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Finding a Gaussian integral

  1. Jan 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the Gaussian integral:

    [itex]I = \int_{-\infty}^{\infty} e^{-x^2-4x-1}dx [/itex]

    (That's all the information the task gives me, minus the [itex]I=[/itex], I just put it there to more easily show what I have tried to do)

    2. The attempt at a solution
    I tried to square [itex]I[/itex] and get a double integral:

    [itex]I^2 = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{(-x^2-4x-1)+(-y^2-4y-1)}dxdy [/itex]

    and then my plan was to convert to polar-coordinates, however, this is my first time ever with double-integrals and/or switching to polarcoordinates, and I am kind of lost because every single example on the internet use the standard [itex]e^{-x^2}[/itex] gaussian function(and it is easy to see [itex]r^2=x^2+y^2[/itex]). Anyone who can push me in the right direction(I'm not even sure what finding the Gassuian integral means(?))?
  2. jcsd
  3. Jan 13, 2012 #2


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    Complete the square. -x^2-4*x-1=(-(x+2)^2+3). Try using that.
  4. Jan 13, 2012 #3
    Tried that earlier on, but didn't get anywhere with it. Been trying it some more now, but I still don't see it though. I don't see the next step, I'm going to try some more though=) Thanks
  5. Jan 13, 2012 #4


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    Next step would be a change of variables, u=x+2. Keep thinking about it.
  6. Jan 13, 2012 #5
    There! At least I got to the same answer as Wolfram Alpha;

    [itex] I = \sqrt{\pi}e^{3}[/itex]

    I hope it is correct. The steps I did after changeing variables [itex]r^2=(x+2)^2 + (y+2)^2[/itex] was substituting


    [itex]\frac{du}{dr}= 2r[/itex]

    [itex]du = 2rdr[/itex]

    [itex]dr = \frac{du}{2r}[/itex]

    And just left the 6-constant alone all until I did the actual integral:

    [itex]2\pi \int_{0}^{\infty} re^{-u+6} \frac{du}{2r}[/itex]

    [itex]\pi \int_{0}^{\infty} e^{-u+6}du = \pi e^6[/itex]

    (I think it's correct)
    Thanks for the help=)
  7. Jan 13, 2012 #6
    wow that's great.
  8. Jan 13, 2012 #7


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    That is great. But you don't have the repeat the polar coordinate trick every time you see a Gaussian integral. After you've done it once, you should just remember (or look up) [itex]\int_{-\infty}^{\infty} e^{-u^2} du=\sqrt{\pi}[/itex]
  9. Jan 13, 2012 #8
    Yeah, the next task was somewhat similar, same objective, and it took only 3 minutes compared to the 3-4 hours of the last one =)

    Now it's on to triple-integrals and what will probably be the most fun weekend since school ended before christmas!

    Again, thanks=)
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