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An antiderivative recovers a function from its local rate of change. For a real-valued function $g$ on an open subset $U$ of the real line, a differentiable function $F$ is an antiderivative whenFor example, $g(x)=x^2$ has the antiderivatives $F_C(x)=x^3/3+C$ for every real additive constant $C$. Their derivatives agree, while their values can differ. On an interval, the constant of integration describes the entire freedom in the answer. ^[ref]

The indefinite-integral notation $\int g(x)dx=F(x)+C$ records this family on an interval. For the illustrated family, the extra condition $F(1)=2$ selects $C=5/3$. Thus reversing differentiation requires both a rate and a reference value if a unique function is wanted.

The fundamental theorem of calculus supplies a construction when $g$ is continuous on an interval $I$. Choose $x_0\in I$ and set $F(x)={\int }_{x_0}^{x}g(t)dt$; then $F^{\prime }=g$. Integrability alone does not guarantee a classical antiderivative. The Heaviside step function is Riemann integrable on bounded intervals, but cannot be a derivative throughout an interval containing its jump: Darboux theorem for derivatives requires every intermediate value. ^[ref]

A bounded function which is $C^1$ away from one point in dimension $n\ge 2$, and whose classical gradient is locally integrable across that point, has that gradient as its weak derivative. Cut a test function off within distance $\epsilon$ of the point. The error from differentiating the cutoff is bounded by $C\Vert u{\Vert }_{\infty }{\epsilon }^{n-1}$ and vanishes. The other terms converge by local integrability. In dimension one the Heaviside step function is a counterexample: its classical derivative off the point is zero, but its distributional derivative is a Dirac delta.

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 $G(t)=H(t)t{e}^{-bt}$.

For $\alpha >0$, the retarded impulse response of $\alpha y^{\prime }+y=f$ is the displayed function, with $H$ the Heaviside step function. The corresponding convolution $y(t)={\alpha }^{-1}{\int }_{-\infty }^t{e}^{-(t-s)/\alpha }f(s)ds$ is the unique solution tending to zero in the remote past whenever the integral exists and has that limit. For example, bounded forcing that tends to zero as $t\to -\infty$ satisfies these conditions. A general forcing need not: $f\equiv 1$ admits no such solution. A unit-duration pulse with $\alpha =1$ gives $y=0$ before zero, $y=1-e^{-t}$ between zero and one, and $y=(1-e^{-1})e^{-(t-1)}$ thereafter.

For functions $f,g$ on $[0,\infty )$ with $e^{-\sigma t}f(t)$ and $e^{-\sigma t}g(t)$ Lebesgue integrable functions, their unilateral convolution, defined almost everywhere if necessary, satisfiesTaking a causal extension by zero of both functions gives the existing whole-line convolution definition. Absolute convergence and Fubini's theorem justify the product rule: put $t=\tau +u$ in the double integral, so $e^{-st}=e^{-s\tau }{e}^{-su}$. For piecewise continuous functions of exponential order $a$ and $b$, the hypotheses hold for $\mathrm{Re}s>\max (a,b)$. The Laplace transform article works out the concrete example $h(t)=e^{-2t}$ and the Heaviside step function input, whose output is $(1-e^{-2t})/2$. ^[ref].

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.

An explicit constructed model separates slow magnetization drift from discrete collective jumps:Here $\Theta$ is the Heaviside step function, $\delta$ the Dirac delta function, $r$ a smooth drift rate, and $t_i$ the jump times. The ideal discontinuities stand for events shorter than the resolved timescale; actual magnetic domain walls move over finite times. This equation stipulates jump sizes rather than deriving them from atomic spin. With $K={\mu }_{0}N{A}_s$ in the preceding geometry, subtracting the smooth background leaves $e_{\mathrm{jump}}=-K{\sum }_i\Delta M_i\delta (t-t_i)$.

Let a stable linear filter have causal impulse response $h(t)$, normalized by $\int h(t)dt=1$, and let $G$ be a constant dimensionless voltage gain. The output is a convolution:A simple low-pass filter has $h(t)=\Theta (t)e^{-t/\tau }/\tau$, with time constant $\tau >0$. Its output pulse has peak proportional to $\Delta M_i/\tau$ and signed area $-GK\Delta M_i$. Integrating the convolution proves area preservation because $\int h=1$; a wider filter response lowers peaks and lengthens tails. This ideal filter models resolution, rather than claiming the cited instrument had exactly this response.

