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Math Help - Today's calculation of integral #3

  1. #1
    Math Engineering Student
    Krizalid's Avatar
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    Today's calculation of integral #3

    Prove that <br />
\left(\frac12\right)!=\frac{\sqrt\pi}2.<br />
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    IM stupid but?

    Isnt this just (1/2)!= \gamma(-1/2)?= \int_{0}{1}{-ln(x)^(1/2) \dx?
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Sorry I meant isnt it just gammafunction(-1/2)?
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  4. #4
    MHF Contributor Mathstud28's Avatar
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    so it is \int_{0}{1}{1/(-ln(x)^(1/2)) \ dx}[/ which is pi^(1/2)*erfi(ln(x)^(1/2))*ln(x)^(1/2)/(-ln(x))^(1/2) evaluated from 0 to one?
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  5. #5
    Behold, the power of SARDINES!
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    Quote Originally Posted by Krizalid View Post
    Prove that <br />
\left(\frac12\right)!=\frac{\sqrt\pi}2.<br />
    \Gamma(1/2)=\int_{0}^{\infty}t^{-1/2}e^{-t}dt

    using the sub u^2=t \mbox{ then } 2udu=dt

    \Gamma(1/2)=\int_{0}^{\infty}u^{-1}e^{-u^2}(2udu)=2\int_{0}^{\infty}e^{-u^2}du

    we could do the same sub with v^2=t and get

    \Gamma(1/2)=2\int_{0}^{\infty}e^{-v^2}dv

    so

    [\Gamma(1/2)]^2=(2\int_{0}^{\infty}e^{-u^2}du)(2\int_{0}^{\infty}e^{-v^2}dv)=4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(u^2+v^2)dudv}

    Convert to polar coordinates the region of integration is the 1st quadrant so we get...

    4 (\int_{0}^{\pi/2}d\theta)(\int_{0}^{\infty}re^{-r^2}dr)=4 \left( \frac{\pi}{2}\right) \left( \frac{1}{2}\right)=\pi

    so

    [\Gamma(1/2)]^2=\pi \iff \Gamma(1/2)=\sqrt{\pi}
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  6. #6
    Math Engineering Student
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    Yeah, that's the standard solution.

    -----

    By considering \int_0^1x^n(1-x)^n\,dx for n\ge0, a direct application of Beta function shows that \int_0^1x^n(1-x)^n\,dx=\frac{(n!)^2}{(2n+1)!}.

    Eventually, for n=\frac12 we have \int_0^1\sqrt{x(1-x)}=\bigg\{\left(\frac12\right)!\bigg\}^2\cdot\fra  c12=\frac\pi8, this 'cause we're integrating a semicircle whose radius is \frac12. The conclusion follows.
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