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Thread: Special Gamma Value

  1. #1
    Rhymes with Orange Chris L T521's Avatar
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    Special Gamma Value

    I'm just wondering, does anybody know of an analytic proof that shows that :

    $\displaystyle \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?

    Or do we just accept that this is true??
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  2. #2
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    I'm just wondering, does anybody know of an analytic proof that shows that :

    $\displaystyle \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?

    Or do we just accept that this is true??
    Ok...I will give this a go...see if ti works

    $\displaystyle \Gamma\bigg(\frac{1}{2}\bigg)=\int_0^1\frac{1}{\sq rt{-\ln(x)}}dx$

    Let $\displaystyle u=\sqrt{-\ln(x)}\Rightarrow{x=e^{-u^2}}$

    so $\displaystyle dx=-2ue^{-u^2}$

    and $\displaystyle u(1)=\sqrt{-ln(1)}=0$

    and $\displaystyle u(0)=\sqrt{-\ln(0)}=\infty$

    So we have

    $\displaystyle \int_{\infty}^{0}\frac{-2ue^{-u^2}}{u}du=2\int_0^{\infty}e^{-u^2}du$

    Now look at your signature
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Ok...I will give this a go...see if ti works

    $\displaystyle \Gamma\bigg(\frac{1}{2}\bigg)=\int_0^1\frac{1}{\sq rt{-\ln(x)}}dx$

    Let $\displaystyle u=\sqrt{-\ln(x)}\Rightarrow{x=e^{-u^2}}$

    so $\displaystyle dx=-2ue^{-u^2}$

    and $\displaystyle u(1)=\sqrt{-ln(1)}=0$

    and $\displaystyle u(0)=\sqrt{-\ln(0}=\infty$

    So we have

    $\displaystyle \int_{\infty}^{0}\frac{-2ue^{-u^2}}{u}du=2\int_0^{\infty}e^{-u^2}du$

    Now look at your signature
    I have never done that before...that is hilarious it came out to be that
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Ok...I will give this a go...see if ti works

    $\displaystyle \Gamma\bigg(\frac{1}{2}\bigg)=\int_0^1\frac{1}{\sq rt{-\ln(x)}}dx$

    Let $\displaystyle u=\sqrt{-\ln(x)}\Rightarrow{x=e^{-u^2}}$

    so $\displaystyle dx=-2ue^{-u^2}$

    and $\displaystyle u(1)=\sqrt{-ln(1)}=0$

    and $\displaystyle u(0)=\sqrt{-\ln(0}=\infty$

    So we have

    $\displaystyle \int_{\infty}^{0}\frac{-2ue^{-u^2}}{u}du=2\int_0^{\infty}e^{-u^2}du$

    Now look at your signature
    hahaha. Thanks for that!
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  5. #5
    Flow Master
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    Quote Originally Posted by Chris L T521 View Post
    I'm just wondering, does anybody know of an analytic proof that shows that :

    $\displaystyle \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?

    Or do we just accept that this is true??
    If you run

    gamma function

    through a famous search engine, you will find more proofs than you've had hot breakfasts.
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  6. #6
    Behold, the power of SARDINES!
    TheEmptySet's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    I'm just wondering, does anybody know of an analytic proof that shows that :

    $\displaystyle \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?

    Or do we just accept that this is true??
    $\displaystyle \Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}t^ {-1/2}e^{-t}dt$

    let $\displaystyle t=u^2 \to dt =2udu$

    $\displaystyle \Gamma\left(\frac{1}{2}\right)=\int_{0}^{\infty}u^ {-1}e^{-u^2}(2udu)=2\int_{0}^{\infty}e^{-u^2}du$

    do the same sub with v gives

    $\displaystyle \left[\Gamma\left(\frac{1}{2}\right) \right]^2=4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(u^2+v^2)}dudv$

    converting to polar coordinates gives

    $\displaystyle 4\int_{0}^{\pi/2}\int_{0}^{\infty}re^{-r^2}drd\theta=\pi$

    $\displaystyle \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$
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  7. #7
    Math Engineering Student
    Krizalid's Avatar
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    Quote Originally Posted by Chris L T521 View Post

    I'm just wondering, does anybody know of an analytic proof that shows that :

    $\displaystyle \Gamma\left(\frac{1}{2}\right)=\sqrt{\pi}$?
    See here.

    Quote Originally Posted by Chris L T521 View Post

    Or do we just accept that this is true??
    Of course not.

    Here's another proof:

    Start by showing for $\displaystyle n\ge0$ that $\displaystyle \int_{0}^{1}{x^{n}(1-x)^{n}\,dx}=\frac{(n!)^{2}}{(2n+1)!}.$ Let $\displaystyle n=\frac12;$ the integral is equal to find the area of a semicircle with radius $\displaystyle \frac12,$ hence $\displaystyle \frac{1}{2}\left[ \left( \frac{1}{2} \right)! \right]^{2}=\frac{\pi }{8}.$

    Now rearrange this to get what you want.

    ----

    Oops, I realized that you want to find $\displaystyle \Gamma \left( \frac{1}{2} \right)$ and I computed $\displaystyle \left( \frac{1}{2} \right)!$ Anyway, the link above I gave you contains another proof.
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  8. #8
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    here's my favorite proof of the so-called Gaussian integral:

    let $\displaystyle f(x)=\left(\int_0^x e^{-t^2}dt \right)^2 + \int_0^1 \frac{e^{-x^2(1+t^2)}}{t^2+1} dt, \ x \geq 0.$ then: $\displaystyle f'(x)=2e^{-x^2}\int_0^x e^{-t^2} dt - 2x \int_0^1 e^{-x^2(1+t^2)}dt.$

    in the first integral let $\displaystyle t=xu$ to get: $\displaystyle f'(x)=2x\int_0^1e^{-x^2(1+ u^2)}du-2x \int_0^1 e^{-x^2(1 + t^2)}dt=0,$ i.e. $\displaystyle f$ is constant.

    hence $\displaystyle f(x)=f(0)=\int_0^1 \frac{1}{1+t^2}dt=\frac{\pi}{4}.$ thus $\displaystyle \lim_{x\to\infty}f(x)=\frac{\pi}{4}.$ call this result (1). on the other hand, from the

    definition of $\displaystyle f$ it's easily seen that $\displaystyle \lim_{x\to\infty}f(x)=\left(\int_0^{\infty} e^{-t^2} dt \right)^2.$ call this (2). now (1) and (2) complete the proof.
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