The nearby positive jumps in the figure have additive negative pulse areas even when their tails overlap. For $\tau =0.50\mathrm{ms}$ the first of the pair still has $e^{-0.2/0.5}=e^{-0.4}\approx 0.67032$ of its initial height when the second arrives. Thresholding can merge those responses into one detected event; limited signal bandwidth also obscures their separation. The total signed area still records the sum if the complete response is integrated, but cannot identify how that sum was partitioned into events. Equal simultaneous jumps are exactly indistinguishable from a single jump of their summed size in this linear measurement. Oppositely signed simultaneous changes can cancel.

Area preservation requires the stated electronics. A zero-DC stage after the low-pass filter has responseConsequently $\int G(q\ast h\ast e_{\mathrm{jump}})(t)dt=0$ over the complete response: a main pulse is followed by a compensating tail. Integrating only a threshold-selected lobe, clipping the record, or saturating the amplifier breaks the simple pulse-area inference. Recovering a magnetization change from such a readout needs the calibrated frequency response and an appropriate reconstruction. Neither the audible crackle nor the displayed voltage is a direct staircase of magnetization.

Consider the one-dimensional diffusion equation on the whole real line,with initial value one for $x<0$ and zero for $x>0$. This is an idealized Heaviside step function profile, not a measured material or an apparatus specification. The heat kernel gives a convolution with the initial profile:The change of variables $s=(x-y)/(2\sqrt{Dt})$ turns the convolution into a Gaussian integral tail. The resulting similarity variable $\eta =x/(2\sqrt{Dt})$ shows that the transition width grows like $\sqrt{Dt}$: multiplying elapsed time by four doubles every fixed-profile distance from the original interface.

At every positive time, $u(0,t)=1/2$, while $u(x,t)\to 1$ as $x\to -\infty$ and $u(x,t)\to 0$ as $x\to +\infty$. For every fixed $x\ne 0$, the limit as $t\to 0^{+}$ returns the specified initial value. The value assigned at the single initial discontinuity does not change this convolution. The error function therefore explains smoothing without pretending that a discontinuous initial profile remains sharp.

Direct differentiation checks the partial differential equation. Writing $\eta =x/(2\sqrt{Dt})$ givesThe model assumes constant diffusion coefficient, no advection, and no finite-domain boundary effects. For an interpretation as heat conduction, the heat equation applies to a normalized temperature difference under its own material assumptions.

A first-order low-pass filter with positive constant gain $K$ and time constant $\tau >0$ obeys the ordinary differential equationThe derivative term limits how quickly the plant output follows a changing plant input. A larger time constant means slower adjustment. This standard form, with $K=1$, is developed in Derek Rowell, MIT 2.004, lecture 20, section 1.1: first-order systems; the following calculation keeps the gain $K$ explicit.

Write $y_0=y(0)$. The Laplace transform of the derivative is $sY(s)-y_0$, soSetting the initial condition $y_0=0$ gives the transfer function $G(s)=K/(\tau s+1)$. Its single simple pole is $s=-1/\tau$, which has negative real part. The unforced response decays, and the input/output map has BIBO stability. The eventual gain for a constant plant input is $G(0)=K$.

For a unit step, represented by the Heaviside step function $u(t)=H(t)$, the Laplace transform is $U(s)=1/s$. From zero initial condition,Take $K=2$ and $\tau =3\mathrm{s}$, with dimensionless plant input and plant output. Then $y(t)=2(1-e^{-t/(3\mathrm{s})})$ rises toward $2$. After one time constant, at $t=3\mathrm{s}$, it reaches $2(1-e^{-1})\approx 1.2642$, or $63.2\%$ of its final value. At $t=9\mathrm{s}=3\tau$, it reaches $2(1-e^{-3})\approx 1.9004$, or $95.0\%$. These percentages describe the rise from zero, rather than an instantaneous jump to the target.

With a nonzero initial condition, the same step instead givesFor $K=2$ and $\tau =3\mathrm{s}$, compare $y_0=0$ with $y_0=1$. Their plant outputs differ by $e^{-t/(3\mathrm{s})}$ despite having the same transfer function and the same plant input. The extra exponential decay term must be supplied from the initial condition; $G$ and $u$ alone cannot determine it.

With the $e^{-ikx}$ convention, exponential damping on the positive half-line gives $1/(a+ik)$. Its even part tends to $\pi \delta (k)$ and its odd part to $-i\mathrm{PV}(1/k)$. Thus the Heaviside step function transform requires a Cauchy principal value, not an ordinary reciprocal function at zero.

