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Let $p(D)$ be a constant-coefficient differential operator, $D=d/dt$, with distinct characteristic roots ${\lambda }_j$ having nonzero real parts. The solution of $p(D)K=\delta$ decaying at both temporal infinities iswhere $H$ is the Heaviside step function. This follows by partial fraction decomposition of $1/p(z)$ and the first-order kernels $e^{\lambda t}H(t)$ for $\mathrm{Re}\lambda <0$, or $-e^{\lambda t}H(-t)$ for $\mathrm{Re}\lambda >0$. Both first-order kernels jump by one and solve $(D-\lambda )K_{\lambda }=\delta$ as distributions.

For example, $p(D)=(D+1)(D-2)(D+2)$ givesIt is continuous together with its first derivative; its second derivative jumps by one, as required by the jump conditions for a third-order Green function. A solution of $p(D)y=f$ is the convolution $K\ast f$ when it exists. If positive-real-part roots occur, decay at both ends generally requires a nonzero negative-time branch: this is the distinction captured by causality versus stability of a Green function. A causal kernel would retain growing positive-time terms and need not have a real-axis Fourier transform.

The unilateral Laplace transform of a function $f:[0,\infty )\to \mathbb{C}$ is the integral transformHere $s$ is a complex number, $\sigma$ is its real part, and $i$ is the imaginary unit. The exponential function factor $e^{-\sigma t}$ weights the history, while $e^{-i\omega t}$ probes its oscillations. The integral starts at zero, so it uses only the history from that starting time onward. This section concerns ordinary functions; their value at the single endpoint does not change the integral. See MIT 18.03SC, definition and elementary examples.

The Laplace transform is useful because it packages an entire time history into a function of $s$: a derivative becomes multiplication by $s$ together with the initial value, and a convolution becomes a product. Thus a linear differential equation can become an algebraic equation that already remembers its initial value problem. The worked example below shows both descriptions of the same response.

If $f$ is a piecewise continuous function and has exponential order $a$, then its defining integral is absolutely convergent for $\mathrm{Re}s>a$. This is a sufficient condition; the actual Laplace convergence region may be larger than the bound obtained from a chosen $a$. ^[ref].

The bound has a short explanation. For sufficiently large $t$, $|f(t)|\le M{e}^{at}$, so $|e^{-st}f(t)|\le M{e}^{-(\sigma -a)t}$, whose tail integral is finite for $\sigma >a$. The finite initial interval is covered by piecewise continuity.

Three calculations make the convergence condition concrete. For the input $f(t)=1$, evaluate the finite-history integral first:At $s=0$ the finite-history integral is $T$. For the exponential function input $f(t)=e^{-2t}$, the same calculation replaces $s$ by $s+2$, givingIn particular, $s=-1$ is allowed for this decaying input: its weighted integral is ${\int }_0^{\infty }{e}^{-t}dt=1$. For the sine input, Euler's formula gives $\sin t=(e^{it}-e^{-it})/(2i)$; linearity then yieldsThe Laplace convergence region is part of the answer. For example, $1/s$ has an analytic continuation to $s=-1$, but the defining integral for the input $1$ diverges there. The sine calculation is also developed in MIT 18.03SC, further elementary transforms.

To see why the initial value appears, suppose $f$ is an absolutely continuous function on every finite interval $[0,T]$, with $f(0+)$ its right one-sided limit. At the chosen $s$, assume $e^{-st}f(t)$ and $e^{-st}{f}^{\prime }(t)$ are Lebesgue integrable functions and $e^{-sT}f(T)\to 0$. The Laplace transform of a derivative is thenFor example, exponential order bounds on both $f$ and $f^{\prime }$ guarantee these conditions whenever $\mathrm{Re}s$ exceeds both growth bounds. The initial value is the lower-end boundary contribution of integration by parts.

