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Thread: help with this :(

  1. #1
    mms
    mms is offline
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    help with this :(

    let $\displaystyle
    f(x) = \left[ {\int\limits_0^x {e^{ - t^2 } dt} } \right]^2 \,\,\,\,\,\,g(x) = \left[ {\int\limits_0^1 {e^{ - x^2 (1 + t^2 )} } \cdot \frac{1}{{(1 + t^2 )}}} \right]^2
    $

    show that $\displaystyle
    f'(x) + g'(x) = 0
    $

    and show that $\displaystyle
    f(x) + g(x) = \frac{\pi }{4}

    $

    finally prove that $\displaystyle
    \int\limits_0^\infty {e^{ - t^2 } dt} = \frac{{\sqrt \pi }}{2}
    $

    i cant seem able to show that f'(x)+g'(x)=0, i think its beacause i cant derivate g(x). Help plz
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    first $\displaystyle g(x)$ was not meant to be defined as written, it's actually, without the square.

    let $\displaystyle h(x,t)=\frac{e^{-x^{2}\left( 1+t^{2} \right)}}{1+t^{2}},$ and $\displaystyle \frac{\partial h}{\partial x}=-2xe^{-x^{2}\left( 1+t^{2} \right)},$ both continuous, then $\displaystyle g'(x)=-2x\int_{0}^{1}{e^{-x^{2}\left( 1+t^{2} \right)}\,dt},$ and $\displaystyle f'(x)=2e^{-x^2}\int_0^x e^{-t^2}\,dt,$ now substitute $\displaystyle t=ux$ in this integral and that shows that $\displaystyle f'(x)+g'(x)=0,$ thus $\displaystyle h(x)=f(x)+g(x)$ is constant and $\displaystyle h(0)=\frac\pi4$ and then $\displaystyle h(x)=\frac\pi4,$ obviously $\displaystyle \lim_{x\to\infty}h(x)=\frac\pi4,$ then $\displaystyle \lim_{x\to\infty}h(x)=\left(\int_0^\infty e^{-t^2}\,dt\right)^2$ and the conclusion follows.
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  3. #3
    Senior Member
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    Mar 2010
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    A very nice way to find standart integral to be $\displaystyle \frac{\sqrt{\pi}}{2}$.
    Could you help to understand that t=ux the sum of derivatives =0.
    Thanks.
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