For electrons directed normally toward an ideal planar collector, a retarding electric potential difference $V_r>0$ imposes the barrier $e{V}_r$. Write its transfer as $T(E)=H(E-e{V}_r)$, where $H$ is the Heaviside step function and $e$ the positive elementary charge. Thus a perfectly sharp loss need not produce a sharp drop in electric current: the drop also depends on where the outgoing energy lies relative to the barrier. For an oblique incident electron at angle $\theta$ to the normal, the same planar model instead tests $E{\cos }^2\theta >e{V}_r$; angular scattering can change collection even at unchanged total kinetic energy.

For example, choose a monoenergetic incoming beam with $u=5.6\mathrm{eV}$, $p=0.75$ and $\Delta =4.9\mathrm{eV}$. After one encounter, $25\%$ have $5.6\mathrm{eV}$ and $75\%$ have $0.7\mathrm{eV}$. A $1.0\mathrm{V}$ retarder collects only $0.25I_0$, whereas a $0.5\mathrm{V}$ retarder collects $I_0$; here $I_0$ is the incident electric current with fixed geometrical acceptance and no background. The mean outgoing kinetic energy is $1.925\mathrm{eV}$, independently checked by $5.6-0.75(4.9)$. These are chosen inputs and calculated fractions, not historical measurements.

More generally, let $f(E;V)$ be the normalized outgoing probability density function at a fixed accelerating electric potential difference $V$, for the normally incident idealization. A retarding scan givesThe second equation follows by differentiating the lower integration limit, with Dirac delta functions understood for sharp beams. It explains why an integrated collector readout can reveal an energy distribution. With angles present it probes the distribution of normal kinetic energy; finite instrumental resolution smooths it, and numerical differentiation amplifies noise. Keep the incoming beam fixed during this scan. A scan of $V$ changes $f$ itself and cannot be differentiated by the same inversion formula.

For zero initial value and slope, the causal Green function is $G_0(x,s)=H(x-s)\sin [k(x-s)]/k$, with $k\ne 0$ and $H$ the Heaviside step function. For zero values at endpoints $a<b$, put $\kappa =|k|$. The Dirichlet Green function isBoth are continuous at the source and have a first-derivative jump of one, giving $L_{x}G=\delta (x-s)$. The Dirichlet inverse exists when $\sin [\kappa (b-a)]\ne 0$; a zero denominator signals a nontrivial homogeneous mode and a Fredholm solvability condition. The endpoint kernel is the zero-conjugation case of the Dirichlet Green function for a conjugated harmonic oscillator. Boundary conditions determine which kernel must be integrated against a forcing term.

For initial data $u_0$ on the real line, Fourier transformation or convolution with the fundamental solution gives

For a bounded function $u_0$, the convolution is well defined at every positive time because the heat kernel has integral one. The error-function diffusion front uses a Heaviside step function initial condition and gives one smoothed interface. The heat equation with interval-indicator initial data instead has two interfaces; subtracting their two error functions produces the finite-interval profile.

For $y^{\prime }+a(x)y=H(x)$ with continuous $a$, the integrating factor $\mu =\exp (\int a)$ gives the displayed solution. It is continuous at the step and solves the differential equation classically on each side; the derivative jumps by one. The value assigned to the Heaviside step function at the single point zero does not affect the integral. A discontinuity in $y$ would instead produce an unintended delta term in its distributional derivative.

For the harmonic oscillator equation $y^{\prime \prime }+{\omega }^{2}y=J\delta (t-t_{\ast })$, with $\omega >0$, the displacement is continuous and the velocity increases by $J$ at the impulse. If the incoming displacement is zero and $J=-y^{\prime }(t_{\ast }-)$, the outgoing displacement and velocity both vanish. The subsequent homogeneous solution is identically zero. Equivalently, the causal forced response $(J/\omega )H(t-t_{\ast })\sin (\omega (t-t_{\ast }))$ cancels the pre-existing oscillation, where $H$ is the Heaviside step function.

Let $\psi$ be an integrable function with at least one vanishing moment. For a unit jump discontinuity $f(x)=H(x-x_0)$, where $H$ is the Heaviside step function, substitution into the coefficient givesThe signed integral of the whole analyzing function is zero, but the jump discontinuity cuts through its support of a function and usually leaves an unbalanced partial integral. At fixed relative position $\theta$ with nonzero partial integral, the coefficient decays as $\sqrt{a}$, slower than the $a^{q+1/2}$ Taylor remainder bound on a smooth region. “Edges survive” means this relative contrast: the normalized coefficients still tend to zero as $a\to 0$.

