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)


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

$Ω (α; β) = \underset{- 1}{\int^{1}} \frac{{(1 + x)}^{2 α - 1} {(1 - x)}^{2 β - 1}}{{(1 + x^{2})}^{α + β}} dx; α; β > 0$ $find a closed form and prove that :$ $Ω (3, 5) > \sqrt{Ω (4, 5) \cdot Ω (3, 6)}$
	Answered by mindispower last updated on 27/Dec/21

	$= \int_{- \frac{π}{4}}^{\frac{π}{4}} \frac{{(\sqrt{2})}^{2 a + 2 b - 2} {c o s}^{2 a - 1} (x - \frac{π}{4}) {c o s}^{2 b - 1} (x + \frac{π}{4})}{{c o s}^{2} (x) {c o s}^{2 b + 2 a - 2} (x)} {c o s}^{2 (a + b)} (x)$ $= 2^{a + b - 1} \int_{- \frac{π}{4}}^{\frac{π}{4}} {c o s}^{2 a - 1} (x - \frac{π}{4}) {c o s}^{2 b - 1} (x + \frac{π}{4} b d t - 2) d x$ $y u \to x + \frac{π}{4}$ $= 2^{a + b - 1} \int_{0}^{\frac{π}{2}} {s i n}^{2 a - 1} (y) {c o s}^{2 b - 1} (y) d y$ $= 2^{a + b - 2} . 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 a - 1} (y) {c o s}^{2 b - 1} (y) d y$ $β (x, y) = 2 \int_{0}^{\frac{π}{2}} {s i n}^{2 x - 1} (t) {c o s}^{2 y - 1} (t) d t b e t a F u n c t i o n n$ $= β (a, b) . 2^{a + b - 2} = 2^{a + b - 2} . \frac{Γ (a) Γ (b)}{Γ (a + b)}$ $Ω (a, b) = 2^{a + b - 2} . β (a, b)$ $Ω (3, 6) = 2^{7} . \frac{2! . 5!}{8!}$ $Ω (4, 5) = 2^{7} . \frac{3! . 4!}{8!}$ $Ω (3, 5) = 2^{6} . \frac{2! . 4!}{7!}$ $\Leftrightarrow . \frac{24}{7!} > 2^{7} . \frac{1}{8!} . \sqrt{(2.5! . 4! . 3!)}$ $24.8 > \sqrt{2.5! . 4.3.2.3.2 .}$ $24.8 > \sqrt{(24) . (2.3.4) . 2.5.3.2}$ $\Leftrightarrow 8 > \sqrt{60} t r u e \Rightarrow Ω (3, 5) > \sqrt{Ω (3, 6) Ω (4, 5)}$
	Commented by HongKing last updated on 28/Dec/21

	$cool my dear Sir thank you so much$
	Commented by mindispower last updated on 29/Dec/21

	$w i t h e p l e a s u r s i r h a v e n i c e d a y$
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