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Math Help - [SOLVED] using beta or gamma function

  1. #1
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    [SOLVED] using beta or gamma function

    I have a problem solving this:

    \int_{-1}^{1} \sqrt {\frac {1+x}{1-x}} dx

    The result is \pi.
    Could someone please help me
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by akane View Post
    I have a problem solving this:

    \int_{-1}^{1} \sqrt {\frac {1+x}{1-x}} dx

    The result is \pi.
    Could someone please help me
    Let I=\int_{-1}^{1}\sqrt{\frac{1+x}{1-x}}dx. Furthermore, let \sqrt{\frac{1+x}{1-x}}=y\implies x=\frac{y^2-1}{y^2+1}\implies dx=\frac{4y}{(y^2+1)^2}. So that I=\int_0^{\infty}\frac{4y^2}{(y^2+1)^2}\text{ }dy. This integral is easy and does come out to \pi. No need for gamma function here. I guess, you could use the reflection formula for that last bit, but that just seems superfluous.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Alternatively, if you are one of "those types" that insists on using the gamma and beta function in everything...

    Define I as before and make the substitution 1+x=2z. We see that: x=2z-1,dx=2\text{ }dz,1-x=2\left(1-z\right) and our limits of integration are x\to-1\implies z\to0,x\to1\implies z\to 1. Therefore, we may finally conclude that I=2\int_0^1 \sqrt{2z}\left(2\left(1-z\right)\right)^{\frac{-1}{2}}\text{ }dz=2\text{B}\left(\tfrac{3}{2},\tfrac{-1}{2}\right)=2\cdot\frac{\pi}{2}=\pi...but that just seems stupid.
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  4. #4
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    \int_{-1}^{1}{\sqrt{\frac{1+x}{1-x}}\,dx}=\int_{-1}^{1}{\frac{\left| 1+x \right|}{\sqrt{1-x^{2}}}\,dx}=\int_{-1}^{1}{\frac{dx}{\sqrt{1-x^{2}}}}+\int_{-1}^{1}{\frac{x}{\sqrt{1-x^{2}}}\,dx}.

    The integral equals \pi, since the second piece is a bounded with two finite discontinuities points, hence integrable and since it's odd over that symmetric interval, that integral is zero.
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  5. #5
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    Quote Originally Posted by Drexel28 View Post
    Alternatively, if you are one of "those types" that insists on using the gamma and beta function in everything...

    Define I as before and make the substitution 1+x=2z. We see that: x=2z-1,dx=2\text{ }dz,1-x=2\left(1-z\right) and our limits of integration are x\to-1\implies z\to0,x\to1\implies z\to 1. Therefore, we may finally conclude that I=2\int_0^1 \sqrt{2z}\left(2\left(1-z\right)\right)^{\frac{-1}{2}}\text{ }dz=2\text{B}\left(\tfrac{3}{2},\tfrac{-1}{2}\right)=2\cdot\frac{\pi}{2}=\pi...but that just seems stupid.
    We have just learnt beta and gamma function so that's the reason for using them, thank you very much
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  6. #6
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     \int_{-1}^1 \sqrt{ \frac{1 + x }{ 1- x } } ~dx


     = \int_0^1 \sqrt{ \frac{1 + x }{ 1- x } } ~dx  + \int_{-1}^0 \sqrt{ \frac{1 + x }{ 1- x } } ~dx

    Substitute  x  = -t in the second integral .

     = \int_0^1 \sqrt{ \frac{1 + x }{ 1- x } } ~dx   + \int_0^1 \sqrt{ \frac{1 -  x }{ 1+ x } } ~dx

     = \int_0^1 \left (  \sqrt{ \frac{1 + x }{ 1- x } } + \sqrt{ \frac{1 - x }{ 1 + x } } \right ) ~dx


     = \int_0^1 \frac{ 1 + x + 1 - x }{\sqrt{1 - x^2 } }~dx

     =  \int_0^1 \frac{2}{\sqrt{ 1 - x^2}}~dx

     = 2 \left [ \sin^{-1}(x) \right ]_0^1

     = 2 \frac{\pi}{2} = \pi
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