Prelude: Suppose I have the following integral $$\int^1_0 dt \frac{t}{1-t}$$ which is divergent. I want to see how the divergence manifests. I can see two approaches
Rewrite the integral as $${\rm lim_{p\rightarrow 0^+}}\int^{1-p}_0 dt \frac{t}{1-t}$$ where $p>0$. I cannot see anything wrong with this approach.
Rewrite the integral as $$\int^1_0 dt {\rm lim_{p\rightarrow 0^+}}\frac{t}{1-t+pt}={\rm lim_{p\rightarrow 0^+}}\int^1_0 dt \frac{t}{1-t+pt}$$
QUESTION: Is the latter actually valid? In other words, do the operations of taking limits and integrating commute?
In both of these methods we obtain the same result.
With the change of variable $t=(1-p)t'$ you get $$ \int^{1-p}_0 \frac{t}{1-t}dt=(1-p)^2\int^{1}_0 \frac{t}{1-t+pt}dt. $$ So your two integrals are not exactly the same, but their ratio tends to $1$ as $p\to0$.