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At an algebraic branch point, a local solution becomes meromorphic after the finite substitution $z-a=t^k$. The displayed convergent expansion has only finitely many negative terms. A square-root branch point corresponds to $k=2$; an essential singularity or logarithmic singularity need not become meromorphic under any such finite substitution.

A branch point is a point around which a sufficiently small loop can change a germ of a holomorphic function under analytic continuation. This is a local condition: a large loop around a regular point might also enclose a distant branch point, so that large loop alone cannot classify the regular point. For the complex logarithm and square root, every nonzero point has a local holomorphic branch, but a loop once around $0$ changes the chosen branch. ^[ref]

For $z=r{e}^{it}$ with fixed $r>0$, start at $t=0$ with values $\sqrt{r}$ and $\log r$. Their continuations areOne counterclockwise turn gives $w(2\pi )=-\sqrt{r}$ and $L(2\pi )=\log r+2\pi i$. At $r=4$, the square root therefore changes from $2$ to $-2$, while the logarithm changes from $\log 4$ to $\log 4+2\pi i$. Two turns restore the square root to $2$; no nonzero number of turns restores the logarithm. This distinguishes a two-sheeted algebraic branch point from a logarithmic singularity.

A pole need not be a branch point: $1/z$ is unbounded near $0$, yet $1/(r{e}^{it})=r^{-1}{e}^{-it}$ returns to its starting value after one turn. Conversely, the square-root values tend to $0$ at their branch point. The issue is whether a single holomorphic choice survives going around the point, rather than whether the values become large. On the Riemann sphere, infinity is also a branch point of both $\sqrt{z}$ and $\log z$: in the local coordinate $\zeta =1/z$, these are ${\zeta }^{-1/2}$ and $-\log \zeta$, whose continuations around $\zeta =0$ change sign and subtract $2\pi i$, respectively. ^[ref]

The integral representation of the modified Bessel function of the second kind gives $K_0(x)={\int }_0^{\infty }{e}^{-x\cosh t}dt$ for $x>0$. Taking the Fourier transform of the even continuation and setting $u=\sinh t$ gives $2{\int }_0^{\infty }(1+k^2+u^2{)}^{-1}du=\pi /\sqrt{1+k^2}$. The large-wavenumber terms $\pi /|k|-\pi /(2|k{|}^3)$ are also obtained from the local logarithmic singularity $-\log |x|-(x^2/4)\log |x|$ using the Fourier transform of the absolute logarithm.

For $b>0$, the Poisson equation $\Delta V=q{\delta }^{(2)}(x,y)$ in $-b<x<b$ with homogeneous Neumann boundary conditions on both edges admits the even-in-$y$ method of images solution in the title, where $r_n^2=(x-2nb{)}^2+y^2$. All image strengths are $q$. Pairing the images at $\pm 2nb$ and subtracting their constant logarithms makes each summand $O(n^{-2})$, giving local uniform convergence away from the source. The unrenormalized sum of logarithms does not converge.

Differentiating givesThe corresponding $x$-derivatives cancel in reflected pairs on $x=\pm b$. For $y>0$, integration across the strip yields a telescoping arctangent sum:One closed form, up to an additive constant, is $V=(q/4\pi )\log [\cosh (\pi y/b)-\cos (\pi x/b)]$. Its near-source logarithmic singularity has coefficient $q/(2\pi )$. Conditions at large $|y|$ or a symmetry class must be specified for a uniqueness claim; side Neumann boundary conditions alone allow additional homogeneous harmonic functions.

Suppose $p(x)=c(x-a)+O((x-a{)}^2)$ with $c>0$, and the other coefficients in $(p{y}^{\prime }{)}^{\prime }+(\lambda w-q)y=0$ extend continuously to $a$, with $w(a)>0$. A bounded eigenfunction has $(p{y}^{\prime }{)}^{\prime }=O(1)$, hence $p{y}^{\prime }=C+O(x-a)$. A nonzero $C$ would give $y^{\prime }\sim C/[c(x-a)]$ and an unbounded logarithmic singularity. Thus $C=0$, and the regular branch has bounded first derivative. The second local branch carries a logarithm. The title condition excludes that branch.

For two regular eigenfunctions, $p(\overline{u}{v}^{\prime }-\overline{u^{\prime }}v)\to 0$. This removes the singular-end contribution to the Lagrange identity for a Sturm-Liouville operator. Together with a separated self-adjoint condition at the other endpoint, it selects the regular self-adjoint differential operator realization. The precise operator domain uses the maximal differential-operator domain and the no-log endpoint condition; boundedness of classical eigenfunctions is its convenient local expression, not a replacement for specifying an operator domain.

Write $K_0(z)=-F(z)\log z+G(z)$ locally, with analytic $F,G$ and $F(0)=1$. The coefficient of $\log z$ in the Modified Bessel differential equation requires $z^2{F}^{\prime \prime }+z{F}^{\prime }-z^{2}F=0$. If $F=\sum a_m{z}^m$, its recurrence is $m^2{a}_m=a_{m-2}$, with $a_1=0$. HenceThe analytic part is even: an odd-power term in $G$ would produce an uncancelled odd-power term under the same differential operator, and the recurrence removes all such terms. Insert a smooth even cutoff equal to one near zero. The smooth part and the exponentially decaying tails give no algebraic large-wavenumber terms after repeated integration by parts. Therefore the logarithmic singularities determine the requested terms. For $k\ne 0$,Thus the Fourier transform of the modified Bessel function K0 has expansionAn independent check follows from the integral representation of the modified Bessel function of the second kind. Integrating $e^{-|x|\cosh t}$ first and then setting $u=\sinh t$ gives $2{\int }_0^{\infty }du/(u^2+1+k^2)=\pi /\sqrt{1+k^2}$, with exactly these first two terms.

