Regularization of functional determinant over an Instanton background

I am reading the paper "ABC of instantons" and meet some problems at section 8. I simplify this problem a little bit as follows.

First, we have a Euclidean path integral like
begin{equation}
Z=int mathcal{D}A; {rm e}^{-S},~~~S=int d^4x mathcal{L}_0(A^{a}_{mu})
end{equation}
An instanton is a solution of equation of motion that makes $S$ finite.

Now expand this action at the instanton solution $A^{ins}$ up to 2nd order:
begin{equation}
A=A^{ins}+a,~~~S=S(A^{ins})+int d^4 x ~a^{j}_{mu}hat{L}^{jk}_{mu nu}(A^{ins})a^{k}_{nu}.
end{equation}
Here $hat{L}^{jk}_{mu nu}(A^{ins})$ is an operator depending on $A^{ins}$. One also needs to add a gauge-fixing term and ghosts to the action $S$, these are
begin{equation}
Delta S=int d^4 x a^{j}_{mu}Deltahat{L}^{jk}_{mu nu}(A^{ins})a^{k}_{nu}
end{equation}
for gauge-fixing and
begin{equation}
Delta S_{gh}=int d^4x bar{Phi}^a hat{L}^{ab}_{gh}Phi^b
end{equation}
for ghost.

Combining everything, one has
begin{equation}
Z=e^{-S(A^{ins})} det(hat{L}+hat{Delta L})^{-1/2} det(hat{L}_{gh})
end{equation}
Now since the operator $hat{L}+hat{Delta L}$ has zero modes (eigenfunction of vanishing eigenvalue), the expression $ det(hat{L}+hat{Delta L})^{-1/2}$ is ill-defined. This paper claims we have to regularize it with a cutoff $M^2$ (eq 74):
begin{equation}
bigg[frac{det(hat{L}+hat{Delta L})}{det(hat{L}+hat{Delta L}+M^2)}bigg]^{-1/2} frac{det(hat{L}_{gh})}{det(hat{L}_{gh}+M^2)}
end{equation}
My question is: How does this cutoff come into the current calculation? I know the infinity of $det(hat{L}+hat{Delta L})^{-1/2}$ is from the integral
begin{equation}
int dc exp[-frac{1}{2}lambda c^2]
end{equation}
for $lambda=0$. But how is the cutoff introduced and how does it work?