Consider the initial value problemWrite $Y=\mathcal{L}\{y\}$. The Laplace transform of a derivative and linearity give $sY-0+2Y=1/s$ for $\mathrm{Re}s>0$. A partial fraction decomposition therefore givesWe recognize the two elementary Laplace transforms already computed. Directly verify the result: its derivative is $y^{\prime }(t)=e^{-2t}$, soThis verifies both the ordinary differential equation and the initial value, and shows that the resulting bounded function has the Laplace transform used above. The response rises from zero toward $1/2$: at that value the input $1$ balances the term $2y$. The general method is described in MIT 18.03SC, solving initial value problems.

The same solution is a convolution. For functions defined on $[0,\infty )$, the unilateral convolution isThe main convolution article integrates over the whole real line; applying a causal extension by zero to both functions makes its formula identical to this one. Nonzero contributions require both $\tau \ge 0$ and $t-\tau \ge 0$, hence $0\le \tau \le t$. If $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ are Lebesgue integrable functions on $[0,\infty )$, the convolution theorem for Laplace transforms statesFor piecewise continuous functions of exponential order $a$ and $b$, it holds throughout $\mathrm{Re}s>\max (a,b)$. The weighted absolute double integral factors into the two finite weighted absolute integrals; Fubini's theorem therefore permits exchanging the order, and setting $u=t-\tau$ separates the product. ^[ref].

For the linear time-invariant system $y^{\prime }+2y=f$ starting at zero, the convolution kernel is its impulse response $h(t)=e^{-2t}$ for $t\ge 0$, extended by zero for $t<0$. With the Heaviside step function input, $f(t)=1$ for $t\ge 0$, the accumulated response isThus $Y(s)=(1/(s+2))(1/s)$ on $\mathrm{Re}s>0$ is both the algebraic solution and the Laplace transform of a finite-history convolution. Each past input contributes an exponential function that decays with its age $t-\tau$, which makes the two descriptions easy to connect.

Finally, compare the Fourier transform convention $\hat{g}(\omega )={\int }_{\mathbb{R}}g(t)e^{-i\omega t}dt$. If $f_0$ is the causal extension by zero of $f$, define $g_{\sigma }(t)=e^{-\sigma t}{f}_0(t)$. Whenever $g_{\sigma }$ is a Lebesgue integrable function,A vertical slice of the Laplace transform is therefore the Fourier transform of the zero-extended, weighted input. Positive $\sigma$ damps its positive-time history. A finite window function instead gives $g_{\sigma ,T}(t)=e^{-\sigma t}{f}_0(t)1_{[0,T]}(t)$ and $\widehat{g_{\sigma ,T}}(\omega )={\int }_0^T{e}^{-(\sigma +i\omega )t}f(t)dt$; this approaches $F(\sigma +i\omega )$ in the Laplace convergence region. Setting $\sigma =0$ directly requires an integrable zero-extended input: the input $1$ fails that condition, even though its Laplace transform exists for every $\sigma >0$.

The history of the Laplace transform spans several stages. Schiff's historical summary notes that Laplace's 1812 treatise acknowledged earlier work on integral transforms and describes later development of the modern Laplace transform method. ^[ref].

For real $\alpha$, the displayed improper integral converges and is an even function of $\alpha$. For $\alpha >0$, differentiation under the integral sign and partial fraction decomposition giveFor $\alpha \ne 1$, use $x^2/[(1+{\alpha }^2{x}^2)(1+x^2)]=[1/(1+x^2)-1/(1+{\alpha }^2{x}^2)]/({\alpha }^2-1)$. Integrate the resulting rational functions and extend to $\alpha =1$ by continuity. Integrating in $\alpha$ and using $E(0)=0$ gives $E(\alpha )=\pi \ln (1+\alpha )$ for $\alpha \ge 0$; evenness gives the formula for negative parameters. Thus $E^{\prime }(\alpha )=\pi \mathrm{sgn}(\alpha )/(1+|\alpha |)$ away from zero. At zero the one-sided derivatives are opposite, so $E$ is continuous but not differentiable. Substituting $\alpha =0$ into the pointwise differentiated integrand would therefore be an invalid interchange.

