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Thread: Beta function proof

  1. #1
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    Beta function proof

    Prove that $\displaystyle \int_{0}^{\pi} sin^n(x) ~dx = B( \frac{n+1}{2} , \frac{1}{2} )$
    where n is a real number ( guys you can prove whether n can be complex ... )


    Then use this result to solve a Physics problem :
    Caculate the time taken for a particle to travel from the top of a unit semicircle to the bottom . ( under gravitational force )
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by simplependulum View Post
    Prove that $\displaystyle \int_{0}^{\pi} sin^n(x) ~dx = B( \frac{n+1}{2} , \frac{1}{2} )$
    where n is a real number ( guys you can prove whether n can be complex ... )


    Then use this result to solve a Physics problem :
    Caculate the time taken for a particle to travel from the top of a unit semicircle to the bottom . ( under gravitational force )
    you know that

    $\displaystyle \beta (p,q) = 2\int_{0}^{\frac{\pi}{2}} sin^{2p-1}x cos^{2q-1}x dx $

    you have

    $\displaystyle \int_{0}^{\pi} sin^nx dx = \int_{0}^{\frac{\pi}{2}} sin^nx dx + \int_{\frac{\pi}{2}}^{\pi} sin^nx dx $

    the first one equal

    $\displaystyle \left(\frac{{\color{red}2}}{2}\right) {\color{red}\int_{0}^{\frac{\pi}{2}} sin^n x dx}=\frac{\beta(\frac{n+1}{2} , \frac{1}{2})}{2} $

    since the red term equal beta and you have $\displaystyle 2p-1 = n \Rightarrow p=\frac{n+1}{2}$ and $\displaystyle 2q-1 = 0 \Rightarrow q=\frac{1}{2}$

    the second integral

    $\displaystyle \int_{\frac{\pi}{2}}^{\pi} sin^nx dx $ let $\displaystyle x=y+\frac{\pi}{2} \Rightarrow dx=dy $

    $\displaystyle \int_{0}^{\frac{\pi}{2}} sin^n(y+\frac{\pi}{2}) dy =\int_{0}^{\frac{\pi}{2}} \left[siny(cos(\frac{\pi}{2}))+cosy(sin(\frac{\pi}{2}))\ right]^{n} dy $

    $\displaystyle \Rightarrow \left(\frac{2}{2}\right)\int_{0}^{\frac{\pi}{2}} (cosy)^{n} dy =\frac{\beta(\frac{n+1}{2} , \frac{1}{2} )}{2} $

    find the sum of the first and the second integarl you will have the answer
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  3. #3
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    Hello,

    Consider this formula of the beta function :

    $\displaystyle \beta(x,y)=2\int_0^{\pi/2} \sin^{2x-1}\theta\cos^{2y-1}\theta ~d\theta$ (which you can get by a substitution $\displaystyle t=\sin^2\theta$ in the common formula of the beta function)

    Then $\displaystyle \beta\left(\frac{n+1}{2},\frac 12\right)=2\int_0^{\pi/2} \sin^n\theta ~d\theta$

    And since $\displaystyle \sin\left(\tfrac\pi 2-x\right)=\sin\left(\tfrac\pi 2+x\right) ~\dots$
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  4. #4
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    Quote Originally Posted by simplependulum View Post
    Prove that $\displaystyle \int_{0}^{\pi} sin^n(x) ~dx = B( \frac{n+1}{2} , \frac{1}{2} )$
    where n is a real number ( guys you can prove whether n can be complex ... )


    Then use this result to solve a Physics problem :
    Caculate the time taken for a particle to travel from the top of a unit semicircle to the bottom . ( under gravitational force )
    Assuming the particle slides frictionlessly under gravity down the semicircle, the particle will leave the circle after the motion has gone through a polar angle equal to $\displaystyle \theta = \cos^{-1} \frac{2}{3}$. So I'm not sure how the particle can end up at the bottom of the semicircle ....
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