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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

    can be evaluated to the value of $\displaystyle \pi$ 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

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

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

    Here's my attempt:

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

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

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

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

    $\displaystyle = I^2$.

    You can also say


    Therefore

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



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

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

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

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

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


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

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

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

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


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

    Here's my attempt:

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

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

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

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

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

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

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

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

    $\displaystyle du = 2r\,dr$.


    So the integral becomes

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

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

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

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

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

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

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

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

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

    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.

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

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

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

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

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

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

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

    $\displaystyle = 2\pi$.


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


    We have shown that

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

    So that means

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


    Therefore

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

    which is what we require for any probability density function.
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