= Solution
Let $d=U'(0)>0$, $h=U'''(0)$, $p=\phi_0(0)\ne0$, and $q=\phi_0'(0)$. At a regular neutral <Rayleigh equation for inviscid shear flow> mode, multiply the equation by $U-c_0$ and evaluate it at the <critical layer in a shear flow>. The term multiplying $\phi_0''-k_0^2\phi_0$ vanishes, leaving $U''(0)p=0$. Therefore $U''(0)=0$. If this condition failed, the coefficient $U''/(U-c_0)$ would have a simple pole and a nonzero $p$ would force a logarithmically singular slope instead of a regular <eigenfunction>.
\b[There is a sign error in the printed first-order equation.] Direct expansion of the original <Rayleigh equation for inviscid shear flow> gives
$$
L_0\phi_1=H\phi_0,\qquad L_0=\frac{d^2}{dy^2}-k_0^2-\frac{U''}{U-c_0},\qquad H=2k_0k_1+\frac{c_1U''}{(U-c_0)^2}.
$$
The coefficient on the left must have a minus sign. Indeed, $1/(U-c)=1/(U-c_0)+\varepsilon c_1/(U-c_0)^2+\cdots$, and moving both first-order parameter corrections to the right gives exactly this equation. The plus sign printed there is incompatible with the original equation and its neutral <eigenfunction>. The later matching and integral formulas follow from the corrected equation. The PDF also has $y^3h/6$ in the <Taylor expansion> of $U$, and $U'''(0)$ in the requested logarithmic coefficient; these are mistranscribed in the converted TeX.
Near zero, $U-c_0=dy+hy^3/6+\cdots$ and $U''=hy+O(y^2)$. Thus $U''/(U-c_0)\to h/d$, whereas $U''/(U-c_0)^2=h/(d^2y)+O(1)$. The singular part of the first-order equation is
$$
\phi_1''\sim\frac{c_1hp}{d^2y}.
$$
Integrating twice on either side gives
$$
\phi_1(y)=a+\beta_\pm y\log|y|+b_\pm y+\cdots,\qquad \boxed{\beta_+=\beta_-=\beta=\frac{c_1hp}{d^2}.}
$$
The constant $a$ is common because the inner solution is continuous. The coefficient necessarily also contains $p$: rescaling the neutral <eigenfunction> rescales its correction. One may choose the normalization $p=1$ if a formula only in terms of the flow derivatives and $c_1$ is wanted.
The <outer expansion> of $1/(U-c)$ fails when $|dy|$ is comparable with $\varepsilon|c_1|$. Therefore the <distinguished limit> is
$$
\boxed{y=\varepsilon\eta,\qquad\ell=1.}
$$
In this <inner variable>, the original equation becomes
$$
\Phi_{\eta\eta}=\varepsilon^2\left(k^2+\frac{U''(\varepsilon\eta)}{U(\varepsilon\eta)-c}\right)\Phi.
$$
At orders one and $\varepsilon$, its right side is zero. Matching the <outer expansion> therefore selects $\Phi_0=p$ and $\Phi_1=q\eta+a$. At order $\varepsilon^2\log\varepsilon$, $\theta_2''=0$. The outer term $\varepsilon\beta y\log|y|$ becomes $\varepsilon^2\beta\eta(\log\varepsilon+\log|\eta|)$, so the required <switchback term> is $\theta_2=\beta\eta$. At order $\varepsilon^2$,
$$
\Phi_2''=p k_0^2+\frac{ph\eta}{d\eta-c_1}.
$$
Integrating, or using the supplied hint, gives
$$
\boxed{\Phi_2=\frac p2\left(k_0^2+\frac hd\right)\eta^2+\frac{phc_1}{d^3}\left[(d\eta-c_1)\log(d\eta-c_1)-(d\eta-c_1)\right]+B\eta+C.}
$$
The constants $B,C$ are homogeneous matching freedoms; they do not affect the slope jump. The quadratic term agrees with $\phi_0''(0)=p(k_0^2+h/d)$, another check on the sign of $L_0$.
For $\operatorname{Im}c_1>0$, $d\eta-c_1$ runs below the real axis. Choose the continuous <branch of the complex logarithm> there. At positive infinity its argument tends to zero; at negative infinity it tends to $-\pi$. Thus the linear parts of the two large-$|\eta|$ expansions of $\Phi_2$ are
$$
\begin{aligned}
\eta\to+\infty:\quad&\beta\eta\log|\eta|+\bigl[B+\beta(\log d-1)\bigr]\eta,\\
\eta\to-\infty:\quad&\beta\eta\log|\eta|+\bigl[B+\beta(\log d-1)-i\pi\beta\bigr]\eta.
\end{aligned}
$$
These expressions omit the already matched common quadratic term and terms only logarithmic or constant in $\eta$. Matching the linear coefficients to $b_\pm$ proves the <neutral-mode critical-layer matching> relation
$$
\boxed{b_+-b_-=i\pi\beta=\frac{i\pi c_1U'''(0)\phi_0(0)}{U'(0)^2}.}
$$
The sign comes from passing below the pole; taking $\operatorname{Im}c_1<0$ would reverse it.
Finally use the <Wronskian> $W=\phi_1\phi_0'-\phi_0\phi_1'$. Since $L_0\phi_0=0$ and $L_0\phi_1=H\phi_0$, subtraction gives $W'=-H\phi_0^2$. The wall <boundary conditions> give $W(a)=W(b)=0$, so
$$
\left(\int_a^{-\delta}+\int_\delta^b\right)H\phi_0^2\,dy=W(\delta)-W(-\delta).
$$
This is the requested excised-interval identity. As $\delta\downarrow0$, the equal logarithmic slope terms cancel and
$$
W(\delta)-W(-\delta)\longrightarrow-p(b_+-b_-)=-\frac{i\pi c_1hp^2}{d^2}.
$$
The remaining real singular integral is interpreted as a <Cauchy principal value>, consistent with excision on both sides of the <critical layer in a shear flow>. Define
$$
I=\int_a^b\phi_0^2dy,\qquad J=\operatorname{PV}\int_a^b\frac{U''\phi_0^2}{(U-c_0)^2}dy,\qquad M=\frac{hp^2}{d^2}.
$$
Then the <neutral Rayleigh-mode dispersion correction> is
$$
\boxed{2k_0k_1I=-c_1(J+i\pi M),\qquad c_1=-\frac{2k_0k_1I}{J+i\pi M},}
$$
where the last form assumes a nonzero denominator. For real $k_1$, $\operatorname{Im}c_1=2k_0k_1I\pi M/(J^2+\pi^2M^2)$, so the assumed growing branch must have the corresponding sign. If $h=0$, the displayed logarithmic coefficient and jump vanish; further degeneracy can require higher-order analysis. No nonzero growth rate follows from that degenerate first-order calculation alone.
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