= Solution
Introduce a constant <filament bending modulus> $B$, which sets the mechanical force scale. The filament has <intrinsic curvature of an elastic filament> $\kappa_0(s)$, and its actual <signed curvature> in a small-slope <Monge representation> is $\kappa\simeq\zeta_{xx}$. To quadratic order one may identify material <arc length> $s$ with $x$ in $\kappa_0$, while
$$
\mathcal L-L=\frac12\int_0^L\zeta_x^2dx+O(\zeta_x^4).
$$
At fixed positive extension force, the mechanical potential is bending energy minus $FL$. Replacing $L$ by $\mathcal L-(\mathcal L-L)$ leaves, up to a force-dependent constant,
$$
\boxed{\mathcal H[\zeta]=\frac12\int_0^L
\left[B(\zeta_{xx}-\kappa_0)^2+F\zeta_x^2\right]dx}.
$$
Its first variation is
$$
\delta\mathcal H=\int_0^L[B(\zeta_{xx}-\kappa_0)_{xx}-F\zeta_{xx}]\delta\zeta\,dx
+\left[B(\zeta_{xx}-\kappa_0)\delta\zeta_x+
\{F\zeta_x-B(\zeta_{xxx}-\kappa_{0,x})\}\delta\zeta\right]_0^L.
$$
The <higher-order Euler-Lagrange equation> is therefore
$$
\boxed{B\zeta_{xxxx}-F\zeta_{xx}=B\kappa_{0,xx}}.
$$
For a <Fourier transform> relation, take a periodic bulk segment, or a long-filament bulk approximation with negligible endpoint contributions; these are necessary <boundary conditions> because the question specifies no filament end conditions. Using $f(x)=\int\widehat f(q)e^{iqx}dq/(2\pi)$, every nonzero <wavenumber> obeys
$$
(Bq^4+Fq^2)\widehat\zeta(q)=-Bq^2\widehat\kappa_0(q),\qquad
\boxed{\widehat\zeta(q)=-\frac{B\widehat\kappa_0(q)}{F+Bq^2}}.
$$
A zero displacement mode is an arbitrary translation. In a periodic representation a constant intrinsic-curvature mode contributes a constant bending cost but no bulk slope, so it does not contribute to the following deficit. The <Parseval identity> gives
$$
\boxed{\mathcal L-L=\frac{B^2}2\int\frac{dq}{2\pi}
\frac{q^2|\widehat\kappa_0(q)|^2}{(F+Bq^2)^2}}.
$$
For clarity about finite normalization, on a periodic interval of length $L$ define $\kappa_n=L^{-1}\int_0^L\kappa_0(x)e^{-iq_nx}dx$, $q_n=2\pi n/L$. Then the equivalent exact quadratic-order formula is
$$
\mathcal L-L=\frac{LB^2}2\sum_{n\ne0}
\frac{q_n^2|\kappa_n|^2}{(F+Bq_n^2)^2}.
$$
This is <force-extension of an intrinsically curved filament>, determined by its particular fixed shape rather than by a thermal orientation entropy.
For an ensemble with <quenched intrinsic curvature>, first minimize each realization and then average the last expression, replacing $|\kappa_n|^2$ by its ensemble mean. For stationary bulk disorder, let $S_\kappa(q)$ be the <power spectrum>, normalized so that its correlation is $\langle\kappa_0(x)\kappa_0(x+r)\rangle=\int S_\kappa(q)e^{iqr}dq/(2\pi)$. The measurable mean deficit per unit projected length is
$$
\boxed{\frac{\langle\mathcal L-L\rangle}{L}
=\frac{B^2}2\int\frac{dq}{2\pi}\frac{q^2S_\kappa(q)}{(F+Bq^2)^2}}.
$$
If the curvature is smooth enough that the second spectral moment is finite, one can take the large-force limit under this integral, obtaining the <smooth-curvature high-force extension law>
$$
\boxed{\langle\mathcal L-L\rangle\sim\frac{B^2}{2F^2}
\int_0^L\langle(\kappa_{0,x})^2\rangle dx}.
$$
For a stationary ensemble the integral is $L\langle(\kappa_{0,x})^2\rangle$. Equivalently, the bulk slope is $\zeta_x\simeq-(B/F)\kappa_{0,x}$; a plot of the mean deficit against $F^{-2}$ measures the mean squared intrinsic-curvature gradient, if $B$ is known. \b[The $F^{-2}$ law is a smooth-curvature bulk result, not a universal law for every random filament.] For example, ideal white intrinsic-curvature disorder with $S_\kappa(q)=S_0$ has an infinite second spectral moment; retaining the full integral instead gives $\langle\mathcal L-L\rangle/L=S_0\sqrt B/(8\sqrt F)$ in that ideal continuum model.