For every real parameter $a$,where $F(0)=0$. For $a>0$, differentiation under the integral sign and a partial fraction decomposition giveSolving this first-order linear differential equation yields $F(a)=a[\pi \ln (1+a)+C]$. To fix the constant, dominated convergence gives ${\lim }_{a\to 0^{+}}F(a)/a={\int }_0^{\infty }\ln (x^2)/(1+x^2)dx=0$; the last absolutely convergent integral changes sign under $x=1/t$. Thus $C=0$. Oddness in $a$ supplies the result for negative parameters. The initial value $F(0)=0$ alone does not determine $C$, since every solution $a[\pi \ln (1+a)+C]$ tends to zero as $a\to 0^{+}$.

Sometimes a sum does not initially look like a difference of neighboring entries. A partial fraction decomposition can reveal the cancellation. For the original examplesubtract the two reciprocals:ThusFor $N=3$, direct addition gives $1/4+1/28+1/70=3/10$. For $N=6$, the six positive entries add to $6/19\approx 0.315789$. The partial sums approach $1/3$, and the exact series remainder isThis formula answers an accuracy question as well as a summation question. To make the remaining positive series remainder smaller than $0.001$, solve $1/[3(3N+1)]<1/1000$. The result is $9N+3>1000$, so the smallest permitted integer is $N=111$. At $N=110$, the error is $1/993>0.001$; at $N=111$, it is $1/1002<0.001$.

Choose a positive integer-valued random variable with probability mass functionThese masses are positive. They sum to one because the partial fraction decompositionmakes their sum a telescoping series with value $2/(1\cdot 2)=1$. Its expectation is another telescoping series:By contrast, since $k+1\le 2k$ and $k+2\le 3k$ for $k\ge 1$, its second moment satisfiesThe comparison uses the divergent harmonic series. To check the variance directly, for $k\ge 4$ we have $(k-2{)}^2\ge k^2/4$, so the corresponding tail of $\mathbb{E}[(X-2{)}^2]$ also diverges. ThusHere “unbounded” means infinite variance for this single probability distribution.

At a simple root of a polynomial $x_i$ of $\omega$, differentiating the product gives ${\omega }^{\prime }(x_i)={\prod }_{k\ne i}(x_i-x_k)\ne 0$. Consequently the Lagrange basis polynomials can be writtenDivide the polynomial interpolation identity from part (a) by $\omega (x)$, at points outside the nodes. This gives the partial fraction decompositionEvery denominator of $\omega$ is simple, and the rational function is proper because $\mathrm{deg}p<\mathrm{deg}\omega$, so no polynomial part is missing.

The partial fraction decomposition is $f(z)=1/(z-i)+1/(z+i)$. Put $w=z-1$ and expand each reciprocal by a geometric series:Adding gives the Taylor seriesEquivalently, since $1\pm i=\sqrt{2}{e}^{\pm i\pi /4}$, its coefficient is $(-1{)}^n{2}^{(1-n)/2}\cos ((n+1)\pi /4)$. The first terms are $1-\frac{1}{2}{w}^2+\frac{1}{2}{w}^3-\frac{1}{4}{w}^4+\cdots$.

The nearest nonremovable poles are $i$ and $-i$, both at distance $\sqrt{2}$ from the center. The function is holomorphic throughout the open disk of that radius and cannot be holomorphic across either pole. Therefore the largest radius of convergence is $r=\sqrt{2}$.

For a first-order system, it is important to specify the frame in which its coefficients are written. In the simple-pole convention intended here, a Fuchsian linear differential system has, at a finite singular point $a$,The residue is the matrix $R_a={\lim }_{z\to a}(z-a)A(z)$, computed entry by entry. Such a point is a regular singular point: on a sector, the coefficient bound $\Vert A(z)\Vert \le C/|z-a|$ and integration along rays give at most power growth of solutions. Intrinsically, a regular singular point means moderate power growth of all solutions on sectors, allowing logarithmic factors. This intrinsic definition is more general than requiring a simple coefficient pole in an arbitrary meromorphic frame.

