- #1
kassem84
- 13
- 0
Hello,
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.
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.