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Integral of e raised to a power (a function)

  1. May 14, 2013 #1
    I was trying to understand something about this:

    1. The problem statement, all variables and given/known data

    [tex]\int_{-\infty}^\infty \mathrm{e}^{-2x^2}\,\mathrm{d}x[/tex]


    3. The attempt at a solution

    According to the solution given, this can be set up where you say:

    let

    [tex]I = \Big(\int_{-\infty}^\infty \mathrm{e}^{-2x}\,\mathrm{d}x\Big)[/tex]

    and

    [tex]I^2 = I \cdot\ I = \Big(\int_{-\infty}^\infty \mathrm{e}^{-2x}\,\mathrm{d}x\Big)\Big(\int_{-\infty}^\infty \mathrm{e}^{-2y}\,\mathrm{d}x\Big)[/tex]

    Well, OK, I am just trying to understand the logic of this step. It seems just completely out of the blue for me and arbitrary. What was done here? The next step says that you have a double integral where e is raised to the (x2+y2) power, and you then convert to polar coordinates. That last bit actually makes sense to me. It's getting there I seem to stumble on.
     
  2. jcsd
  3. May 14, 2013 #2

    haruspex

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    The second integral should be dy, of course.
    Yes, the method of solving this integral is quite brilliant, and it does seem out of the blue. But are you just asking how anyone thought of this, or is there a step you do not follow?
     
  4. May 14, 2013 #3
    I'm not following a step. OK, I can see setting up an integral of some kind like a u-substitution, but what was done here just baffles me. It's nice that it is brilliant, but what did the person do? What is going on here?
     
  5. May 14, 2013 #4

    Dick

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    If you want to understand the proof of http://en.wikipedia.org/wiki/Gaussian_integral you'd better go back to your source for the proof. You've got some stuff wrong. An integral like e^(-2x) won't even converge.
     
  6. May 14, 2013 #5
    That is what I am trying to figure out. This problem is on a practice exam and if there is a mistake in the solution, fine, but that is what I am trying to figure out. But being the Vector Calc student here, I can't necessarily tell.

    So let's try this: is there a mistake in this solution? A typo in the original problem? (I looked again and aside from mistaking the dx in the second bit where I outline the solution, it is exactly as presented).

    Is there another solution method that is more common or something?

    I saw the proof of integrating e^(x^2) and that's all well and good, does this fall in the same category? Is this technique wrong because the power raised is -2x^2?

    If I have something wrong then tell me what it is, please -- I want to understand this but I'm coming at this as someone who is utterly baffled by this solution and I want to understand what I am seeing.
     
  7. May 14, 2013 #6

    Dick

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    If the problem is to integrate e^(-2x^2) you can really figure that out if you've already done e^(-x^2). My problem here starts with $$I = \Big(\int_{-\infty}^\infty \mathrm{e}^{-2x}\,\mathrm{d}x\Big)$$
    That integral doesn't even converge.

    If this is a solution presented on a practice exam, somebody was really sloppy. Or did you mean:
    $$I = \Big(\int_{-\infty}^\infty \mathrm{e}^{-2x^2}\,\mathrm{d}x\Big)$$
     
  8. May 14, 2013 #7
    (Sigh) ANd the problem is????

    I know y'all want people to understand this stuff, and work through things, but I feel like you're being a bit mysterious. "It doesn't converge" means what, exactly? (I am not always up on my mathematical terminology in certain contexts). If I know what's wrong with the solution (as presented) that would go a long way. But right now I can't even begin to figure out what's wrong.

    It's nice that *you* know they were "sloppy," but remember I am not going to know that, right? So if I can tease out what the solution SHOULD look like that would help.

    Pretend you're dealing with the dumbest math student in the class :-)
     
  9. May 14, 2013 #8

    Dick

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    "Not converge" means it's not finite. It doesn't exist. The 'I' integrals in the 'solution' part don't have anything to do with the original integral. Did you already do the proof that $$\Big(\int_{-\infty}^\infty \mathrm{e}^{-x^2}\,\mathrm{d}x\Big)=\sqrt \pi$$?
     
  10. May 14, 2013 #9
    No, we didn't do that proof in class, and I had to look it up with the link. And even knowing that, it doesn't necesasarily tell me what happens when I multiply the -X^2 by 2. (I suppose I could, by using the first method in the Wikipedia link, set up the integral as

    [tex]\int_{-\infty}^\infty\int_{-\infty}^\infty \mathrm{e}^{-(\sqrt2x^2+\sqrt2y^2}\,\mathrm{d}x\mathrm{d}y[/tex]

    but I have no idea if that's allowed, as it were).
     
  11. May 14, 2013 #10

    Dick

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    If you define ##I = \Big(\int_{-\infty}^\infty \mathrm{e}^{-2x^2}\,\mathrm{d}x\Big)## then I*I is just,
    [tex]\int_{-\infty}^\infty\int_{-\infty}^\infty \mathrm{e}^{-2(x^2+y^2)}\,\mathrm{d}x\mathrm{d}y[/tex]

    You should be able to use the technique in the Wikipedia link on that.
     
  12. May 14, 2013 #11
    OK, now it's starting to make a bit more sense. If I ignore the solution that was posted by the instructor :-)

    I'll have to ask him about that tho.
     
  13. May 14, 2013 #12

    Mark44

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    Since you're working with integrals with infinite limits, your book should have discussed the concept of convergence. Some integrals of this type converge (result in a finite number) and some diverge (are unbounded).
     
  14. May 14, 2013 #13
    @Mark44 -- it is probably in there and I forgot about it, understand I am taking up vector calculus after an absence from academia of two decades. I re-studied my basic calc before going in (basically I went to the beginning of my math book from college and worked my way through it chapter by chapter). But even so, sometimes there's terminology I haven't heard in a while, or simple rules and identities I might miss. (I spend a lot of time on one problem because I didn't "know" that sin (arctan x) has a specific identity associated with it, and ended up sort of re-deriving it... a mess all around)

    I'm actually doing ok in the class, BTW. But it's been a slog getting caught up again, and I want to do well.
     
  15. May 14, 2013 #14

    Ray Vickson

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    If you know the value of
    [tex] I = \int_{-\infty}^{\infty} e^{-x^2} \, dx [/tex]
    then it is easy to get the value of
    [tex] J = \int_{-\infty}^{\infty} e^{- a x^2} \, dx [/tex]
    for any value of ##a > 0.## This just involves a change of variables to ##y = \sqrt{a} x##; that changes ##e^{-ax^2}## to ##e^{-y^2},## and you just need to figure out how to relate ##dx## to ##dy##.
     
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