For $A(z)=R/z$, a fundamental matrix on a chosen complex logarithm branch isDifferentiating the matrix exponential verifies ${\Phi }^{\prime }=R\Phi /z$. A counterclockwise circuit replaces $\log z$ by $\log z+2\pi i$, soThis is the residue and monodromy of a constant Fuchsian system. A change of solution basis conjugates $M$. Its eigenvalues are $e^{2\pi i\lambda }$ for the residue eigenvalues $\lambda$; a nontrivial Jordan block supplies the corresponding logarithmic terms in $z^R$. The monodromy does not uniquely recover $R$, since integer shifts of residue eigenvalues can leave their exponentials unchanged.

The coordinate-independent coefficient is the matrix-valued one-form $A(z)dz$. With $t=1/z$, its coefficient isBeing Fuchsian at infinity means $A(z)=O(z^{-1})$, while being ordinary there means $A(z)=O(z^{-2})$. If the finite simple poles are $a_1,\dots ,a_s$, subtract their principal parts:Each entry of $H$ is entire and tends to zero at infinity. Liouville's theorem therefore gives $H=0$. Expanding this partial fraction decomposition at infinity yieldsThus the sum of all residues, including the residue at infinity of the one-form, is zero. If infinity is ordinary, $R_{\infty }=0$ and the finite residues alone sum to zero. Treating $A$ merely as a scalar function of $z$, without transforming $dz$, would give the wrong infinity convention.

Put $\omega =e^{2\pi i/3}$. The possible finite singular points of $(B+zC)/(1-z^3)$ are $1,\omega ,{\omega }^2$, andA point $\zeta$ is removable exactly when $B+\zeta C=0$ as a matrix. At infinity,which is holomorphic at $t=0$. Consequently the actual set is $S=\{\zeta :{\zeta }^3=1,B+\zeta C\ne 0\}$, with $\{1,\omega ,{\omega }^2\}$ the generic set. The finite residues sum to zero, as follows also from ${\sum }_{{\zeta }^3=1}{\zeta }^{-1}={\sum }_{{\zeta }^3=1}{\zeta }^{-2}=0$.

Conversely, suppose a Fuchsian linear differential system has its finite singular points among these three roots and has no singularity at infinity. Then $Q(z)=(1-z^3)A(z)$ extends to an entire matrix function, and ordinariness at infinity gives $Q(z)=O(z)$. By the Cauchy estimate, each entry has zero derivatives of order two or higher, so $Q(z)=B+zC$. The same argument applies to any subset $S$, with additional cancellations at the omitted roots. This proves the requested normal form.

There is a genuine convention limitation: the normal form is not true under intrinsic regular singularity alone in an unrestricted frame. For example, if $N\ne 0$ and $N^2=0$, then $A(z)=N/(z-1{)}^2$ has fundamental matrix $I-N/(z-1)$. Its solutions have only power growth, so $1$ is intrinsically a regular singular point, and infinity is ordinary. Nevertheless its coefficient has a double pole and cannot equal $(B+zC)/(1-z^3)$. The simple-pole convention makes the preceding residue and normal-form assertions correct.

Use normal modes $e^{ikx+st}$, so $s=-i\omega$ in the convention $e^{ikx-i\omega t}$. For a translation-invariant, well-posed linear initial value problem on the whole line, localized data are propagated by a causal Green function. The Fourier transform in space and Laplace transform in time give an inverse integral whose denominator is the dispersion relation $D(k,s)$. The initial temporal contour is a Bromwich line to the right of every real-wave-number temporal singularity.

The Briggs-Bers technique lowers that temporal contour while deforming the spatial contour to avoid its poles. A spatial pinch point occurs when two spatial branches originating on opposite sides of the original Fourier contour collide and prevent further deformation. Algebraically a simple candidate satisfies $D=D_k=0$, with $D_s\ne 0$ and $D_{kk}\ne 0$; these equations alone do not establish the pinch topology. A same-side collision is a false spatial saddle in quartic dispersion. The opposite-side requirement is also explicit in the primary numerical study of pinch-point detection.

