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Integrating problem

  1. Jan 22, 2012 #1

    georg gill

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

    Integrate:

    [tex]\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}e^{-2\sqrt{x^2+y^2+z^2}}dxdydz[/tex]

    hint: use spherical integration



    2. Relevant equations

    [tex]p=\sqrt{x^2+y^2+z^2}[/tex]

    [tex]dV=dp d\phi p sin\phi p d\theta[/tex]

    3. The attempt at a solution

    [tex]\int_{0}^{\infty}\int_{0}^{\pi}\int_{0}^{2\pi}e^{-2p} p^2 sin\phi d\theta d\phi dp[/tex]

    look at integration for p since it is the most hard one first:

    Integration by parts:

    [tex]sin\phi\int_{0}^{\infty}e^{-2p} p^2 dp[/tex]

    [tex]\int_{0}^{\infty}e^{-2p} p^2 dp=[\frac{e^{-2p}}{-2} p^2+\int_{0}^{\infty}e^{-2p}pdp]^{\infty}_0[/tex] (I)

    [tex]\int_{0}^{\infty}e^{-2p}pdp=[\frac{e^{-2p}}{-2}p+\int_{0}^{\infty}e^{-2p}dp]^{\infty}_0=[\frac{e^{-2p}}{-2}p+\frac{e^{-2p}}{-2}]^{\infty}_0[/tex]

    and (I) becomes:

    [tex][\frac{e^{-2p}}{-2} p^2+\frac{e^{-2p}}{-2}p+\frac{e^{-2p}}{-2}]^{\infty}_0[/tex]

    but here i get infinity in numerator and denumerator for [tex]p=\infty[/tex]. How do I solve this?
     
  2. jcsd
  3. Jan 22, 2012 #2
    If you get infinity in the numerator as well as in the denumerator you should check which one goes to infinity fastest. In this case, p^2 approaches infinity much slower then e^(2p), so your fraction will become zero for p=infinity.
     
  4. Jan 22, 2012 #3

    Dick

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    Right. You can verify this by using l'Hopital's rule on p^2/e^(2*p).
     
  5. Jan 22, 2012 #4

    georg gill

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    I thought L'hopitals said what the limit was and that the limit said what value it was approaching as it went to that value. I just get confused in the accuracy of L'hopitals. Is it totally accurate in defining value of fraction?
     
  6. Jan 22, 2012 #5

    Ray Vickson

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    Your sentence "I thought L'hopitals said what the limit was and that the limit said what value it was approaching as it went to that value" is 100% incomprehensible. I don't know why you are worried about the "accuracy" of L'Hopital's rule: it is a rigorously proven theorem, not just some rule of thumb or heuristic; when it is applicable at all it is absolutely accurate, always.

    RGV
     
  7. Jan 22, 2012 #6

    georg gill

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    Good to know. I guess my worries was that a limit gives value as it approaches a given x. If this function was not continuous then then it could have had a different value for x which is here infinity. But L'hopitals assumes continuity since it assumes that function is differentiable

    [tex][\frac{e^{-2p}}{-2} p^2+\frac{e^{-2p}}{-2}p+\frac{e^{-2p}}{-2}]^{\infty}_0[/tex]

    [tex]\frac{e^{-2\infty}}{-2} (\infty)^2+\frac{e^{-2\infty}}{-2}\infty+\frac{e^{-2\infty}}{-2}-\frac{e^{-20}}{-2} 0^2+\frac{e^{-2\cdot 0}}{-2}0+\frac{e^{-2\cdot 0}}{-2}[/tex]

    [tex]-\frac{e^{-2\cdot 0}}{2}[/tex]

    Is it right that this integral could be negative? It is easier to see when integrals should be negative when they are in the plain. Is there any way of telling when it should be negative in the space?
     
  8. Jan 22, 2012 #7

    Dick

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    It shouldn't be negative. You are integrating a positive function. It's negative because you are missing a sign. Check it again. And it's not a good idea to substitute 'infinity' into an expression f(p). What you mean is 'lim p -> infinity f(p)'. And I think you also may have dropped a factor of 1/2 on one of the terms.
     
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