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Question Number 162062 by HongKing last updated on 25/Dec/21

Ω(α;β) =∫_( -1) ^( 1) (((1+x)^(2𝛂-1) (1-x)^(2𝛃-1) )/((1+x^2 )^(𝛂+𝛃) )) dx ; α;β>0 find a closed form and prove that: Ω(3,5) > (√(Ω(4,5)∙Ω(3,6)))

$$\Omega\left(\alpha;\beta\right)\:=\underset{\:-\mathrm{1}} {\overset{\:\mathrm{1}} {\int}}\:\frac{\left(\mathrm{1}+\mathrm{x}\right)^{\mathrm{2}\boldsymbol{\alpha}-\mathrm{1}} \:\left(\mathrm{1}-\mathrm{x}\right)^{\mathrm{2}\boldsymbol{\beta}-\mathrm{1}} }{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\boldsymbol{\alpha}+\boldsymbol{\beta}} }\:\mathrm{dx}\:;\:\alpha;\beta>\mathrm{0} \\ $$ $$\mathrm{find}\:\mathrm{a}\:\mathrm{closed}\:\mathrm{form}\:\mathrm{and}\:\mathrm{prove}\:\mathrm{that}: \\ $$ $$\Omega\left(\mathrm{3},\mathrm{5}\right)\:>\:\sqrt{\Omega\left(\mathrm{4},\mathrm{5}\right)\centerdot\Omega\left(\mathrm{3},\mathrm{6}\right)} \\ $$

Answered by mindispower last updated on 27/Dec/21

=∫_(−(π/4)) ^(π/4) ((((√2))^(2a+2b−2) cos^(2a−1) (x−(π/4))cos^(2b−1) (x+(π/4)))/(cos^2 (x)cos^(2b+2a−2) (x)))cos^(2(a+b)) (x) =2^(a+b−1) ∫_(−(π/4)) ^(π/4) cos^(2a−1) (x−(π/4))cos^(2b−1) (x+(𝛑/4)bdt−2)dx yu→x+(π/4) =2^(a+b−1) ∫_0 ^(π/2) sin^(2a−1) (y)cos^(2b−1) (y)dy =2^(a+b−2) .2∫_0 ^(π/2) sin^(2a−1) (y)cos^(2b−1) (y)dy β(x,y)=2∫_0 ^(π/2) sin^(2x−1) (t)cos^(2y−1) (t)dt beta Functionn =β(a,b).2^(a+b−2) =2^(a+b−2) .((Γ(a)Γ(b))/(Γ(a+b))) Ω(a,b)=2^(a+b−2) .β(a,b) Ω(3,6)=2^7 .((2!.5!)/(8!)) Ω(4,5)=2^7 .((3!.4!)/(8!)) Ω(3,5)=2^6 .((2!.4!)/(7!)) ⇔.((24)/(7!))>2^7 .(1/(8!)).(√((2.5!.4!.3!))) 24.8>(√(2.5!.4.3.2.3.2.)) 24.8>(√((24).(2.3.4).2.5.3.2)) ⇔8>(√(60)) true⇒Ω(3,5)>(√(Ω(3,6)Ω(4,5)))

