Explicit Gaussian continuation of a free resolvent matrix element

By [F1], $H_0$ is self-adjoint and nonnegative. The Gaussian integral in [F2], applied in each coordinate and combined with Tonelli [F4], gives ${\int }_{{\mathbb{R}}^3}{e}^{-\mathrm{\mid }p{\mathrm{\mid }}^2}dp={\pi }^{3\mathrm{/}2}$, so $\mathrm{\parallel }g\mathrm{\parallel }=1$. Polar coordinates [F3] and the resolvent multiplier therefore give $m(z)=\frac{4}{\sqrt{\pi }}{\int }_0^{\mathrm{\infty }}\frac{r^2{e}^{-r^2}}{r^2-z}dr$ for $\mathrm{Im}z>0$. The denominator is nonzero on the integration ray, and the integral is locally uniformly dominated on the upper half-plane, so it defines the stated resolvent matrix element.

For $\mathrm{Re}a>0$, put $I(a)={\int }_0^{\mathrm{\infty }}{e}^{-r^2}(r^2+a^2{)}^{-1}dr$. When $a>0$ is real, the Laplace representation of $(r^2+a^2{)}^{-1}$ and Tonelli [F4] give $I(a)={\int }_0^{\mathrm{\infty }}{e}^{-a^{2}s}{\int }_0^{\mathrm{\infty }}{e}^{-(1+s)r^2}drds=\frac{\sqrt{\pi }}{2}{\int }_0^{\mathrm{\infty }}\frac{e^{-a^{2}s}}{\sqrt{1+s}}ds$ by [F2]. The substitutions $u=\sqrt{1+s}$ and $v=au$ then give $I(a)=\sqrt{\pi }{e}^{a^2}{\int }_1^{\mathrm{\infty }}{e}^{-a^2{u}^2}du=\frac{\sqrt{\pi }{e}^{a^2}}{a}{\int }_a^{\mathrm{\infty }}{e}^{-v^2}dv=\frac{\pi }{2a}{e}^{a^2}\mathcal{E}(a)$, using the horizontal-ray definition of $\mathcal{E}$ for real $a>0$.

Fix $a_0$ with $\mathrm{Re}{a}_0>0$ and a closed disk $K$ about $a_0$ contained in that half-plane. Choose $\eta >0$ and $A<\mathrm{\infty }$ so that $\mathrm{Re}a\ge \eta$ and $\mathrm{\mid }a\mathrm{\mid }\le A$ on $K$. For $r\ge 0$, $\mathrm{\mid }{r}^2+a^2\mathrm{\mid }=\mathrm{\mid }a+ir\mathrm{\mid }\mathrm{\mid }a-ir\mathrm{\mid }\ge {\eta }^2$. For nonzero complex $\delta$ small enough that $a_0+\delta \in K$, the exact difference quotient of the integrand of $I$ is $\frac{e^{-r^2}}{\delta }(\frac{1}{r^2+(a_0+\delta {)}^2}-\frac{1}{r^2+a_0^2})=-\frac{(2a_0+\delta )e^{-r^2}}{(r^2+(a_0+\delta {)}^2)(r^2+a_0^2)}\mathrm{.}$ For $\mathrm{\mid }\delta \mathrm{\mid }\le 1$, its modulus is at most $(2A+1){\eta }^{-4}{e}^{-r^2}$, an integrable majorant by [F2]. The quotients are measurable in $r$, and tend pointwise to $-2a_0{e}^{-r^2}\mathrm{/}(r^2+a_0^2{)}^2$. Applying [F5] along every sequence of nonzero complex increments tending to zero proves that $I$ is complex differentiable at $a_0$, with derivative the integral of this limit. Thus $I$ is holomorphic on the right half-plane; this conclusion uses the actual difference quotients, not an assumption that dominated convergence is itself a differentiation theorem. For $\mathcal{E}$, put $F(w,t)=e^{-(w+t{)}^2}$. Its exact segment identity is $\frac{F(w+\delta ,t)-F(w,t)}{\delta }={\int }_0^1-2(w+s\delta +t)e^{-(w+s\delta +t{)}^2}ds\mathrm{.}$ For completeness this identity follows by expanding the exponential series on the finite segment, differentiating each polynomial in $s$, and integrating it: the series and its derivative converge uniformly there, since for $\mathrm{\mid }w+s\delta +t\mathrm{\mid }\le B$ they are dominated respectively by $\sum B^{2n}\mathrm{/}n!$ and $\mathrm{\mid }\delta \mathrm{\mid }{\sum }_{n\ge 1}2n{B}^{2n-1}\mathrm{/}n!$, both finite. Hence passage through the finite segment integral is justified by the uniform remainder bound. For $\mathrm{\mid }w\mathrm{\mid }\le R$, $\mathrm{\mid }\delta \mathrm{\mid }\le 1$ and $0\le s\le 1$, write $w+s\delta =u+iv$; then $\mathrm{\mid }u\mathrm{\mid },\mathrm{\mid }v\mathrm{\mid }\le R+1$ and $(t+u{)}^2\ge t^2\mathrm{/}2-u^2$. Consequently the segment derivative is bounded in modulus by $C_R(1+t)e^{-t^2\mathrm{/}2}$, independently of $w,\delta ,s$. This majorant is integrable: $(1+t)e^{-t^2\mathrm{/}4}$ is bounded, and the remaining scaled Gaussian is integrable by [F2]. The original integrand is bounded the same way without the factor $1+t$. The measurable difference quotients therefore have one integrable majorant and tend pointwise, as complex $\delta \to 0$, to $-2(w+t)e^{-(w+t{)}^2}$. Apply [F5] to every such sequence to obtain the complex derivative of $\mathcal{E}$ at every $w$. It is entire. Both sides of step 1.2 are therefore holomorphic on the connected half-plane $\mathrm{Re}a>0$; their agreement on the positive real axis has an accumulation point inside this domain, so [F6] extends the identity throughout the half-plane.

For $\mathrm{Im}z>0$, choose the square root $a=\sqrt{-z}$ with $\mathrm{Re}a>0$. Since $r^2\mathrm{/}(r^2-z)=1+z\mathrm{/}(r^2-z)$, $m(z)=2+\frac{4z}{\sqrt{\pi }}I(a)$; substituting the identity from step 2.1 and $z=-a^2$ gives $m(z)=2-2\sqrt{\pi }a{e}^{-z}\mathcal{E}(a)$. On $D\cap \{\mathrm{Im}z>0\}$, the root $a$ equals $-ik(z)$: both square to $-z$, and $\mathrm{Re}(-ik(z))>0$. Substitution gives $m(z)=2+2i\sqrt{\pi }k(z)e^{-z}\mathcal{E}(-ik(z))$ there. The right hand side is holomorphic throughout $D$, so it is the required local continuation through $E_0$.