mathrm{n}}, \alpha \in \mathrm{R}$, then $\alpha$ equals

Solution:

$I_{m, n}=\int_{0}^{1} x^{m-1} \cdot(1-x)^{n-1} d x$

Put $x=\frac{1}{y+1} \Rightarrow d x=\frac{-1}{(y+1)^{2}} d y$

$1-x=\frac{y}{y+1}$

$\therefore I_{m, n}=\int_{\infty}^{0} \frac{y^{n-1}}{(y+1)^{m+\infty}}(-1) d y=\int_{0}^{\infty} \frac{y^{0-1}}{(y+1)^{n+n}} d y \ldots$ (i)

Similarly $I_{m, n}=\int_{0}^{1} x^{n-1} \cdot(1-x)^{m-1} d x$

$\Rightarrow I_{m, n}=\int_{0}^{\infty} \frac{y^{m-1}}{(y+1)^{m+n}} d y .$

From (i) $\backslash$ (ii)

$2 \mathrm{I}_{\mathrm{m}, \mathrm{n}}=\int_{0}^{\infty} \frac{\mathrm{y}^{\mathrm{n}-1}+\mathrm{y}^{\mathrm{n}-1}}{(\mathrm{y}+1)^{\mathrm{m}+\mathrm{a}}} \mathrm{dy}$

$\Rightarrow 2 I_{m, n}=\int_{0}^{1} \frac{y^{m-1}+y^{x-1}}{(y+1)^{*+\infty}} d y+\int_{1}^{\infty} \frac{y^{m-1}+y^{x-1}}{(y+1)^{m+n}} d y$

Put $y=\frac{1}{z}$ in $I_{2}$

$d y=-\frac{1}{z^{2}} d z$

$\Rightarrow 2 \mathrm{I}_{\mathrm{m}, \mathrm{n}}=\int_{0}^{1} \frac{\mathrm{y}^{\mathrm{m}-1}+\mathrm{y}^{\mathrm{n}-1}}{(\mathrm{y}+1)^{\mathrm{mt}+\mathrm{m}}} \mathrm{dy}+\int_{1}^{0} \frac{z^{\mathrm{m}-1}+\mathrm{z}^{\mathrm{n}-1}}{(\mathrm{z}+1)^{\mathrm{v}+\cdots}}(-\mathrm{dz})$

$\Rightarrow I_{m, n}=\int_{0}^{1} \frac{y^{m-1}+y^{n-1}}{(y+1)^{m+n}} d y \Rightarrow \alpha=1$