Shift the Mellin contour of Definition 5.2.14 downward past the poles t = -i(\lambda + 2n), 0 \le n \le N, and evaluate the residues by (41) and the origin value (42) of Lemma 5.2.19; the justification of the shift is in the proof of Lemma 5.5.3.

Range of y. Put y = \pi e^{2h_1'}r^2, H(u) = \int_0^\infty w(a)a\sinh(ua)\,da (so H(1) = h_1'), and \eta_* = 1 + u_* = (\log\lambda)/(4\lambda). The height u_* tends to -1, the normalized height of the first gamma pole, while the gamma shape parameter \lambda\eta_*/2 = (\log\lambda)/8 still diverges; this makes (70) applicable and keeps y(r_*) logarithmic. Indeed H is odd with bounded derivative near -1 (for fixed \epsilon), so H(u_*) + h_1' = H(-1+\eta_*) - H(-1) = O_\epsilon(\eta_*), and by (44) y(r_*) = \exp\bigl(\psi(\lambda\eta_*/2) + 2H(u_*) + 2h_1'\bigr). The digamma asymptotic \psi(x) \le \log x (Lemma 3.1.6) gives (equation (77)) 0 \le y \le y(r_*) = \tfrac18\log\lambda + O_\epsilon(1). Since e^{-y} can be as small as a negative power of \lambda, the errors must be controlled relative to e^{-y}.

Contour shift. Set N = \lceil\log\lambda\rceil, p = N + \tfrac12, \kappa = 1 + 2p/\lambda. The contour t = s - i(\lambda + 2p) lies strictly between consecutive gamma poles, and p \asymp \log\lambda makes the exponential-series tail negligible on (77). The multiplier e^{\lambda h_\epsilon(t/\lambda)} preserves a uniform decay margin: pA/\lambda = o_\epsilon(1) , so the taper satisfies b(a)e^{2pa/\lambda} \le 1 - c\epsilon on [a_0, A] for large \lambda; for every intermediate height 0 \le q \le \kappa, the positive shell contributes nonpositively to \operatorname{Re}h_\epsilon(s/\lambda - iq) - h_\epsilon(iq), and the negative shell gives \lambda\bigl(\operatorname{Re}h_\epsilon(s/\lambda - iq) - h_\epsilon(iq)\bigr) \le \dfrac{(1-c\epsilon)\lambda}{2}\int_0^\infty\dfrac{1-\cos(as/\lambda)}{a^2}\,da, which equals (1-c\epsilon)\pi|s|/4. The gamma factor supplies the complementary e^{-\pi|s|/4} (by Lemma 3.1.3), so the integrand decays like e^{-c\epsilon|s|} on the vertical sides, permitting the shift of (38) to t = s - i(\lambda+2p), which crosses exactly the poles t = -i(\lambda+2n), 0 \le n \le N.

Residue expansion. The residue formula (41) of Lemma 5.2.17 and the origin value (42) of Lemma 5.2.19 give (equations (78)–(79)) \dfrac{f_+(r)}{f_+(0)} = \sum_{n=0}^N\dfrac{(-y)^n}{n!}A_{\lambda,n} + \mathcal{R}_{\lambda,p}(r), A_{\lambda,n} = e^{\lambda[h_\epsilon(\zeta_n) - h_\epsilon(i)] - 2nh_1'}\,\dfrac{P_+(\zeta_n)}{\beta}, with \zeta_n = -i(1+2n/\lambda) as in (41) and h_\epsilon even,, |A_{\lambda,n} - 1| \ll_\epsilon \dfrac{n(1+n)}{\lambda} for 0 \le n \le N: Taylor expansion at u = 1 gives \lambda[h_\epsilon(i(1+2n/\lambda)) - h_\epsilon(i)] = 2nh_1' + O_\epsilon(n^2/\lambda) and P_+(-i(1+2n/\lambda))/\beta = 1 - \tfrac{2n}{\lambda}(2 + \tfrac{2n}{\lambda})^2/\beta = 1 + O_\epsilon(n/\lambda), uniformly for n \le N, where N^2/\lambda = o(1). For r > 0 the remainder is the integral over the shifted contour, \mathcal{R}_{\lambda,p}(r) = \dfrac{\pi^{\lambda/2+p}r^{2p}}{2\pi f_+(0)}\int_{\mathbb{R}}\Xi(s)\,r^{is}\,ds, \Xi(s) = \pi^{is/2}\Gamma(-p - is/2)e^{\lambda h_\epsilon(s/\lambda - i\kappa)}P_+(s/\lambda - i\kappa). The gamma reflection and product estimates (Lemma 3.1.3) with p \in \mathbb{Z} + \tfrac12 give |\Gamma(-p - is/2)| \ll e^{-\pi|s|/4}/\Gamma(1+p), the shell bound above gives \lambda(\operatorname{Re}h_\epsilon(s/\lambda - i\kappa) - h_\epsilon(i\kappa)) \le (1-c\epsilon)\pi|s|/4, and P_+(s/\lambda - i\kappa)/\beta = O_\epsilon(1 + |s|^3). Integrating in s, and using \lambda[h_\epsilon(i\kappa) - h_\epsilon(i)] = 2ph_1' + O_\epsilon(p^2/\lambda) with (\pi r^2)^pe^{2ph_1'} = y^p and p^2/\lambda = o(1), gives (equation (80)) |\mathcal{R}_{\lambda,p}(r)| \ll_\epsilon y^p/\Gamma(1+p), which extends to r = 0 by continuity.

