The function B(?, ?) is defined by

Chapter 4, Problem 197SE

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QUESTION:

The function \(B(\alpha, \beta)\) is defined by

                           \(B(\alpha, \beta)=\int_{0}^{1} y^{\alpha-1}(1-y)^{\beta-1} d y.\)

a Letting \(y=\sin ^{2} \theta\), show that

                           \(B(\alpha,\beta)=2\int_0^{\pi/2}\sin^{2\alpha-1}\theta\cos^{2\beta-1}\theta\ d\theta.\)

b Write \(\Gamma(\alpha) \Gamma(\beta)\) as a double integral, transform to polar coordinates, and conclude that

                           \(B(\alpha, \beta)=\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}.\)

Equation Transcription:

Text Transcription:

B(alpha, beta)

B(alpha,beta)=integral 0 to 1 y^alpha-1 (1-y)^beta-1 dy.

y=sin^2 theta

B(alpha,beta)=2 integral 0 to pi/r sin^2alpha-1 cos^2beta-1 dtheta.

Gamma(alpha)Gamma(beta)

B(alpha,beta)=Gamma(alpha)Gamma(beta) over Gamma(alpha+beta).

Questions & Answers

QUESTION:

The function \(B(\alpha, \beta)\) is defined by

                           \(B(\alpha, \beta)=\int_{0}^{1} y^{\alpha-1}(1-y)^{\beta-1} d y.\)

a Letting \(y=\sin ^{2} \theta\), show that

                           \(B(\alpha,\beta)=2\int_0^{\pi/2}\sin^{2\alpha-1}\theta\cos^{2\beta-1}\theta\ d\theta.\)

b Write \(\Gamma(\alpha) \Gamma(\beta)\) as a double integral, transform to polar coordinates, and conclude that

                           \(B(\alpha, \beta)=\frac{\Gamma(\alpha) \Gamma(\beta)}{\Gamma(\alpha+\beta)}.\)

Equation Transcription:

Text Transcription:

B(alpha, beta)

B(alpha,beta)=integral 0 to 1 y^alpha-1 (1-y)^beta-1 dy.

y=sin^2 theta

B(alpha,beta)=2 integral 0 to pi/r sin^2alpha-1 cos^2beta-1 dtheta.

Gamma(alpha)Gamma(beta)

B(alpha,beta)=Gamma(alpha)Gamma(beta) over Gamma(alpha+beta).

ANSWER:

Solution:

Step 1 of 2:

          Given the function is defined by

                                ----- (1)

a). To show that

              

     Let , substitute this in equation (1).

           

     Then,

                 

                               

                               

                                 

       Since

                  Hence proved

           


   

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