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If I_{m,n}=\int_{0}^{1}x^{m-1}(1-x)^{n-1}dx, for m,n \geqslant 1, and \int_{0}^{1}\frac{x^{m-1}+x^{n-1}}{(1+x)^{m+n}}dx = \alpha I_{m,n}\alpha \euro R, then \alpha equals _______
Option: 1 1
Option: 2 2
Option: 3 3
Option: 4 4

Answers (1)

best_answer

\\\mathrm{I}_{\mathrm{m}, \mathrm{n}}=\int_{0}^{1} \mathrm{x}^{\mathrm{m}-1}(1-\mathrm{x})^{\mathrm{n}-1} \mathrm{~d} \mathrm{x}=\mathrm{I}_{\mathrm{n}, \mathrm{m}}\\

\\\text{Let }x=\frac{1}{y+1} \Rightarrow d x=-\frac{1}{(y+1)^{2}} d y\\\mathrm{I_{m,n}}=-\int^{0}_{\infty}\frac{1}{(y+1)^{m-1}}\frac{y^{n-1}}{(y+1)^{n-1}} \frac{d y}{(y+1)^{2}}\\\mathrm{I_{m,n}}=\int_{0}^{\infty} \frac{y^{n-1}}{(1+y)^{m+n}} d y

similarly,

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

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

=\int_{0}^{1} \frac{y^{m-1}+y^{n-1}}{(1+y)^{m+n}} d y+\int_{1}^{\infty} \frac{y^{m-1}+y^{n-1}}{(1+y)^{m+n}} d y

First, solve

\\\mathrm{I_1}=\int_{1}^{\infty} \frac{y^{m-1}+y^{n-1}}{(1+y)^{m+n}} d y\\\text{put y = }\frac{1}{t}\Rightarrow dy=-\frac{1}{t^2}\\\mathrm{I_1}=\int^0_1\frac{\left (\frac{1}{t} \right )^{m-1}+\left (\frac{1}{t} \right )^{n-1}}{\left (\frac{1+t}{t} \right )^{m+n}}\left ( -\frac{1}{t^2} \right )dt

\Rightarrow 2 \mathrm{I}_{\mathrm{m}, \mathrm{n}}=\int_{0}^{1} \frac{\mathrm{y}^{\mathrm{m}-1}+\mathrm{y}^{\mathrm{n}-1}}{(1+\mathrm{y})^{\mathrm{m}+\mathrm{n}}} \mathrm{d} \mathrm{y}-\int_{1}^{0} \frac{\mathrm{t}^{\mathrm{n}-1}+\mathrm{t}^{\mathrm{m}-1}}{\mathrm{t}^{\mathrm{m}+\mathrm{n}-2}} \frac{\mathrm{t}^{\mathrm{m}+\mathrm{n}}}{(1+\mathrm{t})^{\mathrm{m}+\mathrm{n}}} \frac{\mathrm{dt}}{\mathrm{t}^{2}}

\Rightarrow \text { Hence } 2 \mathrm{I}_{\mathrm{m}, \mathrm{n}}=2 \int_{0}^{1} \frac{\mathrm{y}^{\mathrm{m}-1}+\mathrm{y}^{\mathrm{n}-1}}{(1+\mathrm{y})^{\mathrm{m}+\mathrm{n}}} \mathrm{dy} \Rightarrow \alpha=1

Posted by

himanshu.meshram

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