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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 $L=D_t^2+2k{D}_t+k^2+{\omega }^2$, with $k>0$ and $\omega >0$, the causal Green function is $G(t)=H(t)e^{-kt}\sin (\omega t)/\omega$, where $H$ is the Heaviside step function. It satisfies $LG=\delta (t)$: it is continuous at zero, with a unit jump in its first derivative. Its smooth positive-time kernel $g$ obeys $g(0)=0,g^{\prime }(0)=1$. The zero-frequency limit is $H(t)t{e}^{-kt}$. Shifting $G$ gives the zero-initial-data response to a delayed Dirac delta function.

With the Fourier transform convention $\hat{f}(\xi )=\int f(t)e^{-i\xi t}dt$, the transformed equation gives $\hat{G}(\xi )=[k^2+{\omega }^2-{\xi }^2+2ik\xi {]}^{-1}$. Its poles $\xi =ik\pm \omega$ lie in the upper half of the complex plane. Inverting by contour integration therefore gives zero for negative time and an exponentially damped oscillation for positive time.

For $L=(D_t+a)(D_t+b)$ with $a>b>0$, $H$ denotes the Heaviside step function. This Green function vanishes at negative time and satisfies $LG=\delta (t)$. Its positive-time part solves the homogeneous ordinary differential equation, is zero at the origin, and has right derivative one. Thus it is continuous with a unit derivative jump, which gives precisely the Dirac delta function in its second distributional derivative.

With the negative-exponential Fourier transform convention,Both poles lie in the upper half-plane. Contour integration closes above for positive time and below for negative time, giving the title expression. When the integral converges, convolution $y(t)={\int }_{-\infty }^{t}G(t-s)f(s)ds$ supplies the response with no freely added homogeneous transient growing toward negative infinity. The continuous limit $a\to b$ is $G(t)=H(t)t{e}^{-bt}$.

Complex analysis studies functions of a complex number, especially holomorphic functions. Its distinctive feature is that a single complex derivative controls variation in every direction of the complex plane. The map $f(z)=z^2$ gives a first picture: straight coordinate lines become curved, yet small perpendicular arrows remain perpendicular wherever $f^{\prime }(z)\ne 0$.

The bottom panels show displacements rather than absolute positions. Multiplication by $f^{\prime }(1+i)=2+2i$ predicts the image of each small arrow; the exact image differs by $h^2$. The origin is exceptional because $f^{\prime }(0)=0$. This distinction between local conformal maps and a critical point is central: complex differentiability imposes strong geometric structure, without making every holomorphic function globally one-to-one.

The remarkable next step is that holomorphic functions are determined inside a circle by their values on its boundary. Precisely, let $R>0$, let $D=\{z:|z-a|<R\}$, and suppose $f$ is a holomorphic function on an open set containing the closed disc $\overline{D}$. Orient its boundary counterclockwise. The Cauchy integral formula states that, for every $w\in D$,The contour integral combines boundary values with a kernel depending on the requested interior point. It is a reconstruction formula, not merely a trick for evaluating an integral. The hypothesis about the closed disc ensures that all boundary values and the contour integral are well defined. ^[ref]

At the centre, substituting $z=a+R{e}^{i\theta }$ cancels the denominator against $dz=iR{e}^{i\theta }d\theta$:Thus the centre value is the angular average of the boundary values. For the polynomial $f(z)=z^2$, $a=1+i$ and $R=1/2$, expand the boundary data asBoth oscillating terms have zero average, leaving $f(a)=2i$. Away from the centre the weights change: on the unit circle,With $f(z)=z^2$ and $w=1/2$, expanding the denominator as a geometric series leaves only its second-power term after integration, giving $f(1/2)=1/4$. This also explains how the Cauchy integral formula recovers an entire Taylor series from boundary data.

When a holomorphic function has isolated singularities, a different boundary calculation detects their residues. Suppose $\Gamma$ is a counterclockwise simple closed contour and $f$ is holomorphic on a neighbourhood of it and its interior except at finitely many interior isolated singularities $a_j$. The residue theorem givesA residue is the coefficient of $(z-a_j{)}^{-1}$ in the Laurent series. A short proof idea is to deform the contour to small circles around the isolated singularities using the Cauchy integral theorem. Integrating the Laurent series on each circle kills every power except $-1$. ^[ref]