Let $P(z)=1/(2\sqrt{\pi }{z}^{1/4})$, choose a consistent branch, and truncate the corrected series after $r=n-1$, with $n=|\sigma |+O(1)$. The leading factorial tail in part (iii) is represented by the same indented integral as in part (ii):Here the divergent sum means its lateral Borel summation, not ordinary summation. To see why the pole approximation controls this remainder, write $a_r={\sigma }^r{Y}_r$ and insert $\Gamma (r)={\int }_0^{\infty }{e}^{-v}{v}^{r-1}dv$ in the leading late terms. Summing the geometric tail creates $1/(1-v/\sigma )$; $v=\sigma t$ produces $I(\sigma ,n)$. The relative $O(r^{-1})$ corrections to the late coefficients give lower-order contributions on its saddle scale. More concretely, the Borel transform of the coefficient series isThe simple pole supplies the leading Stokes phenomenon; its weaker logarithmic singularity supplies the smaller corrections. This is the integral justification for retaining the factorial tail.

For $\arg \sigma =\varphi /\sqrt{|\sigma |}$, expand $\sigma =n+i\varphi \sqrt{n}+O(1)$. Part (i), with $\mu =\varphi$, now givesThis formula uses the upper-pole prescription and the upper Airy function switching ray: near $\arg z=2\pi /3$, take $\arg \sigma =3\arg z/2-\pi$ near zero. The coefficient of the subdominant exponential changes smoothly from zero to $i$ across this ray and has value $i/2$ on it. The angular transition width is $O(|\sigma {|}^{-1/2})=O(|z{|}^{-3/4})$. The conjugate lower ray has the conjugate contour sign, rather than the same $+i$ on both rays. This is error-function smoothing of a Stokes multiplier, not a discontinuity of the analytic Airy function. With the literal printed $r=1$ series, the missing $P(z)e^{\sigma /2}$ would invalidate this remainder formula.

Use the principal complex logarithm in the upper half-plane. Deform the interval upward into a half-strip: the vertical side at zero is traversed upward, and the side at one downward. The top side vanishes exponentially; a small indentation at zero contributes $O(r|\log r|)$ and vanishes. Thus the contour deformation gives the exact representation, for $x>0$,Since $\log (is)=\log s+i\pi /2$, scaling $u=xs$ and using the integral defining Euler's constant givesFor the second endpoint, $\log (1+is)=is+s^2/2+O(s^3)$. Watson's lemma, or integration of these powers against $e^{-xs}$ with an exponentially small tail, yieldsIn particular the required first terms areThe two different endpoint scales come from the logarithmic singularity at zero and the simple zero of the amplitude at one.

For $0<y<\pi$, define $J(y)={\int }_0^{\pi }{u}_y(x,y)dx$. Differentiate under the integral and use Laplace's equation:Thus $J$ is constant. The top Neumann boundary condition gives $J(\pi )=0$, soFor the included endpoint $y^{\ast }=0$, the integral is interpreted as an improper integral at the bottom corners if necessary. The Fourier series constructed in c gives, on $0<x<\pi$,The second series is absolutely and uniformly convergent; the first term has only integrable logarithmic singularities. Its integral is zero since ${\int }_0^{\pi }\log (2\sin x)dx=0$, while every cosine term also has zero integral. This proves the endpoint as well:For completeness, the logarithmic integral follows without any endpoint differentiation. Put $I={\int }_0^{\pi /2}\log (\sin x)dx$ and substitute $x\mapsto \pi /2-x$ to obtain $I={\int }_0^{\pi /2}\log (\cos x)dx$. Adding and using $\sin x\cos x=\sin (2x)/2$ gives $2I=I-(\pi /2)\log 2$, hence $I=-(\pi /2)\log 2$ and the required full-interval result.

Put $w=e^x-1=x+x^2/2+x^3/6+O(x^4)$. The local logarithmic series $\ln (1-w)=-w-w^2/2-w^3/3+O(w^4)$ givesFor the infinite Taylor series, it is necessary to distinguish the real domain of the logarithm, $x<\ln 2$, from its convergence interval about zero. Write $a=\ln 2$. The complex zeros of $2-e^z$ are $a+2\pi ik$ for integers $k$, so the nearest logarithmic singularity to zero is at $a$. The radius of convergence is therefore $a$, giving convergence for $|x|<a$ and divergence for $|x|>a$.

The two endpoints need separate checking. Factor off the nearest singularity:The quotient has a nonzero removable value $2a$ at $z=a$; its other zeros are at $a+2\pi ik$ with $k\ne 0$. Consequently it has a holomorphic logarithm on a zero-free disk larger than $|z|\le a$, which is a simply connected domain, and the Taylor series of $H$ converges absolutely at both real endpoints. At $x=-a$, the remaining series $-{\sum }_{n\ge 1}(-1{)}^n/n$ converges by the alternating series test. At $x=a$, it is $-{\sum }_{n\ge 1}1/n$ and diverges. Thus the full real convergence set isConvergence is absolute in the open interval and conditional at the left endpoint, where the sum equals $\ln (3/2)$. This endpoint distinction is an instance of isolating a logarithmic singularity to test a power-series endpoint.