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Question about the symmetry of integrals

  1. Feb 6, 2014 #1
    O.K. , this question is inspired by a physics class I'm taking where we're working out the expectation values of wave functions, but I think the question really belongs in the math section. Thank you in advance for any help. Here goes nothing...
    We have a function ψ(x,y,z) = x e[itex]\sqrt{}x2 + y2+ z2[/itex]
    Now I want to integrate this over all space. So I switch to spherical polar include the Jacobian and I end up with a separable integral of sinθ cubed form 0 to pi which is zero. No problem. But here's where the question pops up. I go to calculate the same function except this time with a z in front of the exponential, and I get a different answer, because I no longer get the sin^3 when I switch to spherical. It just seems really odd to me. I apologize if my formatting leaves something to be desired I'm new to the site, and any help would be greatly appreciated. Any insight on why a arbitrary change would affect the behavior of the integral?
     
  2. jcsd
  3. Feb 6, 2014 #2

    ShayanJ

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    Gold Member

    In the first case(x in front),the integral is:
    [itex]
    \int_0^\infty \int_0^{2\pi} \int_0^\pi r\sin{\theta}\cos{\varphi} e^r r^2\sin{\theta}d\theta d\varphi dr=\int_0^\infty r^3 e^r dr \int_0^{2\pi} \cos\varphi d\varphi \int_0^\pi \sin^2{\theta} d\theta
    [/itex]
    The [itex] \varphi [/itex] integral is equal to zero so the answer is zero.
    In the second case(z in front),the integral is:
    [itex]
    \int_0^\infty \int_0^{2\pi} \int_0^\pi r\cos{\theta} e^r r^2\sin{\theta}d\theta d\varphi dr=
    \int_0^\infty r^3 e^r dr \int_0^{2\pi} d\varphi \int_0^\pi \cos{\theta}\sin{\theta}d\theta
    [/itex]
    Now the [itex] \theta [/itex] integral is zero and so again the answer is zero.
     
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