The rightmost contributing pinch determines the fixed-position exponential growth of the impulse response, ordinarily with a $t^{-1/2}$ saddle point prefactor. A positive rate is absolute wave-packet instability. If real-wave-number normal modes grow but the impulse response decays at each fixed position, it is convective wave-packet instability; growth can then occur along moving rays. If no temporal normal mode grows, the system is stable, with equality thresholds treated as marginal rather than exponential decay. Along a ray $x=vt$, apply the same argument to $s(k)+ikv$, whose saddle point condition is $s^{\prime }(k)+iv=0$. Thus the laboratory frame is part of the absolute/convective definition.

For the quartic problem the temporal dispersion relation isWrite $a=\mathrm{Re}{a}_1$, $b=\mathrm{Re}{a}_2$, $c=\mathrm{Re}{a}_3$. For $b>0$ the causal Fourier kernel is absolutely convergent for positive time and the finite maximum temporal growth rate isThe usual initial-value framework fails if $b<0$, or if $b=0$ and $a>0$, because arbitrarily large real wave numbers grow arbitrarily fast: no finite Bromwich line can start to their right. This is a high-frequency ill-posedness obstruction, not an absolute instability criterion. If $b=0$ and $a\le 0$, the growth bound is finite, but the dispersive limiting case needs oscillatory Fourier inversion or a limiting-absorption construction when there is no quadratic damping. It must not be justified by the absolutely convergent quartic-diffusion argument. The following explicit calculation first assumes $b>0$ and $a_2\ne 0$.

The double-root candidates areRather than assume all are physical, compute the causal resolvent at the impulse location. Put $S=s+a_3$, and for large $\mathrm{Re}s$ factorThen ${\kappa }_1{\kappa }_2=\sqrt{S/a_2}$ and $({\kappa }_1+{\kappa }_2{)}^2=-a_1/a_2+2\sqrt{S/a_2}$, with sheets selected from that large-$s$ region. Partial fraction decomposition and ${\int }_{\mathbb{R}}(k^2+{\kappa }^2{)}^{-1}dk=\pi /\kappa$ give the quartic impulse-response Laplace resolventAll roots in this expression are continued from the causal region, not selected afresh at a candidate collision. The two possible singularities are now transparent. $S=0$ is a square-root singularity when $a_1\ne 0$. The extra singularity requires $\sqrt{S}=a_1/(2\sqrt{a_2})$. Taking $\sqrt{a_2}$ with positive real part and putting $z=a_1/\sqrt{a_2}$, this lies on the physical $\sqrt{S}$ sheet precisely when $\mathrm{Re}z>0$. If that real part is zero the candidate is on a cut boundary and has nonpositive real $S$, so it cannot lie to the right of $S=0$. For $\mathrm{Re}z<0$, the apparently growing candidate may be on the wrong sheet altogether.

Therefore the physical-sheet growth rate of a quartic impulse response isThese are resolvent branch singularities on the causal continuation, so the formula implements the pinch test instead of merely listing formal saddle point rates. For $a_1=0$, the zero saddle point is fourth-order and the resolvent is proportional to $S^{-3/4}$; its impulse response is proportional to $t^{-1/4}{e}^{-a_{3}t}$ and the same exponential-rate formula applies. At nondegenerate candidates the prefactor is $t^{-1/2}$.

The classification in this diffusive case is now explicit: ${\sigma }_T<0$ gives stability; ${\sigma }_A>0$ gives absolute instability; ${\sigma }_T>0$ with ${\sigma }_A<0$ gives convective instability. When ${\sigma }_T>0$ and ${\sigma }_A=0$, the nondegenerate fixed-position response still decays algebraically, at the marginal absolute/convective threshold. Other equality cases require the stated saddle point prefactors. Temporal instability supplies a growing ray: at a real maximum of $\mathrm{Re}s(k)$, choose $v=-\mathrm{Im}{s}^{\prime }(k)$ so that the moving-ray saddle point is stationary there.