Endpoint conditions also matter. If each end is moment-free and has no transverse applied force, the natural conditions from the variation are
$$
\zeta_{xx}=\kappa_0,\qquad
B(\zeta_{xxx}-\kappa_{0,x})-F\zeta_x=0
$$
at each end. These can produce a <free-end curvature boundary layer> of width $\ell=\sqrt{B/F}$. For a constant <intrinsic filament curvature> $\kappa$, integration of the <higher-order Euler-Lagrange equation> with zero transverse force gives $B(\zeta_x)_{xx}-F\zeta_x=0$ and hence
$$
\zeta_x=\kappa\ell\frac{\sinh[(x-L/2)/\ell]}{\cosh[L/(2\ell)]},\qquad
\mathcal L-L=\frac{\kappa^2\ell^3}2\left[\tanh a-a\operatorname{sech}^2a\right],\quad a=\frac{L}{2\ell}.
$$
The last formula follows by integrating $\sinh^2[(x-L/2)/\ell]$, and tends to $\kappa^2\ell^3/2\propto F^{-3/2}$, despite the vanishing bulk curvature gradient. Thus a finite free-ended measurement must include or suppress endpoint bending before interpreting a bulk $F^{-2}$ fit.
For the <freely jointed chain>, rigid links have independent orientations, so the restoring force is entropic. Let $T$ now denote absolute <temperature>, and $\beta=1/(k_BT)$ with <Boltzmann constant> $k_B$. In the usual three-dimensional chain, a segment at polar angle $\theta$ has force energy $-Fb\cos\theta$. Its angular <canonical partition function> and the full chain <partition function> are
$$
Z_1=2\pi\int_{-1}^1e^{\xi\mu}d\mu
=4\pi\frac{\sinh\xi}{\xi},\qquad
Z_N=Z_1^N,\qquad \xi=\beta Fb.
$$
Differentiating $\log Z_N$ with respect to $\beta F$ yields the exact mean projected displacement
$$
\boxed{\langle L\rangle=Nb\left(\coth\xi-\frac1\xi\right)
=Nb\mathscr L(\xi)}.
$$
Here $\mathscr L$ is the <Langevin function>; this is <force-extension of a three-dimensional freely jointed chain>, in the fixed-force <canonical ensemble>. At high extension, $\coth\xi=1+2e^{-2\xi}+\cdots$, so
$$
\boxed{Nb-\langle L\rangle\sim\frac{Nk_BT}{F},\qquad
F\sim\frac{Nk_BT}{Nb-\langle L\rangle}}.
$$
Thus the smooth quenched-curvature bulk deficit falls as $F^{-2}$, whereas the three-dimensional <freely jointed chain> deficit falls as $F^{-1}$ and depends explicitly on <temperature>. The distinction is mechanical straightening of a fixed intrinsic shape versus orientational <entropy>; thermal bending fluctuations of a stiff filament have not been included in the first calculation.
If the comparison chain is also restricted to two dimensions, it is a <planar freely jointed chain> rather than the three-dimensional model. In this case,
$$
Z_1=\int_0^{2\pi}e^{\xi\cos\theta}d\theta=2\pi I_0(\xi),\qquad
\boxed{\frac{\langle L\rangle}{Nb}=\frac{I_1(\xi)}{I_0(\xi)}}.
$$
The last identity follows directly by differentiating the integral defining $I_0$; $I_0$ and $I_1$ are <Modified Bessel functions of the first kind>. At large $\xi$ the angular weight is proportional to $e^{-\xi\theta^2/2}$, giving $\langle\cos\theta\rangle\simeq1-\langle\theta^2\rangle/2=1-1/(2\xi)$. Consequently $Nb-\langle L\rangle\sim Nk_BT/(2F)$: the same inverse-force exponent, with half the three-dimensional coefficient. Both versions assume inextensible links and forces below the regime where the links themselves stretch.
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