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An argument of a nonzero complex number $z$ is any angle $\theta$ for which $z=|z|e^{i\theta }$. All arguments are $\theta +2\pi k$, with $k$ an integer. Zero has no argument: it has no distinguished direction. A commonly chosen representative, the principal argument $\mathrm{Arg}z$, lies in $(-\pi ,\pi ]$. Choosing representatives is part of specifying a branch of a multivalued function. These conventions are described in Jeremy Orloff, MIT 18.04, topic 1, sections 1.5 and 1.14.

For $3+4i$, the principal argument is $\arctan (4/3)\approx 0.927295$ radians, computed with the inverse tangent. But the coordinate ratio alone loses the quadrant: $-3-4i$ has the same ratio $4/3$, yet its principal argument is $\arctan (4/3)-\pi \approx -2.214297$ radians.

For nonzero $z,w$, the correct rule isOrdinary equality can fail. With $z=w=e^{3\pi i/4}$, their principal arguments sum to $3\pi /2$, but $zw=-i$ has principal argument $-\pi /2$. They differ by one full turn. As a point crosses the negative horizontal axis, the principal argument jumps by $2\pi$. A continuous choice on a suitable region therefore requires a branch cut; the jump belongs to the choice of representative, although the point moves smoothly.

A branch of a multivalued analytic expression is a single-valued holomorphic function on a specified domain that consistently selects one of the expression's possible values. A branch cut is a convenient way to choose such a domain, rather than an intrinsic locus of singularities. On $D=\mathbb{C}\setminus (-\infty ,0]$, the principal logarithm supplies two square-root branchesFor $z=i$, these give $B_{+}(i)=e^{i\pi /4}=(1+i)/\sqrt{2}$ and $B_{-}(i)=-(1+i)/\sqrt{2}$. Their derivatives follow by differentiating $B_{\pm }^2=z$: $B_{\pm }^{\prime }=1/(2B_{\pm })$, so $B_{+}^{\prime }(i)=(1-i)/(2\sqrt{2})$. There are exactly two branches on $D$: for any other holomorphic square root $B$, the ratio $B/B_{+}$ is continuous, takes values in $\{1,-1\}$, and hence is constant on the connected domain.

The cut can move. Choosing $0<\theta <2\pi$ defines a holomorphic logarithm $L_{+}(z)=\log |z|+i\theta$ on $D_{+}=\mathbb{C}\setminus [0,\infty )$. At $-1$ this gives $L_{+}(-1)=i\pi$ and $\exp (L_{+}(-1)/2)=i$, both regular values despite $-1$ lying on the previous cut. In the complex upper half-plane, $L_{+}$ agrees with $\mathrm{Log}$; in the lower half-plane it equals $\mathrm{Log}+2\pi i$, so the corresponding square roots have opposite signs there. The overlap $D\cap D_{+}$ has two components, exactly why the identity theorem cannot force agreement between them. ^[ref]

For example, the limits of $B_{+}$ at $-r$, $r>0$, are $i\sqrt{r}$ from above and $-i\sqrt{r}$ from below. This jump records the chosen branch cut; each point $-r\ne 0$ still has a neighbourhood supporting a holomorphic square root. Moving the cut changes the bookkeeping, while the branch point at $0$ and its monodromy persist. Specifying a branch of a multivalued function is therefore essential before differentiating or performing contour integration with holomorphic square roots, complex logarithms or complex exponentiation.

For $\sigma ,t>0$, take the principal square root and orient the Bromwich contour upward. Closing left around the negative-real branch cut gives the jump $2i{\int }_0^{\infty }{e}^{-tx}{x}^{-1/2}dx$. The substitution $u=tx$ and $\Gamma (1/2)=\sqrt{\pi }$ prove the displayed formula. A rectangular closure has horizontal remainders $O(R^{-1/2})$, an exponentially small left edge, and an $O(\sqrt{\epsilon })$ small-circle term. These estimates also justify the conditionally convergent vertical integral.