$$=\int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} \frac{\left(\sqrt{\mathrm{2}}\right)^{\mathrm{2}{a}+\mathrm{2}{b}−\mathrm{2}} {cos}^{\mathrm{2}{a}−\mathrm{1}} \left({x}−\frac{\pi}{\mathrm{4}}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left({x}+\frac{\pi}{\mathrm{4}}\right)}{{cos}^{\mathrm{2}} \left({x}\right){cos}^{\mathrm{2}{b}+\mathrm{2}{a}−\mathrm{2}} \left({x}\right)}{cos}^{\mathrm{2}\left({a}+{b}\right)} \left({x}\right) \\ $$ $$=\mathrm{2}^{{a}+{b}−\mathrm{1}} \int_{−\frac{\pi}{\mathrm{4}}} ^{\frac{\pi}{\mathrm{4}}} {cos}^{\mathrm{2}{a}−\mathrm{1}} \left({x}−\frac{\pi}{\mathrm{4}}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left(\boldsymbol{{x}}+\frac{\boldsymbol{\pi}}{\mathrm{4}}{bdt}−\mathrm{2}\right){dx} \\ $$ $${yu}\rightarrow{x}+\frac{\pi}{\mathrm{4}} \\ $$ $$=\mathrm{2}^{{a}+{b}−\mathrm{1}} \int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{a}−\mathrm{1}} \left({y}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left({y}\right){dy} \\ $$ $$=\mathrm{2}^{{a}+{b}−\mathrm{2}} .\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{a}−\mathrm{1}} \left({y}\right){cos}^{\mathrm{2}{b}−\mathrm{1}} \left({y}\right){dy} \\ $$ $$\beta\left({x},{y}\right)=\mathrm{2}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {sin}^{\mathrm{2}{x}−\mathrm{1}} \left({t}\right){cos}^{\mathrm{2}{y}−\mathrm{1}} \left({t}\right){dt}\:{beta}\:{Functionn} \\ $$ $$=\beta\left({a},{b}\right).\mathrm{2}^{{a}+{b}−\mathrm{2}} =\mathrm{2}^{{a}+{b}−\mathrm{2}} .\frac{\Gamma\left({a}\right)\Gamma\left({b}\right)}{\Gamma\left({a}+{b}\right)} \\ $$ $$\Omega\left({a},{b}\right)=\mathrm{2}^{{a}+{b}−\mathrm{2}} .\beta\left({a},{b}\right) \\ $$ $$\Omega\left(\mathrm{3},\mathrm{6}\right)=\mathrm{2}^{\mathrm{7}} .\frac{\mathrm{2}!.\mathrm{5}!}{\mathrm{8}!} \\ $$ $$\Omega\left(\mathrm{4},\mathrm{5}\right)=\mathrm{2}^{\mathrm{7}} .\frac{\mathrm{3}!.\mathrm{4}!}{\mathrm{8}!} \\ $$ $$\Omega\left(\mathrm{3},\mathrm{5}\right)=\mathrm{2}^{\mathrm{6}} .\frac{\mathrm{2}!.\mathrm{4}!}{\mathrm{7}!} \\ $$ $$\Leftrightarrow.\frac{\mathrm{24}}{\mathrm{7}!}>\mathrm{2}^{\mathrm{7}} .\frac{\mathrm{1}}{\mathrm{8}!}.\sqrt{\left(\mathrm{2}.\mathrm{5}!.\mathrm{4}!.\mathrm{3}!\right)} \\ $$ $$\mathrm{24}.\mathrm{8}>\sqrt{\mathrm{2}.\mathrm{5}!.\mathrm{4}.\mathrm{3}.\mathrm{2}.\mathrm{3}.\mathrm{2}.} \\ $$ $$\mathrm{24}.\mathrm{8}>\sqrt{\left(\mathrm{24}\right).\left(\mathrm{2}.\mathrm{3}.\mathrm{4}\right).\mathrm{2}.\mathrm{5}.\mathrm{3}.\mathrm{2}} \\ $$ $$\Leftrightarrow\mathrm{8}>\sqrt{\mathrm{60}}\:{true}\Rightarrow\Omega\left(\mathrm{3},\mathrm{5}\right)>\sqrt{\Omega\left(\mathrm{3},\mathrm{6}\right)\Omega\left(\mathrm{4},\mathrm{5}\right)} \\ $$ $$ \\ $$ $$ \\ $$

Commented byHongKing last updated on 28/Dec/21

cool my dear Sir thank you so much

$$\mathrm{cool}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much} \\ $$

Commented bymindispower last updated on 29/Dec/21

withe pleasur sir have nice day

$${withe}\:{pleasur}\:{sir}\:{have}\:{nice}\:{day} \\ $$

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