JEE Main 2021 — Mathematics Calculus

$${I}_{m,n}={\int }_{0}^{1}{x}^{m-1}{(1-x)}^{n-1}dx={I}_{n,m}$$Im,n​=∫01​xm−1(1−x)n−1dx=In,m​

Now Let $x=\frac{1}{y+1}\Rightarrow dx=-\frac{1}{{(y+1)}^{2}}dy$x=y+11​⇒dx=−(y+1)21​dy

So, $\displaystyle {I}_{m,n}=-{\int }_{\infty }^{0}\frac{1}{{(y+1)}^{m-1}}\frac{{y}^{n-1}}{{(y+1)}^{n-1}}\frac{dy}{{(y+1)}^{2}}={\int }_{0}^{\infty }\frac{{y}^{n-1}}{{(1+y)}^{m+n}}dy$Im,n​=−∫∞0​(y+1)m−11​(y+1)n−1yn−1​(y+1)2dy​=∫0∞​(1+y)m+nyn−1​dy

similarly $\displaystyle {I}_{m,n}={\int }_{0}^{\infty }\frac{{y}^{m-1}}{{(1+y)}^{m+n}}dy$Im,n​=∫0∞​(1+y)m+nym−1​dy

Now $\displaystyle 2{I}_{m,n}={\int }_{0}^{\infty }\frac{{y}^{m-1}+{y}^{n-1}}{{(1+y)}^{m+n}}dy$2Im,n​=∫0∞​(1+y)m+nym−1+yn−1​dy

$$={\int }_{0}^{\infty }\frac{{y}^{m-1}+{y}^{n-1}}{{(1+y)}^{m+n}}dy$$=∫0∞​(1+y)m+nym−1+yn−1​dy

$$={\int }_{0}^{1}\frac{{y}^{m-1}+{y}^{n-1}}{{(1+y)}^{m+n}}dy+\underset{\mathrm{substitute}y=\frac{1}{t}}{\underset{⏟}{{\int }_{1}^{\infty }\frac{{y}^{m-1}+{y}^{n-1}}{{(1+y)}^{m+n}}dy}}$$=∫01​(1+y)m+nym−1+yn−1​dy+substitutey=t1​⏟∫1∞​(1+y)m+nym−1+yn−1​dy​​

$$\Rightarrow 2{I}_{m,n}={\int }_{0}^{1}\frac{{y}^{m-1}+{y}^{n-1}}{{(1+y)}^{m+n}}dy-{\int }_{1}^{0}\frac{{t}^{n-1}+{t}^{m-1}}{{t}^{m+n-2}}\frac{{t}^{m+n}}{{(1+t)}^{m+n}}\frac{dt}{{t}^{2}}$$⇒2Im,n​=∫01​(1+y)m+nym−1+yn−1​dy−∫10​tm+n−2tn−1+tm−1​(1+t)m+ntm+n​t2dt​

$\Rightarrow$⇒ Hence $\displaystyle 2{I}_{m,n}=2{\int }_{0}^{1}\frac{{y}^{m-1}+{y}^{n-1}}{{(1+y)}^{m+n}}dy\Rightarrow \alpha =1$2Im,n​=2∫01​(1+y)m+nym−1+yn−1​dy⇒α=1