Question

In $I_{m,n}={\int }_1^0{x}^{m-1}{(1–x)}^{n-1}dx$ for $m,n\ge 1$ and ${\int }_1^0\frac{x^{m-1}+x^{n-1}}{{(1+x)}^{m+n}}dx=\alpha I_{m,n}$, $\alpha$ belongs to $R$, then $\alpha$ = ?

Answer 1

Step 1: Identifying the limits using the given Data :

$I_{m,n}={\int }_1^0{x}^{m-1}{(1–x)}^{n-1}dx....\left(i\right)$

${\int }_1^0\frac{x^{m-1}+x^{n-1}}{{(1+x)}^{m+n}}dx=\alpha I_{m,n}.....\left(ii\right)$

considering

$$\begin{gathered} x=\frac{1}{y+1} \\ \Rightarrow 1-x=\frac{y}{y+1} \\ \Rightarrow dx=\frac{-1}{{(y+1)}^2}dy[\mathrm{differentiating}\text{w.r.t.}x] \end{gathered}$$

When $x=1$ we get $y=0$and when $x=0$we get $y=\infty$

So limit ${\int }_0^1$changes to ${\int }_{\infty }^0$

$$\begin{gathered} \Rightarrow l_{m,n}={\int }_{\infty }^0\frac{y^{n-1}}{{(y+1)}^{m+n}}(-1)dy[\text{by substituting to equation}(i\left)\right] \\ ={\int }_0^{\infty }\frac{y^{n-1}}{{(y+1)}^{m+n}}dy.....\left(iii\right)\left[\therefore {\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx\right] \end{gathered}$$

Step 2: Repeating the same process :

For $I_{m,n}={\int }_1^0{x}^{n-1}{(1–x)}^{m-1}dx$

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

$$\begin{gathered} \Rightarrow 2l_{m,n}={\int }_0^{\infty }\frac{y^{m-1}+y^{n-1}}{{(y+1)}^{m+n}}dy \\ \Rightarrow 2l_{m,n}={\int }_1^{\infty }\frac{y^{m-1}+y^{n-1}}{{(y+1)}^{m+n}}dy+{\int }_0^1\frac{y^{m-1}+y^{n-1}}{{(y+1)}^{m+n}}dy\left[\therefore {\int }_a^{b}f\left(x\right)dx={\int }_c^{b}f\left(x\right)dx+{\int }_a^{c}f\left(x\right)dx\right] \end{gathered}$$

Step 3: Substituting $y=\frac{1}{z}$

$\Rightarrow dy=\frac{-1}{z^2}dz[\text{differentating}y\text{w.r.t}.z]$

When $y=1$ we get $z=1$and when $y=0$we get $z=\infty$

$$\begin{gathered} \Rightarrow 2l_{m,n}={\int }_0^1\frac{y^{m-1}+y^{n-1}}{{(y+1)}^{m+n}}dy+{\int }_1^0\frac{z^{m-1}+z^{n-1}}{{(z+1)}^{m+n}}(-1)dz \\ \Rightarrow 2l_{m,n}={\int }_0^1\frac{y^{m-1}+y^{n-1}}{{(y+1)}^{m+n}}dy+{\int }_0^1\frac{z^{m-1}+z^{n-1}}{{(z+1)}^{m+n}}dz...\left(v\right)\left[\therefore {\int }_a^{b}f\left(x\right)dx=-{\int }_b^{a}f\left(x\right)dx\right] \end{gathered}$$

Again substituting $z=y$and the limit will remain the same

$$\begin{gathered} \Rightarrow l_{m,n}={\int }_0^1\frac{y^{m-1}+y^{n-1}}{{(y+1)}^{m+n}}dy \\ \Rightarrow \alpha =1 \end{gathered}$$

Hence, $\alpha =1$ is the answer.