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Gaussian integral

  1. Jul 26, 2014 #1
    1. The problem statement, all variables and given/known data

    I am asked to evaluate ##\displaystyle\int_{-\infty}^{\infty} 3e^{-8x^2}dx##

    2. Relevant equations

    I know

    ##\displaystyle\int_{-\infty}^{\infty} e^{-x^2}dx = \sqrt{\pi}##


    3. The attempt at a solution
    based on an example in the book it seems a change of variables is what I need to do.

    I'm just not sure how
     
  2. jcsd
  3. Jul 26, 2014 #2
    I figured it out

    ##3\sqrt{\frac{\pi}{8}}##

    still a bit puzzled on how I show that though
     
  4. Jul 26, 2014 #3

    SteamKing

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    Have you tried substitution?
     
  5. Jul 26, 2014 #4

    exo

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    You can try setting it up as a double integral and doing it in polar coordinates.
     
  6. Jul 26, 2014 #5

    LCKurtz

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    Now I'm puzzled. You figured it out but now you don't know how you did it???
     
  7. Jul 26, 2014 #6
    I used what I knew of the gaussian integral

    and when it is of the form ##\displaystyle\int_{-\infty}^{\infty} e^{-ax^2}dx##

    the answer is ##\sqrt{\frac{\pi}{a}}## and the 3 is just a constant so that factored out of the integral

    so ##3\sqrt{\frac{\pi}{8}}##

    I'll try it with polar coordinates though, this way was a bit like "cheating"
     
  8. Jul 26, 2014 #7

    exo

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    Guess I am going to give it a go.You can set it up as

    [itex]\displaystyle\int^{\infty}_{-\infty}\int^{\infty}_{-\infty}e^{-8x^{2}-8y^{2}}dxdy[/itex]

    Now this can be rewritten as

    [itex]\displaystyle\int^{\infty}_{-\infty}e^{-8y^{2}}\int^{\infty}_{-\infty}e^{-8x^{2}}dxdy[/itex]

    If we label the original integral [itex]A[/itex] solving this double integral will give us [itex]A^{2}[/itex], so if we manage to solve it we can just take the square root and get our answer.

    Let's use polar coordinates then our integral becomes

    [itex]\displaystyle\int^{\infty}_{0}\int^{2\pi}_{0}re^{-8r^{2}}d\theta dr = [/itex]

    [itex]= 2 \pi \displaystyle\int^{\infty}_{0}re^{-8r^{2}}dr = [/itex]

    [itex]= \left[ 2 \pi \frac{(-e^{-8r^{2}})}{16} \right]_{0}^{\infty} = \frac{\pi}{8}[/itex]

    So [itex]A = \sqrt{\frac{\pi}{8}}[/itex]
     
    Last edited: Jul 26, 2014
  9. Jul 26, 2014 #8

    HallsofIvy

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    No, using the fact that [itex]\int 3 f(x)dx= 3\int f(x)dx[/itex] is NOT cheating!

    If you are concerned about how they got that [itex]\int_{-\infty}^\infty e^{-ax^2} dx= \sqrt{\pi/a}[/itex]-

    Do you know how to show that [itex]\int_{-\infty}^\infty e^{-x^2}dx= \sqrt{\pi}[/itex]? If not, "set it up as a double integral and change to polar coordinates" as exo suggested. Once you know that, because you have [itex]e^{-ax^2}[/itex] and want [itex]e^{-u^2}[/itex] use the substitution [itex]-u^2= ax^2[/itex] or [itex]u= \sqrt{a} x[/itex].
     
  10. Jul 26, 2014 #9

    I didn't mean cheating was about pulling the constant outside the integral. I meant I'd rather work the problem out properly than use a known formula to get skip to the end. And of course that isn't cheating either, moreso cheating myself of the fun of solving a neat integral.

    I did it using polar coordinates, I got to the same answer.
     
  11. Jul 27, 2014 #10

    SammyS

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    Let's see the details of that!

    I'm intrigued .
     
  12. Jul 27, 2014 #11

    I'll let G = ##\int_{-\infty}^{\infty} 3e^{-8x^2}dx##

    and ##G^2 =\int_{-\infty}^{\infty} \int_{-\infty}^{\infty}9e^{-8(x^2+y^2)}dxdy##

    Changing to polar

    ##\int_{0}^{2\pi} \int_{0}^{\infty}9e^{-8r^2}rdrd\theta##

    the outside integral doesn't rely and is not affect by the one on the inside

    ##\int_{0}^{2\pi} d\theta \int_{0}^{\infty}9e^{-8r^2}rdr##

    I'm going to introduce the limit now, previously left out for expedience

    ##2\pi\Bigg[\lim_{a \to \infty}\int_{0}^{\infty}9e^{-8r^2}rdr\Bigg]##

    ##9*2\pi\Bigg[\lim_{a \to \infty}\int_{0}^{\infty}e^{-8r^2}rdr\Bigg]##

    ##u=8r^2; du=16rdr \rightarrow \frac{du}{16r}=dr##

    ##9*2\pi\Bigg[\lim_{a \to \infty}\int_{0}^{\infty}e^{-u^2}r\frac{du}{16r}\Bigg]##

    ##9*2\pi\Bigg[\lim_{a \to \infty}\frac{1}{16}\int_{0}^{\infty}e^{-u^2}du\Bigg]##

    ##9*2\pi \Bigg[\lim_{a \to \infty}\frac{1}{16}(-e^{-8x^2})\Bigg|_{0}^{a}\Bigg]##

    ##9\Bigg[\lim_{a \to \infty}\frac{\pi}{8}(-e^{-8a^2}+e^{0})\Bigg]##

    ##9\cdot\frac{\pi}{8} = G^2##

    ##3\sqrt{\frac{\pi}{8}}=G##
     
  13. Jul 27, 2014 #12

    HallsofIvy

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    You mean "[itex]\lim_{a\to\infty}\int_0^a[/itex]".

     
  14. Jul 27, 2014 #13
    Oops, yeah thanks. It was the end of the night. Its worked out properly on paper.
     
    Last edited: Jul 27, 2014
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