Here is how a complex pole determines a real improper integral. Set $g(z)=1/(1+z^2)$ and use the contour consisting of the segment from $-R$ to $R$, followed by the upper semicircle $A_R$, with $R>1$. Only the simple pole $i$ lies inside, andThe arc must actually disappear. On $A_R$, the reverse triangle inequality gives $|1+z^2|\ge R^2-1$, and its length is $\pi R$, soThe real integrand is bounded by $x^{-2}$ for $|x|\ge 1$, so both tails converge separately. Taking the limit therefore proves the ordinary improper integralFor example, at $R=10$ the real segment contributes $2\arctan (10)\approx 2.942255$, and the arc contributes $0.199337$, below the bound $10\pi /99\approx 0.317333$. The usefulness of the residue theorem is that the same strategy works for many rational functions whose real antiderivatives are cumbersome: find the enclosed poles, compute their residues, and control the connecting arc. ^[ref]

Historically, Cauchy's 1825 memoir on integrals with imaginary limits develops complex contour integration and its use in transforming and evaluating definite integrals. The opening sections define complex path integrals, discuss changing the path, and examine what happens near singular points. These are recognizable ancestors of the modern link between holomorphic functions, contours and residues. ^[ref]

The Fokas method uses a global relation for a linear boundary value problem between transformed boundary traces, complex spectral symmetries, and contour integration to eliminate unknown traces or construct a numerical boundary scheme. The distinction between the domains of analyticity of the transforms and the decay sectors of spectral exponentials controls valid contour deformations.

In ${\int }_a^{b}f(x)dx$, the function $f$ is the integrand. It is distinct from the integration variable, the bounds and the resulting integral. In contour integration, the transformed integrand includes the parametrization derivative.

For any $0<r<1$, a Laurent series converges uniformly on the circle $|z|=r$, so termwise contour integration is valid. Multiplication by $z^{-k-1}$ and integration gives, for every integer $k$,Indeed, the integral of $z^{j-k-1}$ is $2\pi i$ for $j=k$ and zero otherwise. These coefficients depend only on $f$, so any two Laurent expansions must agree coefficient by coefficient. This proves uniqueness, including all terms of the principal part.

Assume $q(z)=w^{\prime }(z)$ is single-valued and a holomorphic function between $C$ and a sufficiently large enclosing circle $|z|=R$. There are no fluid singularities in that region, so Cauchy's integral theorem permits contour deformation:Use the exterior Laurent series about zero, valid for $|z|$ larger than every finite singularity:It converges uniformly on the chosen circle, permitting termwise contour integration. For integer $n$, parameterization by $z=R{e}^{i\theta }$ givesOnly the $z^{-1}$ coefficient survives, provingThus if $b_1=\alpha +i\beta$, then $\Gamma =-2\pi \beta$ and $M=2\pi \alpha$. A uniform flow contributes a constant to $q$ and changes neither integral. The Laurent series here is an exterior expansion; it need not converge in a punctured neighborhood of zero if other singularities intervene.

The Cauchy integral formula states that if $f$ is holomorphic on a neighbourhood of a simple closed positively oriented contour and its interior, then, for an interior point $z$,Fix $0<r<1$ and apply it on $|\xi |=r$. For $|z|<r$, the geometric seriesconverges uniformly in $\xi$ and locally uniformly in $z$. Indeed, for $|z|\le s<r$ its absolute terms are bounded by $s^n/r^{n+1}$. Since $f$ is bounded on the contour, termwise contour integration is valid, givingTermwise differentiation of this locally convergent power series gives $f^{(n)}(0)=n!c_n$. ThereforeThis also derives the needed Taylor series representation rather than assuming an unjustified interchange at the boundary of the analytic disk.

Use the Half-range Fourier transform and the finite-time spectral boundary transformswhere $g_j={\partial }_x^{j}q(0,\cdot )$ for $j=0,1,2$. The spatial transforms are holomorphic in $\mathrm{Im}k<0$, while these finite-time transforms are entire functions. For now $g_2$ is an unknown boundary trace. Integrating the local relation over the half-strip, using spatial decay at infinity, gives the global relationThe boundary sign follows from integrating the flux derivative: its contribution is minus the flux at zero.

Fourier inversion first gives the initial real-line integral plus a real-line integral of $F$. To put the boundary contribution on its useful complex contour, defineOrient its full boundary with $D_{+}$ on the left. It consists of the real axis from left to right and the upper curve $C:k=u+i\sqrt{1+3u^2}$ from right to left. The real-axis piece must not be discarded for this dispersion. The contour integration representation isTo justify the replacement, the integral of $e^{ikx-wt}F$ along $C$ is zero by closing above $C$. In that region $\mathrm{Re}w\ge 0$, and each boundary term is an integral of $e^{-w(t-s)}$ over $0\le s\le t$ times a polynomial in $k$. This has no growing temporal exponential there; $e^{ikx}$ suppresses the upper closing arc for $x>0$. The Cauchy integral theorem then applies. The figure shows the full contour and, separately, the sectors used in the nonlinear problem.