Three examples show why complex coefficients and sheets matter. For $(a_1,a_2,a_3)=(2,1,1/2)$, ${\sigma }_T={\sigma }_A=1/2$, so the instability is absolute. For $(1+2i,1,1/10)$, ${\sigma }_T=3/20>0$ but ${\sigma }_A=-1/10$, so it is convective even though there is no explicit first-derivative advection term: dispersion carries packets in opposite directions. For $(-2,1,1/2)$, both physical rates are $-1/2$. The formal candidates $k=\pm i$ have $s=1/2$ but fail the physical-sheet condition, so this temporally stable example is not absolutely unstable.

On the admissible boundary $b=0$, $a\le 0$, $a_2\ne 0$, the oscillatory or damped Fourier construction has ${\sigma }_T=-c$. The surviving zero-wave-number stationary response has the same exponential rate $-c$, with a quadratic or quartic algebraic factor; damping/dispersion therefore does not generate a separate convective-growth range there. If $a_2=0$, the problem is second order and its diffusion/dispersive kernel must be used instead of the quartic formula. Application to a finite interval or nonlocalized data also changes the spatial inversion problem, so the whole-line localized-data hypothesis is essential.

Let $Y(s)$ be the Laplace transform of $y$. The initial values give $\mathcal{L}(y^{\prime })=sY$ and $\mathcal{L}(y^{\prime \prime })=s^{2}Y-1$, while $\mathcal{L}(t{e}^{-3t})=(s+3{)}^{-2}$. Therefore, for $\mathrm{Re}s$ sufficiently large,The partial fraction decomposition isInverting the elementary Laplace transforms givesThese coefficients give $y(0)=0$ and $y^{\prime }(0)=1$; the transformed equation, or direct differentiation, verifies the forcing. Uniqueness follows from uniqueness for the corresponding linear initial-value ordinary differential equation.

Use a time Laplace transform and a spatial Green function. The first-order transport term must be retained: the Airy resolvent kernel for ${\partial }_x^3+p$ alone would solve a different equation. The governing linear partial differential equation is the linear dispersive Stokes equation, distinct from the viscous-fluid Stokes equation.

For the time interval of interest, extend the given boundary signal by zero after $T$ and putThe possible discontinuity at $T$ does not affect a solution at $t<T$. Equivalently, use any sufficiently regular continuation beyond $T$: the resulting causal Green function produces the same earlier solution. We seek the usual solution with spatial decay and a time transform, rather than an unrestricted spatially growing solution on the half-line.

Writing $Q(x,p)$ for the time Laplace transform, the Laplace transform of a derivative givesThe corresponding characteristic roots are the three roots of a polynomialFor $\mathrm{Re}p>0$, label the unique root of negative real part by $r_{-}(p)$, and the two roots of positive real part by $r_1(p),r_2(p)$. This spatial root splitting for the dispersive Stokes resolvent is easily checked: if $r=ik$, then $p=-i(k-k^3)$ is imaginary, so no root crosses the imaginary axis as $p$ varies in the right half-plane. For real $p>0$ the cubic is strictly increasing on the real line and has one negative real root, while its conjugate pair has positive real parts. The root count remains the same throughout that connected half-plane. A repeated root would satisfy $3r^2+1=0$ and again put $p$ on the imaginary axis. All roots are therefore simple there, and $r_{-}(p)$ is a nonzero holomorphic function.

Set $d_j=3r_j^2+1$, including $j=-$. Define the Stokes resolvent kernel with advectionThe sign on the negative half-line is essential. To verify the Green function, use the partial fraction decompositionIts expansion at infinity yields the reciprocal-polynomial root identitiesConsequently $R_p$ and $R_p^{\prime }$ are continuous at zero, and $R_p^{\prime \prime }$ has jump one. Each piece solves the homogeneous ordinary differential equation, so the jump conditions for a third-order Green function imply, in the sense of distributional derivatives,Both pieces decay in their respective spatial directions.

