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Question Number 161818 by HongKing last updated on 22/Dec/21

Show that :

Φ = \underset{0}{\int^{1}} \sqrt{\frac{1 - x^{2}}{1 + x^{2}}} dx = \frac{\sqrt{π}}{4} (\frac{Γ (\frac{1}{4})}{Γ (\frac{3}{4})} - 4 \frac{Γ (\frac{3}{4})}{Γ (\frac{1}{4})})

where : Γ - Gamma function

Answered by Lordose last updated on 22/Dec/21

Φ = \int_{0}^{1} \sqrt{\frac{1 - x^{2}}{1 + x^{2}}} dx = \int_{0}^{1} \frac{1 - x^{2}}{\sqrt{1 - x^{4}}} dx

Φ \overset{x = u^{\frac{1}{4}}}{=} \frac{1}{4} \int_{0}^{1} \frac{u^{\frac{1}{4} - 1} - u^{\frac{1}{2} + \frac{1}{4} - 1}}{\sqrt{1 - u}} du

Φ = \frac{1}{4} (\int_{0}^{1} u^{\frac{1}{4} - 1} {(1 - u)}^{\frac{1}{2} - 1} - \int_{0}^{1} u^{\frac{3}{4} - 1} {(1 - u)}^{\frac{1}{2} - 1} du)

Φ = \frac{1}{4} (\frac{Γ (\frac{1}{4}) Γ (\frac{1}{2})}{Γ (\frac{3}{4})} - \frac{Γ (\frac{3}{4}) Γ (\frac{1}{2})}{Γ (\frac{5}{4})})

Γ (1 + x) = x Γ (x)

Φ = \frac{\sqrt{π}}{4} (\frac{Γ (\frac{1}{4})}{Γ (\frac{3}{4})} - 4 \frac{Γ (\frac{3}{4})}{Γ (\frac{1}{4})})

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