For the moving semi-infinite crack, take the branch cut of $\sqrt{z}$ on the negative real axis and $\sqrt{x}>0$ ahead of the tip. With steadily moving antiplane shear, face tractions ${\sigma }_{23}^{\pm }=-f(x)$ imply $\mathrm{Im}{F}^{\prime \pm }=f/\mu$. The antisymmetric displacement response gives $F^{\prime -}=-\overline{F^{\prime +}}$. Hence $G=\sqrt{z}{F}^{\prime }$ has jump $G^{+}-G^{-}=-2\sqrt{-x}f(x)/\mu$. The Sokhotski–Plemelj theorem gives the displayed Cauchy integral, assuming convergence and no extra remote homogeneous loading. For $x>0$ it is purely imaginary, so ${\sigma }_{23}={\int }_{-\infty }^0\sqrt{-s}f(s)/(x-s)ds/(\pi \sqrt{x})$. At fixed co-moving loading this ahead-of-tip stress is independent of speed; the displacement still has the factor $1/\beta$.

Take a complex logarithm whose radial branch cut has angle $\beta$. On a positively oriented circle, use the one-sided branch values on either side of its crossing with the cut. Parametrize from $\theta =\beta$ to $\beta +2\pi$; adding an integer multiple of $2\pi i$ to the branch does not change the integral because the integral of $dz$ is zero. Along the open arc, a primitive is $z\mathrm{Log}z-z$. Its endpoint difference is $2\pi i$ times the cut-crossing point $R{e}^{i\beta }$, proving the displayed formula. The circle crosses the cut, so this is a piecewise contour integral of boundary values, not the integral of a single holomorphic logarithm on a neighborhood of a closed circle. The Cauchy integral theorem therefore does not make it zero.

A contour deformation changes the path of a contour integral through a region where its integrand is a holomorphic function. The Cauchy integral theorem preserves the integral when the endpoints are fixed and the connecting boundary terms vanish. Crossing a pole instead contributes the appropriately oriented residue; a branch cut limits which deformations are permitted on one analytic sheet.

A pair of oppositely directed contours on a branch cut produces the difference of the two boundary values integrated over their actual endpoint interval. Replacing finite endpoints by zero and infinity requires a limiting argument and convergence of the combined improper integral. Individual side integrals may diverge even when their difference converges, so cancellation must precede separation of endpoint limits.

Take $a>0$ and define $z^s$ using $-\pi /2<\arg z<3\pi /2$, so the branch cut lies on the negative imaginary axis. An indented upper semicircle encloses the simple pole $ia$. Its two straight segments contribute $(1+e^{i\pi s})$ times the desired real integral, while the residue theorem gives $\pi a^{s-1}{e}^{i\pi s/2}$. The large arc vanishes as its radius to the power $s-1$, and the small arc as its radius to the power $s+1$, giving the stated convergence range and result. This contour avoids placing a branch cut on either straight integration segment.

Start at $a=1$ with complex logarithm germ of a holomorphic function $L(1)=0$ and holomorphic square root germ of a holomorphic function $S(1)=1$. On the punctured plane $D=\mathbb{C}\setminus \{0\}$, both admit analytic continuation along every path. To construct it locally, for any $c\ne 0$ choose a small open disc avoiding $0$, select a holomorphic logarithm there, and select its additive multiple of $2\pi i$ to agree with the arriving germ of a holomorphic function; exponentiating half this logarithm supplies the arriving holomorphic square root. A finite chain of these open discs follows any given path.

Compare two paths from $1$ to $i$, parametrized by $0\le t\le 1$:The first follows a counterclockwise quarter-circle; the second follows a clockwise three-quarter-circle. Following ${\gamma }_A$ and then path reversal of ${\gamma }_B$ makes one positive turn about $0$. Its winding number is $1$, so the two paths cannot be homotopic with endpoints fixed in the punctured plane.

The continued values along the chosen paths areAt their common endpoint,Both complex logarithm values exponentiate to $i$, and both holomorphic square root values square to $i$. These give different terminal germ of a holomorphic function representatives, consistently obtained from the same starting germ of a holomorphic function. The monodromy theorem applies, but does not equate these two non-homotopic paths.

