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The first of the Gaussian drift primitives, $E_0$, is an even function; the second, $E_1$, is an odd function. Their derivatives follow by the fundamental theorem of calculus. In particular $E_0^{\prime }$ is the Dawson function, which behaves as $1/(2z)+O(z^{-3})$. Since the error function tends to $\pm 1$, the derivative of $E_1$ behaves as $1/(2|z|)+O(|z{|}^{-3})$ at either end. Integration therefore givesAlso $F\sim z+\sqrt{\pi }/4$ on the left and $F\sim \sqrt{\pi }/4$ on the right. The two inner overlaps are consequentlyExpressing the outer expansions in $r=\epsilon z$ requires the left constant $1+A\log \epsilon$ and the right constant $B\log \epsilon$. Equating those constants gives the stated $\alpha ,\beta$. This is exactly how a logarithmic overlap creates a switchback term: $\log |r|=\log \epsilon +\log |z|$. Adding the outer solutions and removing these overlaps leaves the displayed additive composite expansion.

At fixed $r$, the leading equation is $(r^2{f}_0^{\prime }{)}^{\prime }=0$. Its solution compatible with the eventual decaying far field and the boundary value at one is $f_0=1/r$. This decay implies $f=O(ϵ)$ at distances $r=O({ϵ}^{-1})$. Balancing radial derivatives against the linear screening term identifiesThe distant region is governed by the modified Helmholtz equation; its decaying homogeneous profile is $e^{-x}/x$.

For a matched asymptotic expansion, write the fixed-$r$ approximation as $f=f_0+ϵf_1+{ϵ}^2{f}_2+\cdots$, allowing logarithms of $ϵ$ in the coefficients. Successive equations areThe boundary condition imposes $f_0(1)=1$ and $f_1(1)=f_2(1)=0$. Before matching, their integrated forms can be writtenIn particular a pure power series with parameter-independent coefficients will be insufficient: the logarithmic overlap creates a switchback term.

In the distant region set $f=ϵF_0(x)+{ϵ}^2{F}_1(x)+\cdots$. The scaled equation is $f_{xx}+2f_x/x-f=x{f}^3/ϵ$, soDecay and leading matching give $F_0=e^{-x}/x$. The radial modified Helmholtz equation gives the supplied particular integral in terms of the exponential integral:As $x\downarrow 0$, the small-argument expansion of the exponential integral yieldswhere $\gamma$ is the Euler--Mascheroni constant. Substitute $x=ϵr$ to compare the two expansions in $1\ll r\ll {ϵ}^{-1}$:The constant at order $ϵ$ fixes $C_1=-1$. Its $1/r$ coefficient then fixes $\alpha +\frac{1}{2}\log 2=1$, and the constant at order ${ϵ}^2$ fixes $D$. HenceThe required inner expansion at fixed $r$ isAt fixed positive $x=ϵr$, the outer expansion isBoth display every term through the requested order, including the ${ϵ}^2\log ϵ$ switchback term in the fixed-$r$ region.

To form an additive composite expansion, subtract the common overlap from the sum of the inner and outer expressions. Their retained common part isThus one composite is $ϵF_0(ϵr)+{ϵ}^2{F}_1(ϵr)-{ϵ}^2(D+1/2)/r$. The last term can be screened by multiplying it by $e^{-ϵr}$ without changing either retained expansion. This gives a useful exponentially decaying version:Its boundary value is $1+o({ϵ}^2)$. If exact satisfaction of the boundary value is desired, use insteadThe small-argument expansion of the exponential integral shows that this normalized composite has the same two retained expansions; it equals one at $r=1$ and tends to zero at infinity.