Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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


    User Avatar
    Gold Member

    In the first case(x in front),the integral is:
    \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
    The [itex] \varphi [/itex] integral is equal to zero so the answer is zero.
    In the second case(z in front),the integral is:
    \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
    Now the [itex] \theta [/itex] integral is zero and so again the answer is zero.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook