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Use the constants in the claim. Since all critical points of $F$ have modulus at most one, their factorization in $F'$ gives $|a|\le6$ and $|b|\le4$. Every critical value has modulus at most $11$. Since some critical point $w$ has $|w|=1$, the identity $4w^3+2aw+b=0$ gives $4\le2|a|+|b|$. Consequently

$$\lambda_F=\max(|a|^{1/2},|b|^{1/3})>1. \tag{1}$$

For a line $L$ through $\sigma_0$, rotate the target and variable so that its preimage is the zero set of the real part of a centered monic quartic. Its quadratic and linear coefficient moduli are still $|a|$ and $|b|$; an imaginary constant may be dropped. The inverse-branch integral against $|\sigma-\sigma_0|^{-3/4}$ equals its renormalized length, since the two radial truncations differ by $O(r^{-1})$ at infinity. Thus L179 and (1) give

$$\widetilde I(F,L)\le-\lambda_F/2500\le-d. \tag{2}$$

This uses no assumption that the other critical values lie on $L$.

A uniform far-field constant

We first prove that whenever $|a|\le6$, $|b|\le4$, and $|\sigma_0|\le11$,

$$\left|G_F(\sigma)-|\sigma-\sigma_0|^{-3/4}\right| \le K|\sigma-\sigma_0|^{-7/4} \quad\text{if }|\sigma-\sigma_0|\ge T_0. \tag{3}$$

Write $H=F-\sigma_0=z^4+az^2+bz+c$, so $|c|\le11$, and put $\xi=(\sigma-\sigma_0)^{-1/4}$. Its normalized inverse roots solve

$$w^4+a\xi^2w^2+b\xi^3w+c\xi^4=1.$$

For $|\xi|\le1/8$ there is exactly one root $w_\omega(\xi)$ in each disc $|w-\omega|<1/4$, $\omega^4=1$. On its boundary, $|w^4-1|\ge(1/4)(\sqrt2-1/4)^2(7/4)>1/2$, whereas the perturbation has modulus at most $651/4096<1/5$. This is Rouché's theorem. At every root the derivative has modulus at least

$$4(3/4)^3-2(6/64)(5/4)-4/512=185/128>1.$$

The roots therefore depend holomorphically on $\xi$ in that disc. The holomorphic functions

$$h_\omega(\xi)= \frac{4\omega^3}{4w_\omega(\xi)^3+2a\xi^2w_\omega(\xi)+b\xi^3}$$

have modulus at most $4$ there, and direct expansion gives

$$h_\omega=1+\frac{a}{4\omega^2}\xi^2 +\frac{b}{2\omega^3}\xi^3+E_\omega(\xi).$$

Cauchy's coefficient estimate yields $|E_\omega(\xi)|\le32768|\xi|^4$ for $|\xi|\le1/16$. In particular $|h_\omega-1|\le130|\xi|^2<3/4$. For $|u|\le3/4$, $0\le|1+u|-1-\operatorname{Re}u\le2|u|^2$. The displayed quadratic and cubic terms cancel on summing over $\omega^4=1$, so

$$\left|\frac14\sum_{\omega^4=1}|h_\omega|-1\right| \le(32768+2\cdot130^2)|\xi|^4<K|\xi|^4.$$

Since $G_F(\sigma)=|\sigma-\sigma_0|^{-3/4} \tfrac14\sum|h_\omega|$, this proves (3) for $T_0=16^4$.

The circle, its tangent line, and their tails

Choose a unit tangent $e$ at $\sigma_0$ and a signed curvature $\kappa$. The circle, with one point omitted, and its tangent line are parametrized by

$$\sigma_\kappa(x)=\sigma_0+\frac{ex}{1-i\kappa x/2}, \qquad \sigma_L(x)=\sigma_0+ex,\qquad x\in\mathbb R.$$

Here $|d\sigma_\kappa/dx|=(1+\kappa^2x^2/4)^{-1}$ and

$$|\sigma_\kappa-\sigma_0|^{-3/4}|d\sigma_\kappa/dx| =|x|^{-3/4}(1+\kappa^2x^2/4)^{-5/8}. \tag{4}$$

