Is the substitution of the numerator in QFT scattering amplitudes always valid?

[ismaili]

When we are calculating the scattering amplitudes in QFT, we often encounter something like

\int \frac{d^Dp}{(2\pi)^D} \frac{p^\mu p^\nu}{(p^2+\Delta)^n}

and we often make the substitution for the numerator

p^\mu p^\nu \rightarrow \frac{g^{\mu\nu}p^2}{D}

It looks like reasonable but I don't know how to prove it.
However, I wonder if this expression is valid in any situation?
Is it correct under certain regularisation scheme, or it's correct in any case?

 

[The_Duck]

Note that the integral is invariant if you simultaneously Lorentz transform both indices. This means it has to be proportional to g^{\mu\nu}. To determine the proportionality constant you can evaluate

\int \frac{d^Dp}{(2\pi)^D} \frac{p^2}{(p^2+\Delta)^n} = \int \frac{d^Dp}{(2\pi)^D} \frac{(p^0)^2 - (p^1)^2 - (p^2)^2 - (p^3)^2}{(p^2+\Delta)^n} = 4 \int \frac{d^Dp}{(2\pi)^D} \frac{p^0 p^0}{(p^2+\Delta)^n}

 

[ismaili]

Note that the integral is invariant if you simultaneously Lorentz transform both indices. This means it has to be proportional to g^{\mu\nu}. To determine the proportionality constant you can evaluate

\int \frac{d^Dp}{(2\pi)^D} \frac{p^2}{(p^2+\Delta)^n} = \int \frac{d^Dp}{(2\pi)^D} \frac{(p^0)^2 - (p^1)^2 - (p^2)^2 - (p^3)^2}{(p^2+\Delta)^n} = 4 \int \frac{d^Dp}{(2\pi)^D} \frac{p^0 p^0}{(p^2+\Delta)^n}

Thank you very much.
This looks reasonable, so it is valid generically because it's from Lorentz transformation property.
But, this leads to a peculiar result:
I am calculating the self-energy diagram of photon in 2D QED,
The divergent part of the integral is:

\int \frac{d^2p}{(2\pi)^2} \frac{-2p^\mu p^\nu + g^{\mu\nu}p^2}{(p^2+\Delta)^2}

If I put p^\mu p^\nu = p^2/2, the divergent terms cancel each other!
This means I don't need regularisation, which is quite unreasonable.
How could this happen?