\Lambda_0$. For such $\lambda$, $(-\Delta +V-\lambda^2)R_0(\lambda)= I+V R_0(\lambda)$ and by meromorphic continuation, $$R_0(\lambda) = R_V(\lambda)(I+VR_0(\lambda)), \; \lambda \in \Lambda.$$ Thus $R_V(\lambda)$ has a pole if and only if $I+VR_0(\lambda) $ has a zero, and multiplicities agree. Writing $V^{1/2}=V/|V|^{1/2}$ with the convention that $V^{1/2}=0$ outside the support of $V$, we see that $I+VR_0(\lambda)$ has a zero if and only if $I+V^{1/2}R_0(\lambda)|V|^{1/2}$ has a zero. Consequently, $I+V^{1/2}R_0(\lambda)|V|^{1/2}$ is invertible for all but a finite number of points in $\overline{\Lambda_0}$. Thus, if $m\in \Integers$, $\lambda\in \overline{\lambda_0}$, $$\begin{gathered}
I+V^{1/2}R_0(e^{im\pi}\lambda)|V|^{1/2}\\
= (I+V^{1/2}R_0(\lambda)|V|^{1/2})\left(I+(I+V^{1/2}R_0(\lambda)|V|^{1/2})^{-1}
V^{1/2}\left(R_0(e^{im\pi}\lambda)-R_0(\lambda)\right)|V|^{1/2}\right).\end{gathered}$$ But when $d$ is even $$R_0(e^{im\pi}\lambda)-R_0(\lambda)= imT(\lambda)$$ with $$(T(\lambda)f)(x)=\alpha_d \lambda^{d-2}\int_{\Real^d} \int_{\Sphere^{d-1}}
e^{i\lambda (x-y)\cdot \omega} f(y) d\omega \; dy$$ for $f\in L^{2}_{\comp}(\Real^d)$, with $\alpha_d =(2\pi)^{1-d}/2$, [@lrb (1.32)]. Moreover, $V^{1/2}T(\lambda) |V|^{1/2}$ is trace class. Thus, with at most a finite number of exceptions, the poles of $R_V(e^{im\pi}\lambda)$ with $\lambda \in \Lambda_0$ correspond, with multiplicity, to the zeros of $$\label{eq:FmV}
F_{m,V}(\lambda) {\stackrel{\rm{def}}{=}}\det \left(I+im(I+V^{1/2}R_0(\lambda)|V|^{1/2})^{-1}
V^{1/2}T(\lambda) |V|^{1/2}\right)$$ in $\Lambda_0$.

Lower bounds on $F_{m,\pm V}(i\sigma)$ when $V$ has fixed sign
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In the remainder of this paper we assume $d\geq 2$ is even.

Let $V\geq 0$, $V\in L^{\infty}_{\comp}(\Real^d)$. In this section we study the function $F_{m,\pm V}$ from (\[eq:FmV\]). For $\sigma \in \Real_+$, we shall use the shorthand 