Suppose $|\kappa|\le1/T$. On $|x|\ge T$, set $u=1/x$. Then $|\sigma_\kappa-\sigma_0|=(u^2+\kappa^2/4)^{-1/2} \ge2T/\sqrt5\ge T_0$. By (3), the absolute signed tail density in $u$ is at most $K(u^2+\kappa^2/4)^{-1/8}\le K|u|^{-1/4}$. Each of the circle and line tails therefore has integral at most $\tfrac83KT^{-3/4}$. Their total possible contribution to the difference is at most

$$\frac{16}{3}KT^{-3/4}\le\frac d8. \tag{5}$$

For finite circles all local singularities are integrable. Together with (3), this also establishes the absolute convergence of the signed integrals used here, including the line case.

Singular windows without a separation assumption

Write $f_1,f_2,f_3$ for the critical values of $F$, with multiplicity. In $[-T,T]$ remove the set

$$E=\{x:|\sigma_L(x)-f_j|\le2\eta \text{ for at least one }j\}.$$

If $|\kappa|T^2/2\le\eta$, then $|\sigma_\kappa(x)-\sigma_L(x)|\le\eta$ on this whole interval. Thus a removed piece lies within target distance $2\eta$ of some $f_j$ on the line and within distance $3\eta$ on the circle.

For a monic quartic, $|F(z)-f_j|\le r$ implies that $z$ lies within $r^{1/4}$ of one of the four roots of $F-f_j$, counted with multiplicity. This follows immediately by factoring $F-f_j$. The preimage of a target line is a real algebraic curve of degree at most four, and that of a circle has degree at most eight. Euclidean Crofton bounds their length inside a disc of radius $s$ by $4\pi s$ and $8\pi s$, respectively. Covering the three target windows by these twelve root discs gives a bound for the sum of the two removed density integrals:

$$48\pi(2\eta)^{1/4}+96\pi(3\eta)^{1/4} <750\eta^{1/4}=\frac d8. \tag{6}$$

For the numerical inequality one may use $\pi<4$, $2^{1/4}<6/5$, and $3^{1/4}<4/3$. Overlaps of the root discs only increase this upper bound. No critical point or critical value is required to stay separated from another.

The regular part and the reference term

For $x\notin E$ the straight segment between $\sigma_L(x)$ and $\sigma_\kappa(x)$ stays at least $\eta$ from every critical value. Every root above this segment has modulus at most $B=T+12$ by the Cauchy root bound, since $|\sigma|\le T+11$. On that disc, $|F'|\le5B^3$ and $|F''|\le M=13B^2$. The resultant identity

$$\left|\prod_{F(z)=\sigma}F'(z)\right| =256\prod_{j=1}^3|f_j-\sigma|$$

therefore gives $|F'(z)|\ge256\eta^3/(125B^9)>m$ at every root on the segment. Label the four inverse branches along it. Differentiating $1/F'$ along a branch bounds its change by $M|\sigma_\kappa-\sigma_L|/m^3$. Hence the density change is at most $4M|\kappa|T^2/(2m^3)$, and the additional Jacobian change is bounded using $G_F(\sigma_L)\le4/m$ and $1-(1+\kappa^2x^2/4)^{-1}\le\kappa^2x^2/4$. The absolute difference of the two regular density integrals is at most

$$\frac{4M|\kappa|T^3}{m^3}+\frac{2\kappa^2T^3}{m} \le\frac d8+\frac d{16}. \tag{7}$$

The reference terms can be compared over the entire interval, including $E$. From (4) and $1-(1+t)^{-5/8}\le5t/8$,

$$\int_{-T}^T |x|^{-3/4} \left(1-(1+\kappa^2x^2/4)^{-5/8}\right)\,dx \le\frac5{36}\kappa^2T^{9/4}\le\frac d{16}. \tag{8}$$

All inequalities used in (5)--(8), as well as the required bounds $|\kappa|\le1/T$ and $|\kappa|T^2/2\le\eta$, follow from the stated definition of $\kappa_0$. Summing them yields

$$|\widetilde I(F,\Gamma)-\widetilde I(F,L)|\le\frac d2.$$

Equation (2) now gives $\widetilde I(F,\Gamma)\le-d/2=-1/5000$, as claimed.