One may replace $F(k,t)$ by $F(k,T)$ for any finite $T\ge t$. Their difference is a future-time integral over $t<s<T$; its factor $e^{w(s-t)}$ decays inside $D_{+}$, so its integral around $\partial D_{+}$ vanishes. All contour formulas are understood as limits of truncated contours, with standard small displacements if a boundary transform needs an Abel limit. Compatibility of the initial and boundary data controls the corner behavior. This is the two-trace half-line cubic dispersion representation.

Part (i) gives $\mathcal{L}(t)=1/s^2$ and $\mathcal{L}(t{e}^{-t})=1/(s+1{)}^2$. The convolution theorem for Laplace transforms therefore givesRepeated integration by parts yieldsSubstitution gives $f(t)=t-2+(t+2)e^{-t}$, agreeing with the contour integration result.

The ordinary Fourier transform does not exist. There is a real simple pole at $x=-1$, withThe two separate improper integrals diverge logarithmically. Thus no ordinary large-$k$ expansion is defined under the stated integral convention. A Cauchy principal value or a contour prescription would be additional data, and must be stated rather than silently introduced.

For completeness, the symmetric Cauchy principal value has a precise answer. The real pole contributes the principal-value Fourier transform of a real poleThe other poles are $z_{\pm }=(1\pm i\sqrt{3})/2$ with residues $r_{\pm }=1/(3z_{\pm }^2)=(-1\mp i\sqrt{3})/6$. For $k>0$, close the contour integration in the lower half-plane, with clockwise orientation. The nonreal pole contributes $-2\pi i{r}_{-}{e}^{-ik{z}_{-}}$, while the real principal-value pole supplies the half-residue above. For $k<0$, the upper-half-plane contour gives the conjugate result. With $s=\mathrm{sgn}(k)$, the exact principal-value formula for $k\ne 0$ isThese are its two nonzero asymptotic contributions: a nondecaying oscillatory real-pole term and an exponentially small complex-pole term. There is no second nonzero inverse-power term. This formula is qualified by the principal-value choice; it is not the ordinary transform requested in the statement.

Other prescriptions change the leading term. For example, replacing the real pole by its upper or lower boundary value changes the distribution by $\mp i\pi \delta (x+1)/3$, and hence changes the transform by $\mp i\pi e^{ik}/3$. This demonstrates why a pole prescription is essential. The smooth Taylor series at the origin alone would miss the decisive real-pole contribution.

Use the half-line Fourier transform and finite-time boundary transformsThe half-line Fourier transform is analytic for $\mathrm{Im}k<0$ under spatial decay. All spectral integrals below have their usual oscillatory, or vanishing Gaussian damping, interpretation until absolute convergence is established.

The free Schrodinger equation has the local divergence identityIntegrating in $x,t$ gives the global relation for the half-line free Schrodinger equationLet $D_{+}=\{\mathrm{Re}k>0,\mathrm{Im}k>0\}$. Orient its boundary from $i\infty$ down to $0$, and then from $0$ to $+\infty$, so that $D_{+}$ lies on the left. Fourier inversion followed by contour integration in the second quadrant yieldsThis is a complex spectral representation involving the initial trace and both boundary traces. The contour deformation works because each boundary-time integrand contains $e^{-i{k}^2(t-s)}$ with $s\le t$, which decays in the second quadrant, as well as $e^{ikx}$ for $x>0$. This fixes both the quadrant and the orientation; changing either without changing the signs would produce an incorrect representation.

Substitute part (b) into the spectral representation. The terms containing $\psi (ik)$ areThe transform $\psi (ik)$ is analytic for $\mathrm{Im}k>0$. Write $k=a+ib$. In the first quadrant,which decays since $x,y>0$. The Cauchy integral theorem in that quadrant equates the two integrals in the first parentheses. In the second quadrant,which decays since $a<0$, $b>0$, and $y<\ell$. The same contour integration equates the two integrals in the second parentheses. Therefore $\psi (ik)$ makes no contribution. This is upper-quadrant cancellation of reflected boundary transforms, and it uses the whole paired expression rather than attempting to discard an individual integral.