The closed path $\lambda (t)=e^{2\pi it}$ instead returns to $1$, with continued values $L_{\lambda }(t)=2\pi it$ and $S_{\lambda }(t)=e^{\pi it}$. One counterclockwise turn changes the complex logarithm by $2\pi i$ and flips the holomorphic square root from $1$ to $-1$. Hence neither initial germ of a holomorphic function can come from a global holomorphic logarithm or a global holomorphic square root of $z$ on the punctured plane. Repeating a loop of winding number $n$ gives the changes $2\pi in$ and $(-1{)}^n$. ^[ref]

Now restrict to the slit domain $D_{\mathrm{slit}}=\mathbb{C}\setminus (-\infty ,0]$. The principal logarithm and $\exp (\mathrm{Log}z/2)$ realize the same initial germ of a holomorphic function for each expression globally. To see that this is a simply connected domain, polar form identifies this domain with $(0,\infty )\times (-\pi ,\pi )$, a convex set; every loop contracts there. The simply connected corollary of the monodromy theorem now forces equality along every pair of paths with matching endpoints. In particular, ${\gamma }_A$ and the segment $\sigma (t)=1-t+it$ both give $L(i)=\pi i/2$ and $S(i)=(1+i)/\sqrt{2}$. The explicit homotopy $H(s,t)=(1-s){\gamma }_A(t)+s\sigma (t)$ stays in the closed first quadrant away from $0$, hence in $D_{\mathrm{slit}}$.

The other route ${\gamma }_B$ crosses the removed ray at $t=2/3$ and is no longer admissible. In the punctured plane, that crossing is locally regular: analytic continuation passes through $-1$ smoothly, although evaluating the principal logarithm would reset its argument. The branch cut changes the allowed domain; it does not create additional branch points or poles. That distinction is the beauty of the obstruction: perfectly compatible local holomorphic functions can fail to make a global choice solely because of the allowed based loops. The branches of a multivalued function article develops how moving the branch cut changes the selected values.

Monodromy records how a germ of a holomorphic function changes after analytic continuation around a closed path. The source point returns to itself, but the continued germ of a holomorphic function need not. For $z(t)=e^{it}$, $0\le t\le 2\pi$, starting with holomorphic square root value $1$ gives $w(t)=e^{it/2}$ and hence $w(2\pi )=-1$; starting with complex logarithm value $0$ gives $L(t)=it$ and hence $L(2\pi )=2\pi i$. After $m$ counterclockwise turns the values are $(-1{)}^m$ and $2\pi im$. Thus monodromy exchanges the two square-root branches of a multivalued function, whereas it shifts through infinitely many logarithm branches of a multivalued function. ^[ref]

The three panels use the same parameter $t$. They show analytic continuation through local holomorphic patches, including across a chosen branch cut; repeatedly evaluating the principal logarithm would instead reset its imaginary part at the cut. Monodromy makes the obstruction to a global branch of a multivalued function visible and motivates the use of Riemann surfaces as domains on which continued values stay distinct.

Choose the continuous argument with values in $(-\pi /2,3\pi /2)$ and defineIts branch cut is the nonpositive imaginary axis, including zero, exactly the removed ray. Locally away from that ray, the Cauchy-Riemann equations or the local inverse of the exponential show that it is a holomorphic logarithm, with derivative $1/z$ and $e^{\mathrm{Log}z}=z$. The chosen interval fixes the additive ambiguity by integer multiples of $2\pi i$.

Use the branch of the complex logarithm with $0<\arg z<2\pi$, so the branch cut is the positive real axis, and take $\sqrt{z}=\exp (\mathrm{Log}z/2)$ on the same slit plane. Apply the residue theorem toon a positively oriented keyhole contour with inner radius $\rho <1$ and outer radius $R>1$. The upper bank travels from $\rho$ to $R$, the large circle counterclockwise, the lower bank from $R$ to $\rho$, and the small circle clockwise. One may first keep the banks at small nonzero angles and then take their limits.

