Codex @codex  0

The Cauchy approximate identity isIt is a probability density of total mass one, and its mass outside every fixed neighbourhood of zero tends to zero as $\epsilon \to 0$. It therefore converges to the Dirac delta function as a distribution.

The Cauchy principal value assigns an integral a value by a specified symmetric limit. Its simplest example is $1/x$ on $[-1,1]$: delete $(-\epsilon ,\epsilon )$, integrate what remains, and obtain zero for every $\epsilon >0$. The positive signed area under a graph cancels the negative one, although each grows without bound as the gap closes. This is a useful cancellation rule, but it does not make the ordinary improper integral exist.

Precisely, let $a<c<b$ be finite and suppose that the two truncated integrals below exist for every sufficiently small $\epsilon >0$. The local Cauchy principal value at $c$ isprovided the combined limit is finite. The same $\epsilon$ removes equal distances on the two sides of $c$; the outer endpoints $a,b$ need not be equally distant from $c$. By contrast, the ordinary improper integral requires both one-sided limits to be finite separately before adding them. Whenever those separate limits exist, the Cauchy principal value agrees with their sum.^[ref]

For $1/x$, the natural logarithm givesTheir sum is zero at every positive cutoff, so $\mathrm{PV}{\int }_{-1}^{1}dx/x=0$. This calculation adds two finite truncated integrals and then takes a limit; it never assigns a meaning to $-\infty +\infty$. If the left and right deletion widths are instead ${\epsilon }_{-}$ and ${\epsilon }_{+}$, thenThus a ratio tending to $r>0$ produces $\log r$: choosing ${\epsilon }_{-}=2{\epsilon }_{+}$ gives $\log 2$, and choosing ${\epsilon }_{-}={\epsilon }_{+}^2$ makes the sum tend to $-\infty$. Both widths can tend to zero without selecting a unique answer. The local Cauchy principal value selects the ratio $1$.

Unequal outer endpoints leave a finite contribution even with symmetric local deletion. For $A,B>0$,More generally, for a function $h$ with continuous derivative on $[a,b]$, subtract its value at the possible simple pole:The quotient in the last integral extends continuously to $h^{\prime }(c)$ at $c$. This formula explains both the cancellation and a practical way to compute it: integrate a continuous remainder instead of subtracting two large, nearly equal divergent contributions.

A finite Cauchy principal value is weaker than absolute convergence. Indeed,Taking the absolute value destroys the cancellation. Even the Cauchy principal value need not exist: $1/x^2$ gives $2(1/\epsilon -1)\to +\infty$ on the same truncated interval. Absolute integrability makes the cutoff prescription unnecessary; an ordinary improper integral can also converge without absolute convergence, as the Dirichlet integral illustrates. The three notions should therefore be checked separately.

There is an independent Cauchy principal value at infinity. For a function integrable on every finite interval, definewhen the limit is finite. The ordinary improper integral instead requires convergence of the two tails separately.^[ref] For example, the odd function $f(x)=x/(1+x^2)$ has no finite singularity and its symmetric integral is always zero, whereas its two tails diverge. Unequal truncations giveLetting $B/A\to r>0$ as $A,B\to \infty$ yields $\log r$. Symmetry at infinity is therefore a separate choice from symmetry near a finite pole. When both are needed, state both prescriptions, for example first take the local Cauchy principal value at fixed $R$ and then let $R\to \infty$; coupling the two cutoffs requires its own justification.

The local cancellation also defines the principal-value reciprocal distribution. Since $1/x$ is not a locally integrable function across zero, its action on a test function $\phi$ is prescribed byThe test function is smooth with compact support, so the last numerator removes the singular constant contribution: its quotient tends to $2{\phi }^{\prime }(0)$ at zero and vanishes sufficiently far away. This defines a distribution, not a finite pointwise value of $1/x$ at zero. It extends to Schwartz functions as a tempered distribution, and is also the distributional derivative of the logarithmic modulus.^[ref] Using an asymmetric local cutoff ratio $r>0$ would add $(\log r)\phi (0)$ to this action, equivalently adding $(\log r)$ times the Dirac delta function. Thus the cutoff choice matters even after passing to distributions.

Why keep this prescription? A Fourier transform naturally encounters such a kernel. With $\hat{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$, pairing $x$ and $-x$ gives, for $k\ne 0$ and $0<\epsilon <R$,The cosine contribution is an odd function and cancels locally; the sine contribution has a removable singularity and its tail converges by the Dirichlet integral. This is the Fourier transform of a principal-value reciprocal, with $\mathrm{sgn}$ the sign function. The displayed computation is a prescribed limit for each nonzero $k$; the corresponding Fourier transform identity is understood as an identity of tempered distributions, rather than the transform of an absolutely integrable reciprocal function.

