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Double integral to single by magic substitution

  1. Jul 9, 2010 #1
    double integral to single by "magic" substitution

    Hi,

    I have a double (actually quadruple, but the other dimensions don't matter here) integral which looks like this:

    [tex] \iint_0^\infty \frac{d^2 k}{k^2} [/tex]

    Now, someone here told me to replace that with

    [tex] \int_0^\infty \frac{1}{2} 2\pi \frac{d k^2}{k^2} [/tex]

    How and why is this ok? Thanks!

    PS: For anyone wondering where this comes up: it's related to Elementary Particle physics and structure function calculation.
     
  2. jcsd
  3. Jul 9, 2010 #2
    Re: double integral to single by "magic" substitution

    If your first integral is meant to be:

    [tex] I= \int_{-\infty}^\infty \int_0^\infty \frac{\mathrm{d}k_1\, \mathrm{d} k_2}{k_1^2 + k_2^2} [/tex]

    then you *could* use polar coordinates to give:

    [tex] I = \int_0^\pi \int_0^\infty \frac{ r\mathrm{d} r\, \mathrm{d} \theta}{r^2} = \pi \int_0^\infty \frac{ \mathrm{d} r}{r} [/tex]

    which seems to be the same as what you have written. However, the integral doesn't make a lot of sense - it fails to converge (logarithmic singularity). Although perhaps if this is from particle physics, you don't mind this divergence.
     
  4. Jul 10, 2010 #3
    Re: double integral to single by "magic" substitution

    Thanks, that looks good. The integrand isn't complete here (waaaay too big to put here).
     
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