Comparison with e^{-y}. Equations (79), (80), (77) and \sum_{n\ge0}n(1+n)y^n/n! = (y^2+2y)e^y give (equation (81)) e^y\sum_{n=0}^N\dfrac{y^n}{n!}|A_{\lambda,n} - 1| \ll_\epsilon \dfrac{(1+y)^2e^{2y}}{\lambda} \ll_\epsilon \dfrac{(\log\lambda)^2}{\lambda^{3/4}}, e^y|\mathcal{R}_{\lambda,p}(r)| \ll_\epsilon \dfrac{e^yy^p}{\Gamma(1+p)}, e^y\sum_{n>N}\dfrac{y^n}{n!} \le \dfrac{e^yy^{N+1}}{(N+1)!}. For the two tails put L = \log\lambda. Since y \le L/8 + O_\epsilon(1) and p = L + O(1), Stirling's formula gives \log\bigl(e^yy^p/\Gamma(1+p)\bigr) \le (\tfrac18 + 1 - \log 8)L + O_\epsilon(\log L), and the same holds with p replaced by N+1. Since \tfrac18 + 1 - \log 8 < 0, both tails are smaller than the coefficient error in (81). Therefore (78) yields, uniformly on 0 \le r \le r_* (equation (82)), \dfrac{f_+(r)}{f_+(0)} = e^{-y}\Bigl(1 + O_\epsilon\Bigl(\dfrac{(\log\lambda)^2}{\lambda^{3/4}}\Bigr)\Bigr) > 0.

The saddle equation (44) and the digamma asymptotic \psi(\lambda(1+u_0)/2) = \log(d(1+u_0)/4) + o(1) (Lemma 3.1.6).

The two shell contributions in Lemma 5.2.8 and Lemma 5.2.9 give \int_0^\infty w(a)a\sinh(u_0a)\,da \to -\tfrac12\log\tfrac\pi2. Since u_0 \to 1, Lemma 5.1.3 gives the limit 1/\pi.

Set R_{\epsilon,d} = e^{v(u_0)} (Lemma 5.5.4). The Fourier identities and origin values are (40) and (42) of Lemma 5.2.18 and Lemma 5.2.19. Theorem 5.4.2 gives the required exterior signs, and (82) of Lemma 5.5.3 supplies positivity of f_+ on the remaining interval [0, r_*]. Thus (equation (83)) \widehat{f_-} = f_+ > 0, f_-(0) = f_+(0) > 0, \widehat{f_0} = f_0, f_0(0) = 0, f_-(r) < 0 < f_0(r) for r \ge R_{\epsilon,d}. The limits (84) and (85) are Lemma 5.5.4 and Lemma 5.5.5.

Fix 0 < \epsilon < \epsilon_0 and let d be large. By (83) of Theorem 5.3 and Lemma 5.1, the dilation F_{\epsilon,d}(x) = f_-(R_{\epsilon,d}x) is admissible with F_{\epsilon,d}(0)/\widehat{F_{\epsilon,d}}(0) = R_{\epsilon,d}^d, so (equation (86)) \mathrm{LP}_d \le \dfrac{v_d}{2^d}R_{\epsilon,d}^d, i.e. \mathrm{LP}_d^{1/d} \le \dfrac{v_d^{1/d}\sqrt d}{2}\cdot\dfrac{R_{\epsilon,d}}{\sqrt d}. By Lemma 3.1.8 and (84) of Lemma 5.5.4, the right side tends to \tfrac12\sqrt{2\pi e}\,\alpha_\epsilon as d \to \infty. Hence \limsup_d\mathrm{LP}_d^{1/d} \le \tfrac12\sqrt{2\pi e}\,\alpha_\epsilon for every \epsilon, and letting \epsilon \downarrow 0 with (85) of Lemma 5.5.5 gives \tfrac12\sqrt{2\pi e}/\pi = \sqrt{e/(2\pi)}.

Fix 0 < \epsilon < \epsilon_0 and let d be large. Define g_{\epsilon,d,-} = f_+ - f_- and g_{\epsilon,d,+} = f_0 with f_\pm, f_0 from Theorem 5.3. Equation (40) and Fourier inversion give \widehat{g_{\epsilon,d,-}} = -g_{\epsilon,d,-} and \widehat{g_{\epsilon,d,+}} = g_{\epsilon,d,+}. By (83) both vanish at the origin and are strictly positive outside B(0,R_{\epsilon,d}); in particular neither is zero. By Lemma 5.2, \mathsf{A}_\varsigma(d) \le R_{\epsilon,d} < \infty for both signs. Hence \limsup_d\mathsf{A}_\varsigma(d)/\sqrt d \le \lim_d R_{\epsilon,d}/\sqrt d = \alpha_\epsilon by (84), and letting \epsilon \downarrow 0 with (85) of Lemma 5.5.5 gives 1/\pi.