Solution Found!
The function B(?, ?) is defined by
Chapter 4, Problem 197SE(choose chapter or problem)
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