For an original exact comparison, use the smooth polynomial $f(x)=x+x^2$ at $b=0$. The Haar wavelet $h$ and the second-difference analyzing function $\eta$ giveThe linear term cancels for $\eta$; it remains for $h$. For the fixed unit jump discontinuity $H(x)$, choose $b=-a/4$ so that $\theta =1/4$ at every scale. Then

Position matters. If an analyzing function's support of a function lies entirely on one constant side of the jump discontinuity, its coefficient is zero. Even a crossing can cancel: $\eta$ has zero partial integral at $\theta =3/2$. A jump discontinuity on a dyadic interval boundary can likewise escape finer Haar wavelets whose supports do not cross it. These examples describe particular functions and normalizations, not a universal ranking of wavelets for signal processing.

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].

Start with a fundamental solution of the second derivative operator:Here $H$ is the Heaviside step function and the last equality holds in the sense of distributions. The factor $x-\xi$ removes a jump in the function itself, while its first derivative has a unit jump. Adding any homogeneous solution preserves the Dirac delta source.

Choose $y_1(x)=x-2$ and $y_2(x)=x-3$. They are independent homogeneous solutions, with $B_1{y}_1=0$, $B_2{y}_2=0$, $B_1{y}_2=-1$, and $B_2{y}_1=1$. Since $1<\xi <2$, the fundamental solution has $B_1\tilde{g}=1$ and $B_2\tilde{g}=2-\xi$. Imposing the two boundary conditions on the proposed combination therefore gives $-A_2+1=0$ and $A_1+2-\xi =0$. ConsequentlyEquivalently,At $x=\xi$ both expressions equal ${\xi }^2-3\xi +1$ and the right slope minus the left slope is $\xi -(\xi -1)=1$. At the endpoints, $g(1,\xi )=-\xi$ and $g_x(2,\xi )=\xi$, while $g_x(1,\xi )=\xi -1$ and $g(2,\xi )=1-\xi$. These explicitly verify both coupled boundary conditions. The homogeneous equation has only affine solutions; its two boundary conditions give $2a+b=3a+b=0$, hence $a=b=0$, proving uniqueness of the Green function.

The reciprocity defect is $g(x,\xi )-g(\xi ,x)=2(x-\xi )$. It records how the coupled boundary conditions affect self-adjointness, even though the differential expression $D^2$ is formally self-adjoint. The construction is an instance of the boundary-matrix construction of an ordinary differential Green function.

The step in the source must be at zero. We use the Heaviside step function $H(x)=1$ for $x>0$, zero for $x<0$, with its value at zero immaterial here. The printed threshold $1$ is inconsistent with both $H^{\prime }=\delta (x)$ and the stated number density. For example, measuring velocities in fixed units and taking $v_F=2$, the literal printed cutoff fills only $v<1$ and gives one eighth of the stated number density. Its distributional derivative is $\delta (x-1)$, not $\delta (x)$. The zero-threshold convention restores the stated Fermi sphere and its normalization of a probability distribution.

Let $k=|\mathbf{k}|>0$, choose $u=\mathbf{v}\cdot \mathbf{k}/k$, and abbreviate $V_{\alpha }=v_{\mathrm{F}\alpha }$. The three-dimensional phase-space distribution function is $f_{0\alpha }=C_{\alpha }{1}_{v<V_{\alpha }}$, whereIntegrate over the transverse velocity disk to obtain the parallel marginal distributionThere are no edge Dirac delta functions in $F_{\alpha }^{\prime }$ because $F_{\alpha }$ itself vanishes continuously at the endpoints. In particular, the three-dimensional constant occupation does not imply a constant parallel marginal distribution.

Use time dependence $e^{pt+i\mathbf{k}\cdot \mathbf{x}}$, with $p=\gamma -i\omega$ and $\gamma <0$ for damping. The given dielectric function becomesThe Landau contour supplies causal continuation. For a real phase velocity $a=\omega /k>V_{\alpha }$, there is no resonant pole in the populated interval, andHence this species contributes

For comparable species number densities and fixed charge ratios, $m_e\ll m_i$ implies $V_i/V_e=O(m_e/m_i)$ and ${\omega }_{\mathrm{p}i}^2/{\omega }_{\mathrm{p}e}^2=O(m_e/m_i)$. Since $\omega >k{V}_e$, the ions have $s_i\gg 1$ and ${\chi }_i=-{\omega }_{\mathrm{p}i}^2/{\omega }^2+O(k^2{V}_i^2{\omega }_{\mathrm{p}i}^2/{\omega }^4)$. Their response is negligible to leading order in the mass ratio. The real dispersion relation is thereforeThis degenerate-plasma dielectric response treats a completely degenerate, nonrelativistic Fermi gas with classical particle trajectories. It does not include quantum recoil or a collision term. A finite ion response shifts the root slightly; the boxed equation is the stated leading electron approximation.