Generic Feynman parameterisation

[CAF123]

I am in the process of reducing tensor integrals down to a sum of scalar ones with the tensor structure factored out in some basis of decomposition. I am able to write some scalar products in terms of appearing propagators but I encountered one where I have something like $$\int_k \frac{k^2-m^2}{A_1(k) A_2(k) A_3(k) A_4(k)}$$ where $A_i$ are all different propagators depending on k and external momenta.

Now, turns out I can write $k^2 - m^2 = A_3 - f(p_i) - m^2 - 2k \cdot g(p_i)$. The first three terms pose no problem but the k dependence in the last term does. For only this particular term, I was thinking of using Feynman parameters to write my denominator of four terms in terms of one single propagator as follows $$\frac{1}{A_1 A_2 A_3 A_4} \sim \frac{1}{(k^2 - \Delta)^4}$$ and then argue that under the integral $$\int d^dk \frac{k \cdot g(p_i)}{(k^2 - \Delta)^4} = 0$$ from symmetric integration.

So, my question is, without doing the lengthy calculation of feynman parameters is it true that I can write $$\frac{1}{A_1 A_2 A_3 A_4} \sim \frac{1}{(k^2 - \Delta)^4}$$ where the $\sim$ accounts for the three integrations over feynman parameters but otherwise crucially the k dependence is simply as shown?

Thanks!

 

[LightHero]

Yes, it is true that you can write $$\frac{1}{A_1 A_2 A_3 A_4} \sim \frac{1}{(k^2 - \Delta)^4}$$ with the understanding that the $\sim$ accounts for the three integrations over Feynman parameters. The k dependence in the denominator is unchanged after accounting for the Feynman parameters, so this is a valid way of writing the denominator.