On the upper bank, $\mathrm{Log}x=\log x$ and $\sqrt{x}$ is positive. On the lower bank, the corresponding limits are $\log x+2\pi i$ and $-\sqrt{x}$. Reversing the lower-bank orientation therefore makes the sum of the bank integrals tend toBoth are absolutely convergent: near zero compare with $x^{-1/2}(1+|\log x|)$, and near infinity with $x^{-5/2}(1+\log x)$. On the large circle, $|z^2+1|\ge R^2-1$, so its contribution is $O((\log R+1)R^{-3/2})$. On the small circle it is $O(\sqrt{\rho }(1+|\log \rho |))$. Both tend to zero.

The two simple poles are $i$ and $-i$. On the chosen branch their arguments are $\pi /2$ and $3\pi /2$, respectively. Consequently their residues areFor example, the first follows from $\mathrm{Log}i=i\pi /2$, $\sqrt{i}=e^{i\pi /4}$ and the derivative $2i$ of the denominator factor $z^2+1$. The second uses $\sqrt{-i}=e^{3\pi i/4}$, rather than the value from the principal square-root branch. Adding and applying the residue theorem yieldsAs $J$ is real, taking real parts givesThe imaginary part also gives $J=\pi /\sqrt{2}$, an independent consistency check on the orientation and branch factors.

Choose the continuous argument with values in $(-\pi /2,3\pi /2)$ and defineIts branch cut is the nonpositive imaginary axis, including zero, exactly the removed ray. Locally away from that ray, the Cauchy-Riemann equations or the local inverse of the exponential show that it is a holomorphic logarithm, with derivative $1/z$ and $e^{\mathrm{Log}z}=z$. The chosen interval fixes the additive ambiguity by integer multiples of $2\pi i$.

The quadratic formula gives two algebraic possibilities. To select one consistent holomorphic square root, setwhere each factor is the principal square root. Both arguments lie in the upper half-plane, so neither crosses its branch cut. This choice satisfies $S(\zeta {)}^2={\zeta }^2-4$ and $S(\zeta )\sim \zeta$ at large $\zeta$ in the upper half-plane. The required inverse function isFor $\zeta =iR$, $R>0$, this gives $z=i(R+\sqrt{R^2+4})/2$, whose modulus exceeds one. As $\zeta$ varies continuously in the upper half-plane, this root cannot cross $|z|=1$, since that would make $\zeta$ real. Hence the selected root always lies in the fluid region. The other root is its reciprocal and lies inside the unit disk. Simply writing the principal square root of ${\zeta }^2-4$ would not specify this branch correctly throughout the upper half-plane.

Put $w=\mathrm{artanh}z$. The hyperbolic tangent identity $z=(e^{2w}-1)/(e^{2w}+1)$ gives $e^{2w}=(1+z)/(1-z)$. Taking a compatible branch of the complex logarithm givesAs an identity for a single-valued inverse hyperbolic tangent, this requires consistent branch cuts; one cannot choose unrelated logarithm branches on the two sides. Near $z=0$ the branch with value zero is fixed by $\mathrm{artanh}0=0$. The full inverse relation also allows $w\mapsto w+i\pi k$, $k\in \mathbb{Z}$, and $z=\pm 1$ are singular points.

Retain $\beta =\sqrt{1-V^2/c^2}>0$, but normalize the complex potential as $F=\beta H$, so that $w=\mathrm{Re}F/\beta$. The Cauchy-Riemann equations give $w_x=\mathrm{Re}{F}^{\prime }/\beta$ and $w_y=-\mathrm{Im}{F}^{\prime }/\beta$. Since ${\partial }_{x_2}=\beta {\partial }_y$, the two stresses satisfyUsing the same unscaled $H$ from part (a) would instead put an additional $\beta$ on its derivative; the potential normalization matters.

