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Math Help - gamma and e^(-x^2)

  1. #1
    Eater of Worlds
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    gamma and e^(-x^2)

    You know, I was just playing around and noticed something. I am sure you all

    have thought of this before, but I just 'discovered this'. I always integrated

    e^{-x^{2}} by the polar method.

    But if we use gamma it is easy.

    Knowing that {\Gamma}(\frac{1}{2})=\sqrt{\pi}

    we can use the definition:

    {\Gamma}(n)=\int_{0}^{\infty}t^{n-1}e^{-t}dt=2\int_{0}^{\infty}x^{2n-1}e^{-x^{2}}dx

    Therefore,

    \int_{0}^{\infty}e^{-x^{2}}dx=\frac{1}{2}{\Gamma}(\frac{1}{2})=\frac{\s  qrt{\pi}}{2}

    Since e^{-x^{2}} is even, we get

    \int_{-\infty}^{\infty}e^{-x^{2}}dx=\sqrt{\pi}

    I know, I know, it's obvious to you all and older than Moses. But I had not

    thought about using gamma before for this famous integral and it just

    dawned on me. I always just used the old polar thing and never worried

    about another method. Stuck in that rut, I reckon. But, if there are those

    out there who like this, here it is.
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  2. #2
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    Quote Originally Posted by galactus View Post

    But if we use gamma it is easy.

    Knowing that {\Gamma}(\frac{1}{2})=\sqrt{\pi}
    How'd ya prove this little fact?
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  3. #3
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    You want me to prove that?. That is a famous gamma identity, so I did not bother proving every aspect.

    But, if you would like to see it, let me get back. Okey-doke. I always just used it and took it for granted and never bothered proving it.

    But, I am thinking we can play on the old polar thing to show it.

    To be lazy, I would just use {\Gamma}(n){\Gamma}(1-n)=\frac{\pi}{sin({\pi}n)} and let n=1/2.

    I know, you want something more tangible.

    Let's see, we could start with \int_{0}^{\infty}t^{\frac{-1}{2}}e^{-t}dt

    Then, sub in t=y^{2}, \;\ dt=2ydy.

    I will finish later. Got to go for a second.
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  4. #4
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    Well, Kriz, it appears I ended up using the polar thing to do that. But I think you knew that, didn't you?.

    To conclude:

    2\int_{0}^{\infty}y^{-1}e^{-y^{2}}ydy=2\int_{0}^{\infty}e^{-y^{2}}dy

    or 2\int_{0}^{\infty}e^{-x^{2}}dx

    \left[{\Gamma}(\frac{1}{2})\right]^{2}=4\int_{0}^{\infty}\int_{0}^{\infty}e^{-(x^{2}+y^{2})}dxdy

    Now, we use the polar thing. So, I learned myself something with a little prodding from the Krizmeister. Cheers.

    I suppose that infernal polar was hidden there afterall.

    I can not believe I had never bothered doing these before. Oh well, I derived something I should have a long time ago. Especially, since I like the Gamma function so much.
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  5. #5
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    Ohhh, I'm sorry, I misread the question, it's actually calculating again the gaussian integral.

    Cheers and Happy New Year!!!!
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  6. #6
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    Since we're on Gamma, derive this one:

    {\Gamma}(2n)=\frac{1}{\sqrt{\pi}}\cdot 2^{2n-1}\cdot {\Gamma}(n)\cdot {\Gamma}(n+\frac{1}{2})

    This is the duplication formula for Gamma.

    Try using the identity 2sin{\theta}cos{\theta}=sin2{\theta} and put 2{\theta}={\phi}


    Happy New Years
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  7. #7
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    Don't worry, I'm also a good friend of Gamma function.

    But it's actually a bit interesting in finding a closed form for \Gamma\left(n+\frac12\right).
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  8. #8
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    You mean this one:

    {\Gamma}(n+\frac{1}{2})=\frac{1\cdot 3\cdot 5\cdots (\not{2}n-1)}{2^{n}}=\frac{(2n)!}{4^{n}n!}\sqrt{\pi}?.
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  9. #9
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    Yes.
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  10. #10
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    Well, Kriz, let' see. I have been working on it while I was waiting on your reponnse and gong to the bathroom.

    It is getting late, Kriz. I am going to skip some major steps and join you tomorrow. Gotta go. But I have this so far. Need to explain some steps further if need be.

    From B(p,q)=\frac{{\Gamma}(p){\Gamma}(q)}{{\Gamma}(p+q)  }

    Then,

    {\Gamma}(p+\frac{1}{2})=\frac{{\Gamma}(p)\cdot {\Gamma}(\frac{1}{2})}{B(p,1/2)}

    =\frac{(p-1)!\sqrt{\pi}}{\frac{4^{p}(p!)^{2}}{p(2p)!}}

    \boxed{=\frac{(2p)!\sqrt{\pi}}{4^{p}\cdot p!}}

    We need to show that B(p,1/2)=\frac{4^{n}(n!)^{2}}{n(2n)!} which I think can be gotten from \frac{4^{p}}{2}\cdot B(p,p)

    because B(p,p)=\frac{{\Gamma}(p){\Gamma}(p)}{{\Gamma}(2p)}  =\frac{(p-1)!(p-1)!}{(2p-1)!}

    Therefore, we get =\frac{(p!)^{2}\cdot 2p}{p^{2}\cdot (2p)!}

    =\frac{2(p!)^{2}}{p(2p)!}

    B(p,1/2)=\frac{4^{p}}{2}\cdot \frac{2(p!)^{2}}{p(2p)!}

    Plugging this in the above gives:

    \frac{(n-1)!\sqrt{\pi}}{\frac{(p!)^{2}4^{p}}{p(2p)!}}=\frac  {(2p)!\sqrt{\pi}}{4^{p}p!}

    Hope you can follow that. I will touch it up tomorrow.

    It's getting late and my brain is beginning to see p's and q's everywhere.

    Cheers, Kriz
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  11. #11
    Moo
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    The method I really love for calculating this integral is considering the function :

    f(x)=\left(\int_0^x e^{-t^2} ~dt\right)^2+\int_0^1 \frac{e^{-x^2(t^2+1)}}{t^2+1} ~dt

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