= Solution
{c}
Use the total <density of states> $D^*=Am^*/(2\pi\hbar^2)$ and $C=A/(2\pi\hbar)^2$ from part (b). In the paper's double-sum convention, $f(\phi)$ has units of <energy> and scales as $A^{-1}$. Define its first angular moment by
$$
\overline f_1=\int_0^{2\pi}\frac{d\phi}{2\pi}\,f(\phi)\cos\phi.
$$
This convention matters: the <Landau quasiparticle energy functional> written with an explicit $1/A$ instead uses the intensive kernel $\widetilde f=Af$.
At zero <temperature>, the displayed functional is the change in $E-\mu N$; a true <Galilean boost> preserves $N$, so $\Delta\mathcal F=\Delta E$. For a radial displacement $\nu(\theta)$, the first term in the <Landau quasiparticle energy functional> becomes
$$
\begin{aligned}
\Delta\mathcal F_{\rm qp}
&=C\int_0^{2\pi}d\theta\int_{p_F}^{p_F+\nu(\theta)}p\,dp\,\frac{p_F}{m^*}(p-p_F)\\
&=\frac{Cp_F^2}{2m^*}\int_0^{2\pi}\nu(\theta)^2\,d\theta+o(p_0^2).
\end{aligned}
$$
The oriented radial <integral> also works for negative $\nu$: removing occupied levels below the <chemical potential> costs positive <energy>. Substituting $\nu=p_0\cos\theta+O(p_0^2/p_F)$ and using $N=C\pi p_F^2$ gives
$$
\boxed{\Delta\mathcal F_{\rm qp}=\frac{Np_0^2}{2m^*}+o(p_0^2).}
$$
The exact second-order correction to the boundary is unnecessary here because the unperturbed excitation <energy> vanishes on the <Fermi surface>; it changes this term only beyond quadratic order. It must nevertheless be understood when identifying $\Delta\mathcal F$ with a fixed-<particle number> <energy> change.
For the quadratic interaction term, the first-order occupation change alone suffices. The radial <Dirac delta function> integrals give
$$
\Delta\mathcal F_{\rm int}=\frac{C^2p_F^2p_0^2}{2}
\int_0^{2\pi}d\theta\int_0^{2\pi}d\theta'\,
f(\theta-\theta')\cos\theta\cos\theta'+o(p_0^2).
$$
Put $\phi=\theta-\theta'$. The remaining angular <integral> uses
$$
\int_0^{2\pi}\cos(\theta'+\phi)\cos\theta'\,d\theta'=\pi\cos\phi,
$$
so only the first cosine component of the interaction survives. Since $D^*=2\pi Cm^*$,
$$
\begin{aligned}
\Delta\mathcal F_{\rm int}
&=\frac{C^2p_F^2p_0^2\pi}{2}\int_0^{2\pi}f(\phi)\cos\phi\,d\phi+o(p_0^2)\\
&=\frac{Np_0^2}{2m^*}\,D^*\overline f_1+o(p_0^2).
\end{aligned}
$$
Adding both contributions and matching the bare-<mass> result of part (a) yields the <two-dimensional Landau effective-mass relation>:
$$
\boxed{\Delta E=\frac{Np_0^2}{2m^*}\left(1+D^*\overline f_1\right)+o(p_0^2),\qquad
\frac{m^*}{m}=1+D^*\int_0^{2\pi}\frac{d\phi}{2\pi}\,f(\phi)\cos\phi.}
$$
Equivalently, using the <density of states> per unit area $g^*=D^*/A$,
$$
\boxed{\frac{m^*}{m}=1+g^*\int_0^{2\pi}\frac{d\phi}{2\pi}\,\widetilde f(\phi)\cos\phi.}
$$
If dimensionless <Landau parameters> are defined through the <Fourier series> $D^*f(\phi)=F_0+\sum_{\ell\geq1}F_\ell\cos(\ell\phi)$, then $D^*\overline f_1=F_1/2$ and
$$
\boxed{\frac{m^*}{m}=1+\frac{F_1}{2}.}
$$
Defining the cosine expansion with an extra factor of two would instead absorb that factor into $F_1$; the angular-<integral> formula above fixes the convention unambiguously. The physical content is that interaction changes accompanying a moving <Fermi sea> compensate the renormalized <quasiparticle> <velocity>, so the total boost <energy> is governed by the bare <mass>. For an interaction independent of angle, $\overline f_1=0$ and this consistency condition requires $m^*=m$.
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