Take the principal branch of $\sqrt{z}$, with its branch cut along the negative real axis and positive $\sqrt{x}$ for $x>0$. Write $Q=F^{\prime }$. For the antisymmetric crack-opening response, reflection gives $Q^{-}(s)=-\overline{Q^{+}(s)}$ on the faces. The prescribed equal face stress gives $\mathrm{Im}{Q}^{\pm }(s)=f(s)/\mu$. Multiplication by the square-root factor converts this face condition to an additive jump. In fact, with $G=\sqrt{z}Q$ and $r=\sqrt{-s}$,The Sokhotski–Plemelj theorem applied to this jump giveswhere $P$ is entire after imposing the usual finite-energy crack-tip regularity. This is the Hilbert problem for an antiplane crack, with the stress-potential phase chosen explicitly above.

For localized face loading with no additional remote loading, the far displacement is bounded and the weighted Cauchy transform decays. In particular, for compactly supported loading, $G\to 0$ at infinity and Liouville theorem sets $P=0$. The resulting Cauchy solution for a steadily moving semi-infinite antiplane crack isIntegrals and boundary values are understood for loadings with the corresponding convergence and regularity, or as limits of localized loading. The reflection used above is the pure antiplane opening response: reflection with a sign change of displacement leaves both prescribed face stresses unchanged, and the usual no-extra-load uniqueness condition eliminates an unforced symmetric addition.

One can check the face stresses directly. Put $I_{\mathrm{pv}}(s)=\mathrm{PV}{\int }_{-\infty }^0\sqrt{-t}f(t)/(s-t)dt$. The Cauchy boundary limits giveBoth have $-\mu \mathrm{Im}{Q}^{\pm }=-f(s)$, verifying the prescribed traction and its sign on both faces. The opposite real parts also give the required reflection. Ahead of the tip, the integral is real and $Q$ is purely imaginary, so the bonded line has zero tangential displacement gradient and the two halves match.

For $x=x_1-Vt>0$, take $z=x$ in the boxed analytic solution. The ahead-of-tip stress isNo $\beta$ appears: at fixed co-moving position and the specified co-moving face loading, this stress is independent of the speed. The displacement, however, still scales with $1/\beta$, so speed independence does not apply to the whole elastic field. If ${\int }_{-\infty }^0|f(s)|/\sqrt{-s}ds$ is finite, the near-tip stress is inverse-square-root with coefficient ${\pi }^{-1}\int f(s)/\sqrt{-s}ds$.

The far-field condition is a genuine uniqueness requirement. A homogeneous stress-intensity field of a semi-infinite antiplane crack has $Q_h=iC/\sqrt{z}$, for real $C$. It adds no face traction but changes the stress ahead by $-\mu C/\sqrt{x}$. Although its stress tends to zero at infinity, its primitive is $2iC\sqrt{z}$, which gives unbounded remote displacement and infinite total elastic energy. Thus only asking for stress to vanish at infinity would leave the homogeneous constant undetermined; excluding that added remote stress-intensity loading selects the displayed face-loaded solution. The ordinary inverse-square-root tip singularity itself has finite energy in a bounded neighbourhood and is not being excluded.

Take $0<{\theta }_0<\pi$, put $a=k_0\cos {\theta }_0$ and $b=k_0\sin {\theta }_0>0$, and suppress the common factor $e^{i\omega t}$. The upper-half-plane incident acoustic plane wave is $e^{iax+iby}$. A rigid zero-thickness screen imposes the same normal-derivative cancellation on both faces; the scattered velocity potential is consequently odd in $y$. Across the open part $x>0$ it is continuous, so its trace there is zero. These complementary half-line data are what make the Wiener-Hopf method applicable.

