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Math Help - Integration, change of variables

  1. #1
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    Integration, change of variables

    Hi

    I have the standard integral

    {\displaystyle \int}_{-\infty}^{\infty} e^{-x^2} dx = \sqrt{\pi}

    But need to find what

    {\displaystyle \int}_{-\infty}^{\infty} e^{-\alpha x^2} dx is using the expression above

    I have found on the internet that it is \sqrt{\frac{\pi}{\alpha}} but can't see how to get it.

    I am assuming the use of changing the variable but can't see how to go about it.

    Thanks, James
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  2. #2
    MHF Contributor

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    Re: Integration, change of variables

    Try y= \sqrt{2}x.
    Thanks from bobred
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  3. #3
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    Re: Integration, change of variables

    I think he means to substitute y=\sqrt{\alpha}x, which should transform your integral into the standard integral.

    - Hollywood
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  4. #4
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    Re: Integration, change of variables

    Thanks.

    I can see how we get the substitution when we know the integral equals \sqrt{\frac{\pi}{\alpha}}, but how would we go about it not knowing what it equalled?

    Thanks
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  5. #5
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    Re: Integration, change of variables

    \displaystyle \begin{align*} \int_{-\infty}^{\infty}{e^{-\alpha \, x^2}\,dx} &= \int_{-\infty}^{\infty}{e^{-\left( \sqrt{\alpha} \, x \right) ^2 } \, dx} \\ &= \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}{\sqrt{\alpha}\, e^{ -\left( \sqrt{\alpha} \, x \right) ^2 } \, dx } \end{align*}

    Now make the substitution \displaystyle \begin{align*} u = \sqrt{\alpha} \, x \implies du = \sqrt{\alpha} \, dx \end{align*} and note that \displaystyle \begin{align*} u \end{align*} still goes to \displaystyle \begin{align*} -\infty \end{align*} and \displaystyle \begin{align*} \infty \end{align*} when \displaystyle \begin{align*} x \end{align*} does. Then the integral becomes

    \displaystyle \begin{align*} \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}{\sqrt{\alpha}\, e^{-\left( \sqrt{\alpha} \, x \right) ^2 } \, dx} &= \frac{1}{\sqrt{\alpha}}\int_{-\infty}^{\infty}{ e^{-u^2} \, du } \\ &= \frac{1}{\sqrt{\alpha}} \, \sqrt{\pi} \\ &= \frac{\sqrt{\pi}}{\sqrt{\alpha}} \\ &= \sqrt{\frac{\pi}{\alpha}} \end{align*}
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  6. #6
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    Re: Integration, change of variables

    Got it, thanks all.
    James
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