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Thread: Evaluating a double integral to find its signle integral.

  1. #1
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    Evaluating a double integral to find its signle integral.

    Evaluate I=integral[e^(-x^2)dx] from 0 to infinity,
    by computing I^2= double integral[[e^(-x^2)e^(-y^2)dxdy]] both integrals from 0 to infinity. (hint: use appropriate coordinates (i think they mean maybe polar??))

    (hope you can understand the math i wrote) Thanks
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by flawless View Post
    i think they mean maybe polar??
    Yes, the polar coordinates $\displaystyle (r,\,\theta)$ are the appropriate ones. What can you get from it ?
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  3. #3
    Moo
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    Hello,

    Quote Originally Posted by flawless View Post
    Evaluate I=integral[e^(-x^2)dx] from 0 to infinity,
    by computing I^2= double integral[[e^(-x^2)e^(-y^2)dxdy]] both integrals from 0 to infinity. (hint: use appropriate coordinates (i think they mean maybe polar??))

    (hope you can understand the math i wrote) Thanks
    Yes, it's polar coordinates.

    $\displaystyle x=r \cos \theta$
    $\displaystyle y=r \sin \theta$

    Therefore, $\displaystyle dxdy$ becomes $\displaystyle r dr d\theta$ (I know it is the determinant of the Jacobian matrix, but I don't know how you call it).

    --> $\displaystyle I^2=\int_0^\infty \int_0^\infty e^{-x^2-y^2} dx dy=\int_0^{\frac{\pi}{2}} \int_0^\infty r e^{-r^2} dr d \theta=\frac{\pi}{2} \cdot \frac{1}{2}=\frac{\pi}{4}$

    $\displaystyle \implies I=\frac{\sqrt{\pi}}{2}$
    Last edited by Moo; May 18th 2008 at 12:06 AM. Reason: mistakes
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  4. #4
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by Moo View Post
    --> $\displaystyle I^2=\int_0^\infty \int_0^\infty e^{-x^2-y^2} dx dy=\int_0^{{\color{red}2 \pi}} \int_0^\infty r e^{-r^2} dr d \theta=2 \pi \cdot \frac{1}{2}=\pi$
    $\displaystyle x$ and $\displaystyle y$ are positive hence $\displaystyle \theta$ lies in $\displaystyle \left[0,\frac{\pi}{2}\right]$.
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  5. #5
    Moo
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    I forgot it
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  6. #6
    Super Member wingless's Avatar
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    $\displaystyle \int_{0}^{\infty}e^{-x^2}~dx$

    Let's write it two times.

    $\displaystyle \int_{0}^{\infty}e^{-x^2}~dx~ \int_{0}^{\infty}e^{-x^2}~dx$

    Let x be y on the second integral.

    $\displaystyle \int_{0}^{\infty}e^{-x^2}~dx~ \int_{0}^{\infty}e^{-y^2}~dy$

    By the properties of double integral:

    $\displaystyle \int_{0}^{\infty}\int_{0}^{\infty}e^{-x^2-y^2}~dx~dy$

    Let $\displaystyle x = r\cos \theta$ and $\displaystyle y = r\sin \theta$.
    Then $\displaystyle -x^2-y^2 = -r^2$ and $\displaystyle dx~dy = r~dr~d\theta$.

    $\displaystyle \int_{0}^{\frac{\pi}{2}}\int_{0}^{\infty}e^{-r^2}r~dr~d\theta$

    $\displaystyle \int_{0}^{\frac{\pi}{2}}~d\theta~\int_{0}^{\infty} r e^{-r^2}~dr$

    $\displaystyle \frac{\pi}{2} \cdot \frac{1}{2} = \frac{\pi}{4}$

    So,

    $\displaystyle \int_{0}^{\infty}e^{-x^2}~dx~ \int_{0}^{\infty}e^{-x^2}~dx = \frac{\pi}{4}$

    $\displaystyle \int_{0}^{\infty}e^{-x^2}~dx = \frac{\sqrt{\pi}}{2}$
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  7. #7
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    Let $\displaystyle 0<z<1.$ Start by showing that $\displaystyle \Gamma (z)\Gamma (1-z)=\int_{0}^{\infty }{\frac{x^{z-1}}{1+x}\,dx}.$ Set $\displaystyle z=\frac12\implies\Gamma \left( \frac{1}{2} \right)^{2}=\pi \,\therefore \,\Gamma \left( \frac{1}{2} \right)=\sqrt{\pi }.$ Finally $\displaystyle \int_{0}^{\infty }{e^{-x^{2}}\,dx}=\frac{1}{2}\Gamma \left( \frac{1}{2} \right)=\frac{\sqrt{\pi }}{2}$ as required. $\displaystyle \blacksquare$

    (There're may proofs anyway.)
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