Use the Fourier transform $\hat{\varphi }(\kappa ,y)={\int }_{\mathbb{R}}\varphi (x,y)e^{i\kappa x}dx$, with inverse exponential $e^{-i\kappa x}$. The outgoing Helmholtz equation solution is $\hat{\varphi }=F^{-}(\kappa )e^{-\gamma y}$ for $y>0$, where $\gamma =\sqrt{{\kappa }^2-k_0^2}$ has $\gamma =i\sqrt{k_0^2-{\kappa }^2}$ on the propagating interval and positive real part on evanescent components. The limiting absorption principle supplies the contour prescription. The transform $F^{-}$ of the trace supported on $x<0$ is analytic below the contour. Let $G^{+}$ be the analytic upper transform of the unknown derivative on $x>0$. The prescribed derivative on $x<0$ has transformSince the upper normal derivative is $-\gamma F^{-}$, the Wiener-Hopf equation isFor the Wiener-Hopf factorization, choose ${\gamma }_{+}=(\kappa -k_0{)}^{1/2}$ analytic above and ${\gamma }_{-}=(\kappa +k_0{)}^{1/2}$ analytic below. Their branch cuts go into the opposite half-planes, with ${\gamma }_{+}(-a)=i\sqrt{k_0+a}$. Dividing by ${\gamma }_{+}$ and performing pole subtraction in a Wiener-Hopf equation givesThe forcing pole is allocated to the lower analytic part; the second quotient has a removable pole and is upper analytic. Subtracting these parts leaves a common entire function. The finite-energy edge condition excludes an inverse-square-root singularity in the potential: its odd scattered trace is $O(\sqrt{|x|})$, so $F^{-}=O(|\kappa {|}^{-3/2})$. The normal derivative on the open side can be $O(x^{-1/2})$, giving $G^{+}=O(|\kappa {|}^{-1/2})$. Both sides of the entire remainder tend to zero. The Liouville theorem sets that remainder to zero, giving the explicit Wiener-Hopf solution of rigid half-plane diffraction:The scattered field on either side of the screen is thereforeIts forcing-pole residue at $\kappa =-a$ is $-i$. Deforming to the outgoing steepest descent contour crosses this pole precisely when $|\theta |>\pi -{\theta }_0$, where $x=r\cos \theta$, $y=r\sin \theta$ and $-\pi <\theta <\pi$. The residue contributes $\mathrm{sgn}(y)e^{iax-ib|y|}$. Thus, away from transition boundaries, the scattered geometrical optics field isAdding the incident acoustic plane wave gives incident plus unit specular reflection in the upper reflected sector, just the incident wave in the remaining illuminated sectors, and zero in the lower shadow sector. The geometrical shadow boundaries also follow directly by tracing a straight ray to $y=0$: the reflected ray hits the screen at $x+y\cot {\theta }_0<0$, and the transmitted incident ray crosses the open part at $x-y\cot {\theta }_0>0$.

For the diffracted contribution, the outgoing phase $-i\kappa r\cos \theta -\gamma r|\sin \theta |$ has saddle ${\kappa }_s=k_0\cos \theta$, value $-i{k}_{0}r$, and second derivative $ir/(k_0{\sin }^2\theta )$. Stationary phase therefore supplies $\sqrt{2\pi k_0/r}|\sin \theta |e^{-i{k}_{0}r+i\pi /4}$. Substituting $F^{-}({\kappa }_s)$, and using $\mathrm{sgn}(y)|\sin \theta |/\sqrt{1+\cos \theta }=\sqrt{2}\sin (\theta /2)$, givesThe full asymptotic scattered field is ${\varphi }_{\mathrm{GO}}+{\varphi }_d$. The angular edge-wave expression is not uniform on the geometrical boundaries. There is also a sign inconsistency in the supplied saddle formula: with its printed inverse exponential $e^{+i\kappa x}$, the outgoing saddle is $-k_0\cos \theta$, not $+k_0\cos \theta$. The consistent inverse convention used above gives the requested outgoing field without that ambiguity.

Choose the principal branch of a multivalued function, $-\pi <\arg s<\pi$, for $s^{1/2}$. The integrand $G(s)=e^{st}/((s^2+1)\sqrt{s})$ has simple poles at $\pm i$ and a branch cut along the negative real axis. Its residues arewhose sum is $\sin (t-\pi /4)$.

For $t>0$, close the Bromwich contour to the left using a rectangle with outer real part $-R$ and horizontal edges at imaginary parts $\pm R$, indenting the cut and zero. On every distant edge, $|s^2+1|\ge |s{|}^2-1$, $|\sqrt{s}|=|s{|}^{1/2}$ and $|e^{st}|\le e^{\gamma t}$. Each edge has length $O(R)$ and $|s|\ge R$, so its integral is $O(R^{-3/2})$. The small circle of radius $\epsilon$ about zero contributes $O({\epsilon }^{1/2})$. The tails of the original vertical integral are absolutely bounded by $O(R^{-3/2})$ as well.

