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Math Help - improper integral

  1. #1
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    improper integral

    how to evaluate ?
    thankyou!
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  2. #2
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    Hello,
    Quote Originally Posted by iwonder View Post
    how to evaluate ?
    thankyou!
    It's a common technique
    I=\int_{-\infty}^\infty e^{-3x^2} ~ dx

    Hence I^2=\left(\int_{-\infty}^\infty e^{-3x^2} ~ dx\right) \cdot \left(\int_{-\infty}^\infty e^{-3y^2} ~ dy\right)

    I^2=\int_{y=-\infty}^\infty \int_{x=-\infty}^\infty e^{-3x^2} \cdot e^{-3y^2} ~ dx ~ dy

    I^2=\iint_{\mathbb{R}^2} e^{-3(x^2+y^2)} ~ dx ~ dy


    And you're back to a previous integral
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  3. #3
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    and i got the answer for this is pi/3
    however when i submit the answer online, it says the answer
    is not equal to
    plz exaplin
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  4. #4
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    Quote Originally Posted by iwonder View Post

    and i got the answer for this is pi/3
    however when i submit the answer online, it says the answer
    is not equal to
    plz exaplin
    It's I^2, so take the square root !

    I^2= \frac \pi 3 \implies I= \sqrt{\frac \pi 3} (and not -, since the integrand is always positive, its integral is positive)
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  5. #5
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    Wink

    Quote Originally Posted by iwonder View Post
    how to evaluate ?
    thankyou!
    If you know the \Gamma function the solution is very simple.

    Noting that the function is even, we have
    \int\limits_{-\infty}^{\infty}\mathrm{e}^{-3x^{2}}dx\stackrel{y\to\sqrt{3}x}{=}\frac{2\sqrt{3  }}{3}\int\limits_{0}^{\infty}\mathrm{e}^{-y^{2}}dy
    .................. \stackrel{z\to y^{2}}{=}\frac{\sqrt{3}}{3}\int\limits_{0}^{\infty  }z^{-1/2}\mathrm{e}^{-z}dz
    .................... =\frac{\sqrt{3}}{3}\Gamma(1/2)
    .................... =\frac{\sqrt{3\pi}}{3}.

    Also see Gamma function - Wikipedia, the free encyclopedia
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