Gamma and Beta function extravaganza

Calculus Level 3

∫ 0 π 2 csc ⁡ θ n d θ = α π n ⋅ 2 β + 1 n Γ ( γ n ) Γ δ ( 1 ϵ n ) \int_{0}^{\frac {\pi}{2}} \sqrt [n] {\csc \theta}\ d\theta = \dfrac {\alpha \pi n \cdot 2^{\beta+\frac 1n}\Gamma\left(\frac \gamma n\right)}{\Gamma^\delta \left(\frac 1{\epsilon n}\right)}

The equation above holds true for integers α \alpha , β \beta , γ \gamma , δ \delta , and ϵ \epsilon , where β > 0 \beta > 0 . Find α β γ δ π + ϵ \alpha \beta \gamma \delta \pi + \epsilon .

10 π + 14 10\pi + 14 4 e 4e 14 π − 9 14\pi - 9 11 π + 7 11\pi + 7 7 π e 7\pi e 2 π − 2 2\pi - 2

N. Aadhaar Murty by N. Aadhaar Murty , 15, India

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1 solution

Chew-Seong Cheong
Aug 17, 2020

Note: The method used by @N. Aadhaar Murty was correct but computation was incorrect.

I = ∫ 0 π 2 csc ⁡ x n d x = ∫ 0 π 2 sin ⁡ − 1 n x cos ⁡ 0 x d x = 1 2 B ( 1 2 − 1 2 n , 1 2 ) where B ( ⋅ ) is the beta function. = Γ ( 1 2 − 1 2 n ) Γ ( 1 2 ) 2 Γ ( 1 − 1 2 n ) where Γ ( ⋅ ) is the gamma function. = π Γ ( 1 2 − 1 2 n ) 2 ( − 1 2 n ) Γ ( − 1 2 n ) = − π n Γ ( − 1 2 n ) Γ ( 1 2 − 1 2 n ) Γ 2 ( − 1 2 n ) By Legendre duplication formula (see note) = − π n ⋅ 2 1 + 1 n π Γ ( − 1 n ) Γ 2 ( − 1 2 n ) = − 2 1 + 1 n π n Γ ( − 1 n ) Γ 2 ( − 1 2 n ) \begin{aligned} I & = \int_0^\frac \pi 2 \sqrt[n]{\csc x} \ dx \\ & = \int_0^\frac \pi 2 \sin^{-\frac 1n} x \cos^0 x \ dx \\ & = \frac 12 B \left(\frac 12 - \frac 1{2n}, \frac 12 \right) & \small \blue{\text{where }B(\cdot) \text{ is the beta function.}} \\ & = \frac {\Gamma\left(\frac 12 - \frac 1{2n}\right) \Gamma \left(\frac 12 \right)}{2 \Gamma \left(1 - \frac 1{2n}\right)} & \small \blue{\text{where }\Gamma(\cdot) \text{ is the gamma function.}} \\ & = \frac {\sqrt \pi \Gamma\left(\frac 12 - \frac 1{2n}\right)}{2 \left(-\frac 1{2n} \right) \Gamma \left(- \frac 1{2n}\right)} \\ & = - \frac {\sqrt \pi n \blue{\Gamma \left(- \frac 1{2n}\right)\Gamma\left(\frac 12 - \frac 1{2n}\right)}}{\Gamma^2 \left(- \frac 1{2n}\right)} & \small \blue{\text{By Legendre duplication formula (see note)}} \\ & = - \frac {\sqrt \pi n \cdot \blue{2^{1+\frac 1n}\sqrt \pi \Gamma \left(- \frac 1n\right)}}{\Gamma^2 \left(- \frac 1{2n}\right)} \\ & = - \frac {2^{1+\frac 1n} \pi n \Gamma \left(- \frac 1n\right)}{\Gamma^2 \left(- \frac 1{2n}\right)} \end{aligned}

Therefore, α β γ δ π + ϵ = 2 π − 2 \alpha \beta \gamma \delta \pi + \epsilon = \boxed{2\pi - 2}

Note: Legendre duplication formula

π Γ ( 2 z ) = 2 2 z − 1 Γ ( z ) Γ ( z + 1 2 ) Putting z = − 1 2 n π Γ ( − 1 n ) = Γ ( − 1 2 n ) Γ ( 1 2 − 1 2 n ) 2 1 + 1 n ⟹ Γ ( − 1 2 n ) Γ ( 1 2 − 1 2 n ) = π ⋅ 2 1 + 1 n Γ ( − 1 n ) \begin{aligned} \sqrt \pi \Gamma(2z) & = 2^{2z-1}\Gamma(z) \Gamma \left(z + \frac 12\right) & \small \blue{\text{Putting }z = - \frac 1{2n}} \\ \sqrt \pi \Gamma \left(-\frac 1n \right) & = \frac {\Gamma \left(-\frac 1{2n} \right)\Gamma \left(\frac 12 -\frac 1{2n} \right)}{2^{1+\frac 1n}} \\ \implies \Gamma \left(-\frac 1{2n} \right)\Gamma \left(\frac 12 -\frac 1{2n} \right) & = \sqrt \pi \cdot 2^{1+\frac 1n} \Gamma \left(-\frac 1n\right) \end{aligned}

References:

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