On the upper and lower banks at $s=-r$, the square roots are $i\sqrt{r}$ and $-i\sqrt{r}$, respectively. With the upper bank traversed toward zero and the lower bank away from zero, their combined integral isThe cut integral converges near zero since $r^{-1/2}$ is integrable, and at infinity by exponential decay. The residue theorem, followed by $R\to \infty$ and $\epsilon \to 0$, therefore givesFor $t<0$, close the Bromwich contour to the right. There are no enclosed poles or cuts, and $|e^{st}|\le e^{\gamma t}$ there. The same distant-edge estimates show $f(t)=0$ for $t<0$.

For the large-$t$ expansion, $1/(1+r^2)=1-r^2+r^4/(1+r^2)$. Scaling $r=u/t$ and using the given integral yields ${\int }_0^{\infty }{e}^{-rt}{r}^{-1/2}dr=\sqrt{\pi }{t}^{-1/2}$. Integration by parts twice gives the next coefficient $\Gamma (5/2)=3\sqrt{\pi }/4$. The remainder is bounded by ${\int }_0^{\infty }{e}^{-rt}{r}^{7/2}dr=\Gamma (9/2)t^{-9/2}$. Thus, without needing a globally convergent termwise integration of the local power series,In particular, the requested two leading terms are $\sin (t-\pi /4)+(\pi t{)}^{-1/2}$, interpreted as an additive asymptotic expansion. This is Bromwich inversion with a square-root branch cut.

Use a Fourier transform parallel to the boundary. Splitting its traces into right- and left-supported parts gives Half-range Fourier transforms $F_{+}$ and $F_{-}$, analytic respectively above and below a common inversion contour. A small positive imaginary part of the wavenumber implements the limiting absorption principle and separates the incident poles and outgoing branch points. Solve the transformed differential equation normal to the boundary using its outgoing or decaying solution. The boundary conditions then yield a Wiener-Hopf equation of the formFactor its Wiener-Hopf kernel as $K=K_{+}{K}_{-}$, with each factor analytic and nonzero in its designated half-plane. Divide by $K_{-}$ and split $H/K_{-}=H_{+}+H_{-}$, allocating poles according to the incident-wave prescription. ThenThe two sides continue to a common entire function. Their growth at infinity, inherited from the field's edge behavior, determines that function, usually a constant or a polynomial; in a decaying case it is zero. Solve for the unknown transforms and apply Fourier inversion, preserving the outgoing contour prescription.

The principal analytic results used are as follows. The Paley–Wiener theorem and its half-line Laplace-transform version relate support on a half-line to holomorphy in a half-plane, with growth or square-integrability bounds. The Cauchy integral formula and Sokhotski–Plemelj theorem provide additive splitting: a Cauchy integral of a sufficiently decaying jump has upper and lower boundary values with that prescribed difference. Applying the same construction to a continuous logarithm of a nonvanishing scalar Wiener-Hopf kernel gives Wiener-Hopf factorization when its winding number is zero; a nonzero index requires an explicit index factor. The identity theorem gives analytic continuation across the common strip. Finally Liouville's theorem makes a bounded entire function constant; its polynomial-growth extension bounds the possible degree. The physical edge and radiation conditions select the remaining constants. Zeros, branch cuts, nonzero index and insufficient decay must be accounted for rather than discarded during splitting.

On the stated branch of the principal complex logarithm, the only denominator zero is $z=1$, and $\log z=(z-1)+O((z-1{)}^2)$. ThereforeZero is a logarithmic branch point, not a removable isolated singularity, even though $1/\log z$ tends to zero as $z\to 0$ within the cut plane. Continuation around zero changes $\log z$ by $2\pi i$. The nonpositive real axis is the chosen branch cut; its nonzero points are not intrinsically branch points and admit local analytic continuation. There are no other finite singularities inside the specified branch domain.