 Solution

Define the nested integralThe order-$s^2$ equation in the inner expansion is ${\mathcal{L}}_0{Y}_2=2z{Y}_1$. The preceding result decomposes its forcing into $2z^{2}E+2bzE-2bz+2bzg$. The particular solutions combine toFor example, ${\mathcal{L}}_{0}H=E$ follows directly by differentiating the nested integral, so the equation can also be verified without relying on the supplied particular-solution formulas. At zero $Y_{2,p}=b^2$; add $-b^2+DE$ to impose the left boundary condition. As $z\to \infty$ this givesThe outer expansion in the overlap, derived in the previous section, requires the remaining constant to be $-\frac{1}{2}\log s$. Thus $D=C+b^2-\frac{1}{2}\log s$. The logarithmic matching constant is an error-function logarithmic switchback.

The complete answer through order $\epsilon$ in the inner expansion isHere $E=\mathrm{erf}z$, $g=e^{-z^2}$ and $b=2/\sqrt{\pi }$. This expression is zero at $z=0$, and its large-$z$ expansion agrees with $1+x+x^2/2-(\epsilon /2)\log x$. Together with $y_{\mathrm{out}}=e^x[1-(\epsilon /2)\log x]$, it covers both requested regions. The overlap uses $1\ll z\ll {\epsilon }^{-1/2}$, with the usual termwise ordering restrictions when matching higher terms.

If a single continuous approximation is desired, an additive composite expansion isThe apparent logarithmic divergence at $x=0$ cancels: its residual is $-(\epsilon /2)(e^x-1)\log x\to 0$. The composite therefore extends continuously to the left endpoint and satisfies the right boundary condition to the requested order.