Question Number 162062 by HongKing last updated on 25/Dec/21
$$\Omega\left(\alpha;\beta\right)\:=\underset{\:-\mathrm{1}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}\boldsymbol{\alpha}-\mathrm{1}} \:\left(\mathrm{1}-\mathrm{x}\right)^{\mathrm{2}\boldsymbol{\beta}-\mathrm{1}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\boldsymbol{\alpha}+\boldsymbol{\beta}} }\:\mathrm{dx}\:;\:\alpha;\beta>\mathrm{0} \\ $$$$\mathrm{find}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{and}\:\mathrm{prove}\:\mathrm{that}: \\ $$$$\Omega\left(\mathrm{3},\mathrm{5}\right)\:>\:\sqrt{\Omega\left(\mathrm{4},\mathrm{5}\right)\centerdot\Omega\left(\mathrm{3},\mathrm{6}\right)} \\ $$
Answered by mindispower last updated on 27/Dec/21
$$=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}{a}+\mathrm{2}{b}−\mathrm{2}} {cos}^{\mathrm{2}{a}−\mathrm{1}} \left({x}−\frac{\pi}{\mathrm{4}}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left({x}+\frac{\pi}{\mathrm{4}}\right)}{{cos}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}{b}+\mathrm{2}{a}−\mathrm{2}} \left({x}\right)}{cos}^{\mathrm{2}\left({a}+{b}\right)} \left({x}\right) \\ $$$$=\mathrm{2}^{{a}+{b}−\mathrm{1}} \int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} {cos}^{\mathrm{2}{a}−\mathrm{1}} \left({x}−\frac{\pi}{\mathrm{4}}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left(\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{4}}{bdt}−\mathrm{2}\right){dx} \\ $$$${yu}\rightarrow{x}+\frac{\pi}{\mathrm{4}} \\ $$$$=\mathrm{2}^{{a}+{b}−\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{a}−\mathrm{1}} \left({y}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left({y}\right){dy} \\ $$$$=\mathrm{2}^{{a}+{b}−\mathrm{2}} .\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{a}−\mathrm{1}} \left({y}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left({y}\right){dy} \\ $$$$\beta\left({x},{y}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{x}−\mathrm{1}} \left({t}\right){cos}^{\mathrm{2}{y}−\mathrm{1}} \left({t}\right){dt}\:{beta}\:{Functionn} \\ $$$$=\beta\left({a},{b}\right).\mathrm{2}^{{a}+{b}−\mathrm{2}} =\mathrm{2}^{{a}+{b}−\mathrm{2}} .\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)} \\ $$$$\Omega\left({a},{b}\right)=\mathrm{2}^{{a}+{b}−\mathrm{2}} .\beta\left({a},{b}\right) \\ $$$$\Omega\left(\mathrm{3},\mathrm{6}\right)=\mathrm{2}^{\mathrm{7}} .\frac{\mathrm{2}!.\mathrm{5}!}{\mathrm{8}!} \\ $$$$\Omega\left(\mathrm{4},\mathrm{5}\right)=\mathrm{2}^{\mathrm{7}} .\frac{\mathrm{3}!.\mathrm{4}!}{\mathrm{8}!} \\ $$$$\Omega\left(\mathrm{3},\mathrm{5}\right)=\mathrm{2}^{\mathrm{6}} .\frac{\mathrm{2}!.\mathrm{4}!}{\mathrm{7}!} \\ $$$$\Leftrightarrow.\frac{\mathrm{24}}{\mathrm{7}!}>\mathrm{2}^{\mathrm{7}} .\frac{\mathrm{1}}{\mathrm{8}!}.\sqrt{\left(\mathrm{2}.\mathrm{5}!.\mathrm{4}!.\mathrm{3}!\right)} \\ $$$$\mathrm{24}.\mathrm{8}>\sqrt{\mathrm{2}.\mathrm{5}!.\mathrm{4}.\mathrm{3}.\mathrm{2}.\mathrm{3}.\mathrm{2}.} \\ $$$$\mathrm{24}.\mathrm{8}>\sqrt{\left(\mathrm{24}\right).\left(\mathrm{2}.\mathrm{3}.\mathrm{4}\right).\mathrm{2}.\mathrm{5}.\mathrm{3}.\mathrm{2}} \\ $$$$\Leftrightarrow\mathrm{8}>\sqrt{\mathrm{60}}\:{true}\Rightarrow\Omega\left(\mathrm{3},\mathrm{5}\right)>\sqrt{\Omega\left(\mathrm{3},\mathrm{6}\right)\Omega\left(\mathrm{4},\mathrm{5}\right)} \\ $$$$ \\ $$$$ \\ $$
Commented by HongKing last updated on 28/Dec/21
$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$
Commented by mindispower last updated on 29/Dec/21
$${withe}\:{pleasur}\:{sir}\:{have}\:{nice}\:{day} \\ $$