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Math Help - Is this function integrable?

  1. #1
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    Is this function integrable?

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

    If so, how can you integrate it?

    Also, what about \int_0^\infty e^{-x^2}
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  2. #2
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    Quote Originally Posted by paupsers View Post
    \int_{0}^{\infty}x^2e^{-x^2}

    If so, how can you integrate it?
    For this I would try integration by parts twice.

    \int x^2e^{-x^2}~dx
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  3. #3
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    A function is integrable \neq is possible to find it's primitve o it's exact numerical value (using elemental functions).
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    Quote Originally Posted by pickslides View Post
    For this I would try integration by parts twice.

    \int x^2e^{-x^2}~dx
    I've tried that and I still haven't been able to solve it, I just end up going in circles. Any more advice?
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  5. #5
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    Quote Originally Posted by paupsers View Post
    I've tried that and I still haven't been able to solve it, I just end up going in circles.
    Can you post your workings.
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  6. #6
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    Quote Originally Posted by pickslides View Post
    Can you post your workings.
    \int x^2e^{-x^2}
    u=x^2
    dv=e^{-x^2}dx
    du=2xdx
    v=xe^{-x^2}+2\int x^2e^{-x^2}dx

    I've also tried it with letting u=e^{-x^2}
    and I end up getting stuck as well.
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    Quote Originally Posted by paupsers View Post
    \int x^2e^{-x^2}
    u=x^2
    dv=e^{-x^2}dx
    du=2xdx
    v=xe^{-x^2}+2\int x^2e^{-x^2}dx

    I've also tried it with letting u=e^{-x^2}
    and I end up getting stuck as well.
    Parts won't work this time, my apologies

    Wolfram Mathematica Online Integrator
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  8. #8
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    You can do this integral using doble integrals, or some know values of the gamma function.
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  9. #9
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    this can be done.

    Choose
    u=x
    dv=xe^{-x^2}dx

    Notice that then v= \frac{-e^{-x^2}}{2}

    You will then notice that the uv term of the integration by parts is 0 (remember that for definite integrals, this term gets evaluated at the main integral's bounds, i.e., \infty and 0.) What you're left with is
    \frac{1}{2}\int_0^\infty e^{-x^2}


    A rather cool method for that one will use polar coordinates. here's a link:
    Gaussian integral - Wikipedia, the free encyclopedia
    (note the limits of integration for the Gaussian integral: You will need to divide by 2 (justified as the integrand is an even function)).
    Last edited by vince; February 7th 2010 at 02:29 PM. Reason: added words to describe better
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  10. #10
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    Quote Originally Posted by vince View Post
    this can be done.

    Choose
    u=x
    dv=xe^{-x^2}dx

    Notice that then v= \frac{-e^{-x^2}}{2}

    You will then notice that the uv term of the integration by parts is 0 (remember that for definite integrals, this term gets evaluated at the main integral's bounds, i.e., \infty and 0.) What you're left with is
    \frac{1}{2}\int_0^\infty e^{-x^2}


    A rather cool method for that one will use polar coordinates. here's a link:
    Gaussian integral - Wikipedia, the free encyclopedia
    (note the limits of integration for the Gaussian integral: You will need to divide by 2 (justified as the integrand is an even function)).
    Hmmm, I don't understand why the uv term is 0.
    If u = x and v = -(1/2)e^(-x^2), why does multiplying them make it 0?
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    Oops, I just saw that you edited your post to explain that part. Thanks!!
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  12. #12
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    Quote Originally Posted by paupsers View Post
    Oops, I just saw that you edited your post to explain that part. Thanks!!

    you may be compelled to click 'thanks'...im a thanks hog :]
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  13. #13
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    Use the fact that  \int_{0}^{\infty} e^{-x^{2}} \ dx = \frac{\sqrt{\pi}}{2}

    Make the substitution  x^{2}= au^{2}

    which implies that  2xdx=2audu

    so dx=\frac{au}{x} dx = \frac{ax}{\sqrt{a}x}dx = \sqrt{a} dx

    then   \int^{\infty}_{0} e^{-x^{2}} \ dx = \sqrt{a} \int^{\infty}_{0} e^{-au^{2}} du = \frac{\sqrt{\pi}}{2}

    and  \int^{\infty}_{0} e^{-au^{2}} \ du =  \frac{1}{2} \sqrt{\frac{\pi}{a}}

    notice that  - \int^{\infty}_{0} \frac{\partial}{\partial a} \ e^{-au^{2}} \ du =  \int^{\infty}_{0} u^{2} e^{-au^{2}} \ du

    now swtich the order of integration and differentiation (which in this case is allowed)

     - \frac{\partial}{\partial a} \int^{\infty}_{0}  e^{-au^{2}} \ du = - \frac{\partial}{\partial a} \ \frac{1}{2} \sqrt{\frac{\pi}{a}} =  \ \frac{\sqrt{\pi}}{4}a^{-3/2}

    so  \int^{\infty}_{0} u^{2} e^{-au^{2}} \ du = \frac{\sqrt{\pi}}{4}a^{-3/2}

    letting a=1

     \int^{\infty}_{0} u^{2} e^{-u^{2}} \ du = \frac{\sqrt{\pi}}{4}
    Last edited by Random Variable; February 7th 2010 at 04:04 PM.
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  14. #14
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    I=\int_0^{\infty}x^2e^{-x^2}dx

    Set x^2=t then I=\dfrac{1}{2}\int_0^{\infty}\sqrt{t}e^{-t}=\dfrac{1}{2}\Gamma(\dfrac{1}{2})=\dfrac{\sqrt{\  pi}}{4}
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