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Integration in d-dimension

  1. Oct 7, 2011 #1
    As you may know in the context of dimensional regularization, integration is performed in d-dimension where d can take non-integer values. For example:
    [itex]\int d^{d}q f(q^2)[/itex]=S[itex]_{D}[/itex][itex]\int_{0}^{∞}q^{q-1}f(q^2)dq[/itex]
    My questions are:
    1) Is the integration in d-dimension performed is well defined in the mathematical sense when integration is done for: 1<|q|<∞ rather than 0<q<∞:
    [itex]\int_{|q|>1} d^{d}q f(q^2)[/itex]=S[itex]_{D}[/itex][itex]\int_{1}^{∞}q^{q-1}f(q^2)dq[/itex]

    2)Are tadpole mass-less integrals still equal to zero in DR?
    [itex]\int_{|q|>1} d^{d}q (q^2)^{β}[/itex]= 0 for β=1,2,...

    Thanks in advance.
    Best regards.
  2. jcsd
  3. Oct 7, 2011 #2
    With regards to 1) I think that the integral should be fine. Just imagine
    f(q2) = θ(q2-1)g(q2)

    But as for 2) and the massless tadpoles, no, you can no longer consistently set them to zero.

    Normally, for the integral over all q, you can argue for its vanishing in two ways:

    a) Dimensional analysis: The only dimensional quantity is q, with [q]=1. So The integral has dimension d + 2β. Since you integrate out q, no dimensional quantities remain, so the only consistent result is that the integral must be 0.
    b) The integral does not converge for any particular choice of d. But splitting the integral in half (or wherever) and choosing different d's for each half so that it converges in both UV and IR, then recombining the integral, the UV and IR poles cancel and the result is zero.

    With your integral, both arguments don't work. For a), your "1" actually has the dimensions of mass (since otherwise |q|>1 would not make sense) and so the argument fails.
    For b), you no longer have an IR limit, so there is no cancellation with the UV.

    Hope that helps,
  4. Oct 7, 2011 #3
    Thanks Simon for your answers.
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