In Im-n-10xm-11-xn-1dx- for m-n - 1 and -10xm-1-xn-11-xm-ndx- Im-n- - then - equals

I_m,n=∫^1_0x^m 1.(1 x)^n 1dx Put x=1y+1⇒ dx= 1(y+1)^2dy 1 x=yy+1 ∴ I_m,n=∫^0_∞y^n 1(y+1)^m+n( 1)dy =∫^∞_0y^y 1(y+1)^m+ndy...(1) Similarly, I_m,n=∫^1_0x^n 1.(1 x ...

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Byju's Answer

Standard XII

Mathematics

Finding Integral Part of Numbers of the Form a^b Where a Is Irrational

In Im,n=∫10xm...

Question

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

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Solution

$I_{m, n} = \int_{0}^{1} x^{m - 1} . (1 - x)^{n - 1} d x$
Put $x = \frac{1}{y + 1} \Rightarrow d x = \frac{- 1}{(y + 1)^{2}} d y$
$1 - x = \frac{y}{y + 1}$
$∴ I_{m, n} = \int_{\infty}^{0} \frac{y^{n - 1}}{(y + 1)^{m + n}} (- 1) d y$
$= \int_{0}^{\infty} \frac{y^{y - 1}}{(y + 1)^{m + n}} d y . . . (1)$
Similarly,
$I_{m, n} = \int_{0}^{1} x^{n - 1} . (1 - x)^{m - 1} d x$
$\Rightarrow I_{m, n} = \int_{0}^{\infty} \frac{y^{m - 1}}{(y + 1)^{m + n}} d y . . . (2)$
From (1) & (2)
$2 I_{m, n} = \int_{0}^{\infty} \frac{y^{m - 1} + y^{n - 1}}{(y + 1)^{m + n}} d y$
$\Rightarrow 2 I_{m, n} = \int_{0}^{1} \frac{y^{m - 1} + y^{n - 1}}{(y + 1)^{m + n}} d y I_{1} + \int_{1}^{\infty} \frac{y^{m - 1} + y^{n - 1}}{(y + 1)^{m + n}} d y I_{2}$

Put $y = \frac{1}{z}$ in $I_{2}$
$d y = - \frac{1}{z^{2}} d z$
$\Rightarrow 2 I_{m, n} = \int_{0}^{1} \frac{y^{m - 1} + y^{n - 1}}{(y + 1)^{m + n}} d y + \int_{1}^{0} \frac{z^{m - 1} + z^{n - 1}}{(z + 1)^{m + n}} (- d z)$
$\Rightarrow I_{m, n} = \int_{0}^{1} \frac{y^{m - 1} + y^{n - 1}}{(y + 1)^{m + n}} d \Rightarrow α = 1$

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In Im,n=∫10xm−1(1−x)n−1dx, for m,n≥1 and ∫10xm−1+xn−1(1+x)m+ndx=αIm,n, α∈R, then α equals

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