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Math Help - Double Integration

  1. #1
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    Double Integration

    Any help with the following problem is appreciated (I have no idea how to approach it):

    Consider the integral I= \int^{\infty}_{-\infty} e^{-x^2} dx. Express I^2 as a double integral involving x and y.
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    Quote Originally Posted by demode View Post
    Any help with the following problem is appreciated:

    Consider the integral I= \int^{\infty}_{-\infty} e^{-x^2} dx. Express I^2 as a double integral involving x and y.
    It's well known that

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

    So if I = \sqrt{\pi}, surely I^2 = \pi...
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    Quote Originally Posted by Prove It View Post
    It's well known that

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

    So if I = \sqrt{\pi}, surely I^2 = \pi...
    But the problem says "express I^2 as a double integral". How do we need to express it like that?
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    I= \int_{-\infty}^\infty e^{-x^2}dx

    I= \int_{-\infty}^\infty e^{-y^2}dy since that is just the same integral with "dummy variable" y rather than x.

    Then I^2= \left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty}^\infty e^{-y^2}dy\right)

    by Fubini's theorem, that is the same as
    I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-x^2}e^{-y^2} dy dx

    I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-(x^2+y^2)} dy dx
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    Quote Originally Posted by HallsofIvy View Post
    I= \int_{-\infty}^\infty e^{-x^2}dx

    I= \int_{-\infty}^\infty e^{-y^2}dy since that is just the same integral with "dummy variable" y rather than x.

    Then I^2= \left(\int_{-\infty}^\infty e^{-x^2}dx\right)\left(\int_{-\infty}^\infty e^{-y^2}dy\right)

    by Fubini's theorem, that is the same as
    I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-x^2}e^{-y^2} dy dx

    I^2= \int_{x= -\infty}^\infty \int_{y= -\infty}^\infty e^{-(x^2+y^2)} dy dx
    It should also be pointed out that this double integral

    I^2= \int_{-\infty}^\infty \int_{-\infty}^\infty  e^{-(x^2+y^2)} \,dy\, dx

    can be evaluated to the value of \pi by converting to polars...
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    Quote Originally Posted by Prove It View Post
    It should also be pointed out that this double integral

    I^2= \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2+y^2)} \,dy\, dx

    can be evaluated to the value of \pi by converting to polars...
    How did you evaluate it to the value of \pi?

    Here's my attempt:

    \int_{0}^\infty [e^{-(r^2cos^2 \theta+r^2sin^2 \theta)}]r \,dr

    = \lim_{k \to \infty}\int_{0}^k re^{-r^2} \,dr

    = \frac{-1}{2} \lim_{k \to \infty} e^{-r^2} \,dr

    But how does this equal to \pi?
    Last edited by demode; May 1st 2010 at 06:26 PM.
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    You know that

     = I^2.

    You can also say


    Therefore

    I^2 = \pi which means I = \sqrt{\pi}.



    Another handy application is using the Gaussian Integral to compute \Gamma\left(\frac{1}{2}\right).

    Since e^{-x^2} is an even function, that means

    \int_{-\infty}^{\infty}{e^{-x^2}\,dx} = 2\int_0^{\infty}{e^{-x^2}\,dx}.

    Using the substitution x =t^{\frac{1}{2}} yields

    -x^2 = -t and dx = \frac{1}{2}t^{-\frac{1}{2}}


    So that the integral 2\int_0^{\infty}{e^{-x^2}\,dx} becomes

    2\int_0^{\infty}{\frac{1}{2}\,e^{-t}\,t^{-\frac{1}{2}}\,dt}

     = \int_0^{\infty}{e^{-t}\,t^{-\frac{1}{2}}\,dt}

     = \Gamma\left(\frac{1}{2}\right).


    So that means \Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}.
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    Quote Originally Posted by demode View Post
    How did you evaluate it to the value of \pi?

    Here's my attempt:

    \int_{0}^\infty [e^{-(r^2cos^2 \theta+r^2sin^2 \theta)}]r \,dr

    = \lim_{k \to \infty}\int_{0}^k re^{-r^2} \,dr

    = \frac{-1}{2} \lim_{k \to \infty} e^{-r^2} \,dr

    But how does this equal to \pi?
    It was a double integral, remember?

    I^2= \int_{\theta= 0}^{ 2\pi} \int_{r= 0}^\infty r e^{-r^2}drd\theta = 2\pi \int_{r= 0}^\infty r e^{-r^2}dr

    Let u= r^2 so that du= \frac{1}{2}r dr and 2du= rdr. The integral becomes
    4\pi\int_0^\infty e^{-u}du.
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    Quote Originally Posted by HallsofIvy View Post
    It was a double integral, remember?

    I^2= \int_{\theta= 0}^{ 2\pi} \int_{r= 0}^\infty r e^{-r^2}drd\theta = 2\pi \int_{r= 0}^\infty r e^{-r^2}dr

    Let u= r^2 so that du= \frac{1}{2}r dr and 2du= rdr. The integral becomes
    4\pi\int_0^\infty e^{-u}du.
    Actually, I believe you'll find that

    du = 2r\,dr.


    So the integral becomes

    2\pi \int_{r= 0}^\infty {r\,e^{-r^2}\,d} = \pi \int_{r = 0}^{\infty}{e^{-r^2}\,2r\,dr}

     = \pi \int_{r = 0}^{\infty}{e^{-u}\,du}
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    Prove It,

    We know that I=\int^{\infty}_{-\infty}e^{-x^2}= \sqrt{\pi}

    If I want to integrate the density function of the normal distribution:

    \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}dx

    using the value of I, what would be a suitable change of variable in this case?
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  12. #12
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    Quote Originally Posted by demode View Post
    Prove It,

    We know that I=\int^{\infty}_{-\infty}e^{-x^2}= \sqrt{\pi}

    If I want to integrate the density function of the normal distribution:

    \int_{-\infty}^{\infty} \frac{1}{\sqrt{2 \pi}}e^{\frac{-x^2}{2}}dx

    using the value of I, what would be a suitable change of variable in this case?
    You can use the exact same change of variable as given to integrate the Gaussian function.

    I^2 = \left(\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx}\right)^2 = \left(\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx}\right)\left(\int_{-\infty}^{\infty}{e^{-\frac{1}{2}y^2}\,dy}\right)

     = \int_{-\infty}^{\infty}{\int_{-\infty}^{\infty}{e^{-\frac{1}{2}(x^2 + y^2)}\,dy}\,dx}

     = \int_0^{2\pi}{\int_0^{\infty}{e^{-\frac{1}{2}r^2}\,r\,dr}\,d\theta}

     = 2\pi \int_0^{\infty}{r\,e^{-\frac{1}{2}r^2}\,dr}

     = 2\pi \int_{-\infty}^0{e^u\,du}

     = 2\pi \left[e^0 - e^{-\infty}\right]

     = 2\pi \left[1 - 0\right]

     = 2\pi.


    Since I^2 = 2\pi that means I = \sqrt{2\pi}.


    We have shown that

    \int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx} = \sqrt{2\pi}

    So that means

    \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}{e^{-\frac{1}{2}x^2}\,dx} = \frac{1}{\sqrt{2\pi}}\sqrt{2\pi}.


    Therefore

    \int_{-\infty}^{\infty}{\frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}x^2}\,dx} = 1

    which is what we require for any probability density function.
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