Can you switch integrals if you don't care about the infinite part? So, schematically, suppose one is doing some real integral
$$\mathcal{I}=\int_a^\Lambda \int_b^cf(x,y)\,dx\,dy,$$
where $\Lambda$ is some cutoff that one would like to take to infinity and suppose the inner integral is finite and $\int_b^c|f(x,y)|\,dx<\infty$ for any $y$. Further suppose that the whole integral diverges as $\Lambda\to\infty$ (for example, the case I have in mind is $\mathcal{I}\sim \alpha\Lambda + \beta\log\Lambda+$ finite). Obviously, switching the order of integration is not okay. However, if I am interested only in the non-divergent part of the integral, is switching the order of integration allowed? Or are there conditions under which it is allowed if I only want the part that is finite as $\Lambda\to\infty$?

Without going too far afield, e.g., theorems of Tonelli and Fubini,  if $f$ is continuous on a bounded rectangle $[a,b]\times [c,d]$, then it must hold that
$$\int_a^b \int_c^d f(x,y) \, dy \, dx =\int_c^d \int_a^b f(x,y) \, dx \, dy $$
This can be proved easily using the uniform continuity of $f$ on the compact rectangle.
A counterexample when $f$ is not continuous is
$$ \int_0^1\int_0^1 \frac{x-y}{(x+y)^3} \, dx\, dy = -\frac{1}{2}, \quad   \int_0^1 \int_0^1 \frac{x-y}{(x+y)^3} \, dy\, dx = \frac{1}{2}$$
The integrals are evaluated easily by noticing that
$$\frac{x-y}{(x+y)^3} = -\frac{\partial}{\partial x} \left( \frac{x}{(x+y)^2} \right) =\frac{\partial}{\partial y} \left( \frac{x}{(x+y)^2} \right),$$
and, hence,
$$ \int_0^1\int_0^1 \frac{x-y}{(x+y)^3} \, dx\, dy =  -\int_0^1\int_0^1\frac{\partial}{\partial x} \left( \frac{x}{(x+y)^2} \right) \,dx\,dy \\ = -\int_0^1\frac{1}{(1+y)^2} \, dy = \left.\frac{1}{1+y} \right|_0^1 = - \frac{1}{2}$$