Now defineIt satisfies the transformed inhomogeneous equation. The difference $Q-V$ solves the homogeneous linear ordinary differential equation; spatial decay leaves only $e^{r_{-}x}$. Imposing the Neumann boundary condition fixes its coefficient, giving the half-line Stokes solution with a boundary derivativeNo division by a vanishing root is involved, since $r_{-}=0$ would require $p=0$. Moreover the remaining boundary trace is explicitly known from $q_0$:Thus introduce the completely explicit initial-data kernelFor any upward Bromwich contour $\mathrm{Re}p=\sigma >0$ to the right of any transform singularities, the requested integral representation isThe contour integral is understood in the standard Bromwich inversion formula sense, with truncation limits if it is not absolutely convergent. Every term is expressed in the supplied data and algebraic characteristic roots; there are no unknown Dirichlet or second-derivative boundary traces.

The transformed equation verifies the linear partial differential equation and the initial term, while taking the spatial derivative gives $Q_x(0,p)=G(p)$. Bromwich inversion formula therefore supplies the prescribed derivative for $0<t<T$ and the initial value for $x>0$. For zero initial and boundary data, the transformed solution is a decaying homogeneous mode with zero boundary derivative, hence vanishes; this proves uniqueness in the stated transform/decay class. A change of the boundary continuation after $T$ has transform $e^{-pT}$ times another causal transform and contributes only after $T$, by the Laplace transform time-shift rule. The printed dot on $q_0$ is interpreted as its derivative in its sole variable $x$, so the stated corner compatibility is $q_0^{\prime }(0)=g_1(0)$. Additional differentiability at the corner requires the corresponding higher-order compatibility, but is not needed for the representation at positive $x,t$.

The ordinary saddle-point approximation with a nearby pole loses uniformity when the pole lies within the width of the saddle's Gaussian function profile. The distinguished limit isUse the exact quadratic phase coordinate $s=2i\sqrt{k}(u-1)$ and defineOn the descending contour, $s$ runs along the real axis from $-\infty$ to $+\infty$. The transformed differential has the useful partial fraction decompositionThe first term contains the nearby pole; the second is regular at the saddle and contributes $O(e^{-k}{k}^{-1/2})$. Approaching from below the descending contour means $\mathrm{Im}{s}_d<0$. The residue theorem then gives the Gaussian saddle-pole transitionThe exponent is $-s^2/4$, as in the original PDF; the converted TeX's $-s^2/2$ is a transcription error. The exact identity $k\varphi (z_0)=-k-s_d^2/4$ recovers the pole exponential in the requested expression.

The Gaussian pole integral expresses the bracket as $i\pi w(s_d/2)$, where $w$ is the Faddeeva function. Thus a convenient uniform leading answer isThe error estimate applies for bounded scaled pole position on the indicated side. In the overlap $1\ll |s_d|\ll \sqrt{k}$ below the real axis, the integral contributes $-2\sqrt{\pi }/s_d$ to leading order while the residue remains $2\pi i{e}^{-s_d^2/4}$. Since $s_d\sim \sqrt{k}(z_0-i)$, this reproduces the near-saddle limit of the separate saddle and residue terms. Boundary values on the descending contour are limits of this combined expression; the isolated real-axis pole integral itself needs a Cauchy principal value prescription.

A Laurent series about $a$ is an expansionthat converges to $f$ on an annulus $r<|z-a|<R$.

The partial fraction decomposition isFor $0<|z|<1$, expand both denominators by the geometric series:This is actually a Taylor series at zero, since the apparent puncture at zero contains no singularity.

For $1<|z|<2$, the $z+2$ term still expands in nonnegative powers, while the quadratic term must be expanded in negative powers:Hence

The partial fraction decomposition isFor $0<|z|<2$, expand the second term as a geometric series:Hence