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

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Definite Integrals

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

, for

m, n \geq 1

, and

\int_{0}^{1} \frac{x ^{m - 1} + x ^{n - 1}}{( 1 + x ) ^{m + n}} d x = α I_{m, n}, α \in R

, then

α

equals.....

Answer: 1

The Beta integral has the standard alternative form

\int_{0}^{\infty} \frac{x ^{m - 1}}{( 1 + x ) ^{m + n}} d x = B (m, n) = I_{m, n}

Split it at

x = 1

and substitute

x = \frac{1}{t}

in the part over

[1, \infty)

, where

d x = - \frac{d t}{t ^{2}}

and the limits

x = 1

x \to \infty

become

t = 1

t \to 0

\int_{1}^{\infty} \frac{x ^{m - 1}}{( 1 + x ) ^{m + n}} d x = \int_{0}^{1} t^{1 - m} \cdot \frac{t ^{m + n}}{( 1 + t ) ^{m + n}} \cdot \frac{d t}{t ^{2}} = \int_{0}^{1} \frac{t ^{n - 1}}{( 1 + t ) ^{m + n}} d t

Adding the two pieces,

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

α = 1

JEE Main · 2021 · 26 Feb 2021 · Shift 2 · Integer answer