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Math Help - Integral

  1. #1
    MHF Contributor Mathstud28's Avatar
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    Integral

    Will someone check my work....I think this is right

    \int_0^{\infty}e^{-a^2x^2}dx,\forall{a}>0

    First Let\text{ }u=a^2x^2\Rightarrow{dx=\frac{1}{2a\sqrt{u}}}

    and u(0)=a^2\cdot{0^2}=0

    and u(\infty)=a^2\cdot{\infty}=\infty

    So this would be \frac{1}{2a}\int_0^{\infty}\frac{e^{-u}}{\sqrt{u}}du=\frac{1}{2a}\int_0^{\infty}u^{\fra  c{-1}{2}}e^{-u}du=\frac{\Gamma\bigg(\frac{1}{2}\bigg)}{2a}=\fra  c{\sqrt{\pi}}{2a}


    This doesnt seem right...I mean I think I did my calculations right but this was in a very simple book and this seems uncharcteristic of such a book ....

    Could someone verify?
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    Will someone check my work....I think this is right

    \int_0^{\infty}e^{-a^2x^2}dx,\forall{a}>0

    First Let\text{ }u=a^2x^2\Rightarrow{dx=\frac{1}{2a\sqrt{u}}}

    and u(0)=a^2\cdot{0^2}=0

    and u(\infty)=a^2\cdot{\infty}=\infty

    So this would be \frac{1}{2a}\int_0^{\infty}\frac{e^{-u}}{\sqrt{u}}du=\frac{1}{2a}\int_0^{\infty}u^{\fra  c{-1}{2}}e^{-u}du=\frac{\Gamma\bigg(\frac{1}{2}\bigg)}{2a}=\fra  c{\sqrt{\pi}}{2a}


    This doesnt seem right...I mean I think I did my calculations right but this was in a very simple book and this seems uncharcteristic of such a book ....

    Could someone verify?
    \int_{0}^{\infty}e^{-a^2x^2}\,dx \ \forall \ a>0
    let \ u=a^2x^2 \implies du=2a^2x\,dx\implies dx=\frac{du}{2a^2\sqrt{\frac{u}{a^2}}}\implies dx=\frac{du}{2a\sqrt{u}}

    u(0)=0 \ and \ u(\infty)=\infty.

    \therefore \frac{1}{2a}\int_{0}^{\infty}e^{-u}u^{-\frac{1}{2}}\,du=\frac{\Gamma\left(\frac{1}{2}\rig  ht)}{2a}=\frac{\sqrt{\pi}}{2a}.

    I get what you get.
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Chris L T521 View Post
    \int_{0}^{\infty}e^{-a^2x^2}\,dx \ \forall \ a>0
    let \ u=a^2x^2 \implies du=2a^2x\,dx\implies dx=\frac{du}{2a^2\sqrt{\frac{u}{a^2}}}\implies dx=\frac{du}{2a\sqrt{u}}

    u(0)=0 \ and \ u(\infty)=\infty.

    \therefore \frac{1}{2a}\int_{0}^{\infty}e^{-u}u^{-\frac{1}{2}}\,du=\frac{\Gamma\left(\frac{1}{2}\rig  ht)}{2a}=\frac{\sqrt{\pi}}{2a}...

    I get what you get.
    I didnt think this was right because this was juxtaposed beside this integral

    \int_2^3\frac{dx}{1-\cos(x)}


    So I was like...hmm
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