The terms involving $c$ cancel by the same two quadrant arguments with the analytic factors $k\pm \gamma$ and $\psi$ omitted. There is a sign defect in the printed reconstruction prefactor. The sides specified in part (a) form a clockwise boundary, so the Cauchy integral theorem gives a negative reconstruction sign. Indeed, integrating each spectral ray first yields $i/(z-\zeta )$, and hence the sum of the three ray integrals is $i{\oint }_{\mathrm{clockwise}}{q}_z(\zeta )/(z-\zeta )d\zeta =-2\pi q_z(z)$. Thus the printed positive prefactor reconstructs $-q_z$. The requested cancellation above holds with either overall sign, but the actual data-dependent derivative isDecay at infinity fixes the integration constant when recovering $q$ from its Wirtinger derivative.

For completeness, the missing condition at infinity and the unrestricted printed $\gamma$ deserve an explicit solvability of a decaying Robin strip check. Let $Y_n$ be orthonormal eigenfunctions from a Sturm-Liouville problem of $-d^2/d{y}^2$ with $Y_n^{\prime }(0)=\gamma Y_n(0)$ and $Y_n^{\prime }(\ell )=-\gamma Y_n(\ell )$, and eigenvalues ${\lambda }_n$. With $g_n={\int }_0^{\ell }g{Y}_{n}dy$, separation of variables givesIndeed, each coefficient obeys $q_n^{\prime \prime }-{\lambda }_n{q}_n=0$ and $q_n^{\prime }(0)=g_n$. A positive eigenvalue has one decaying exponential; a zero eigenvalue has only affine solutions and a negative eigenvalue only oscillatory solutions, neither of which decays unless its coefficient vanishes. This also proves uniqueness in the decay class and justifies reflection symmetry there. For $\gamma >0$, all eigenvalues are positive. For $\gamma =0$, the constant eigenfunction requires ${\int }_0^{\ell }gdy=0$; the apparent zero of $D$ at the origin is then removable. For $\gamma <0$, compatibility with every nonpositive eigenmode is necessary. These restrictions cannot be inferred from smoothness and reflection symmetry alone. The printed boundary value problem without any far-field condition permits additional growing or nondecaying homogeneous solutions, whereas the spectral transforms select the decaying branch.

Jordan lemma states that if $a>0$ and $g$ is analytic in the upper half-plane outside a fixed circle, withthen on the upper semicircle $C_R$,Indeed, $|e^{iaR{e}^{i\theta }}|=e^{-aR\sin \theta }$. Using symmetry and $\sin \theta \ge 2\theta /\pi$ on $0\le \theta \le \pi /2$ gives

The residue of $f$ at an isolated singularity $a$ is the coefficient of $(z-a{)}^{-1}$ in its Laurent series, equivalentlyIf $f(z)=g(z)/(z-a{)}^k$, the Taylor expansion of the analytic function $g$ shows that

For the real integral, apply contour integration in the upper half-plane toJordan's lemma removes the semicircular arc. The only enclosed singularity is the double pole at $z=i$, whose residue isThe residue theorem therefore givesIts real part vanishes because $x^3\cos x/(1+x^2{)}^2$ is an odd function, while its imaginary part is the requested integral. Hence

Differentiating under the contour integration sign givesSince ${\partial }_t{e}^{zt}=z{e}^{zt}$, integration by parts turns this intoThe integral vanishes whenHence, up to an irrelevant multiplicative constant,The remaining endpoint condition isThus each open end of $\gamma$ must lie where $\mathrm{Re}(zt-1/t)\to -\infty$; a closed contour also makes the boundary term vanish when the integrand is single-valued along it.

Consider the meromorphic functionClose the contour with a large semicircle in the upper half-plane, indenting above the three real pole $-1,0,1$. The large-arc integral vanishes by Jordan's lemma, while the upper-half-plane indentation rule givesThe three residue areTheir sum is $\cos 1-1$, so the residue theorem yields

The real part of the integrand is an odd function, so its full-line principal value is zero. Its imaginary part,is an even function. Taking imaginary parts and halving the full-line integral therefore gives

After integration by parts, the quadratic action isWith the pole prescription appropriate to the conventions in the question, its inverse kernel isFor $t>0$ close the contour in the lower half-plane and for $t<0$ close it in the upper half-plane. The enclosed pole in each case giveswhich equivalently satisfies $({\partial }_t^2+{\omega }^2)D(t)=i\delta (t)$.

The source-dependent Gaussian functional integral is evaluated by translating the integration variable by the classical sourced solution. Completing the square givesChanges in the sign of the source term or of the path-integral phase move factors of $i$ between $D$ and the exponent but leave the contraction rules equivalent.