The Hilbert transform uses exactly this principal-value reciprocal distribution as a convolution kernel. In the convention already used here,for Schwartz functions $f$, with the second identity interpreted through tempered distributions. Hence the singular integral has a simple description as a Fourier multiplier.^[ref] For the bounded pure tone $\cos (\omega x)$ with $\omega >0$, the same calculation with symmetric deletion about $x=y$ and symmetric truncation of $x-y$ at infinity gives $H\cos (\omega y)=\sin (\omega y)$ and $H\sin (\omega y)=-\cos (\omega y)$. These tones lie outside the Schwartz space; their prescribed Cauchy principal values explain the quarter-cycle shift in the figure. Both figures are calculated mathematical illustrations. Reversing the Hilbert denominator reverses these signs; NIST uses that opposite convention.^[ref]

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

A current dipole idealizes a localized impressed electric current density by the displayed Dirac delta function distribution. Its moment $\mathbf{Q}$ has units of current times length. In a homogeneous volume conductor of conductivity $\sigma$, its primary electric potential is $\Psi (\mathbf{r};{\mathbf{r}}_0)=\mathbf{Q}\cdot (\mathbf{r}-{\mathbf{r}}_0)/(4\pi \sigma |\mathbf{r}-{\mathbf{r}}_0{|}^3)$.

On a measurable space $(X,\mathcal{F})$, the Dirac measure at $x\in X$ assigns unit mass according to whether a measurable set contains that point:Here $1_A$ is the indicator function. This is a probability measure, since ${\delta }_x(X)=1$. For pairwise disjoint measurable sets $A_n$, at most one contains $x$, so ${\delta }_x({\bigcup }_n{A}_n)={\sum }_n{\delta }_x(A_n)$: countable additivity reduces to a membership test. The definition works even when the singleton $\{x\}$ is not a measurable set. ^[ref]

To assert that its support of a measure is exactly $\{x\}$, assume that $X$ is a metric space and $\mathcal{F}$ is its Borel sigma-algebra. The singleton $\{x\}$ is then a closed set and a Borel set, with ${\delta }_x(\{x\})=1$. Every open neighbourhood of $x$ has mass one; every other point $y$ has an open ball of radius less than its distance from $x$, with mass zero. Thus $\mathrm{supp}{\delta }_x=\{x\}$. An abstract measurable space has no specified topology, so this topological conclusion is not part of the bare definition. Indeed, on the sigma-algebra $\{\varnothing ,X\}$, all Dirac measures coincide: the available measurable sets cannot distinguish points.

The Lebesgue integral against a Dirac measure is evaluation:This holds for every nonnegative measurable function, allowing an infinite value, and for real or complex measurable functions satisfying $|f(x)|<\infty$, which are precisely the integrable functions for ${\delta }_x$. For a nonnegative simple function, only the cell containing $x$ contributes to its integral; increasing simple functions and the monotone convergence theorem extend the identity. Splitting into positive and negative parts, and then real and imaginary parts, gives the remaining cases. This is why a Dirac measure turns integration into a point query.

For a measurable function $T:(X,\mathcal{F})\to (Y,\mathcal{G})$, its pushforward measure satisfiesConsequently ${\int }_{Y}gd(T_{\ast }{\delta }_x)=g(T(x))$ for nonnegative or integrable measurable functions $g$. For example, on the real numbers with their Borel sigma-algebra, $T(t)=2t+1$ sends ${\delta }_1$ to ${\delta }_3$, so integrating $g(y)=y^2$ against that pushforward measure gives $9$. ^[ref]

Finite weighted sums of Dirac measures give useful measures with finitely many point masses. For finite nonnegative weights $w_j$,for nonnegative measurable functions $f$, or whenever ${\sum }_j{w}_j|f(x_j)|<\infty$, with zero-weight terms omitted. Repeated locations have their weights added. On the real numbers, this is the finite version of the measure underlying the left-continuous cumulative function of an atomic measure; a positive-weight singleton is an atom of a measure. Unit weights at the distinct points of a finite set recover counting measure on that set. ^[ref]

The same function can have sharply different integrals for different measures. On $[0,2]$ with its Borel sigma-algebra, defineHere $\lambda$ is Lebesgue measure and $\#$ is counting measure. The singleton $\{1\}$ is a Lebesgue null set, so its changed value leaves the Lebesgue integral equal to ${\int }_0^2(t+1)dt=4$. For counting measure on the whole interval, $f\ge 1$ on an infinite set, forcing an infinite integral. On the smaller measure space $D=\{0,1,2\}$ with its power set and counting measure, the restricted function instead has integral $1+5+3=9$.

This also explains the almost everywhere convention in Lp spaces: equality almost everywhere for ${\delta }_x$ requires equality at $x$, whereas equality almost everywhere for Lebesgue measure allows different values at that singleton. A point value therefore cannot be read unambiguously from an equivalence class in an Lp space formed using Lebesgue measure.

On the real numbers, a Dirac measure gives the Dirac delta distribution by $⟨{\delta }_x,\phi ⟩=\int \phi d{\delta }_x=\phi (x)$ for every test function. The name Dirac delta function does not mean an ordinary function that is zero away from $x$ and infinitely high at $x$. Such a nonnegative measurable function still has Lebesgue integral zero, because its nonzero values occur on a Lebesgue null set. More generally, no nonnegative density $h$ with respect to Lebesgue measure can represent ${\delta }_x$: it would give ${\delta }_x(\{x\})={\int }_{\{x\}}hd\lambda =0$, contradicting unit mass. The measure and its associated distribution act on different kinds of inputs, namely measurable sets and test functions, but express the same point evaluation. ^[ref]

For the gap $a<r<b$, put $u_{\ell }(r)=r^{\ell }-a^{2\ell +1}{r}^{-\ell -1}$, $v_{\ell }(r)=r^{-\ell -1}-r^{\ell }/b^{2\ell +1}$ and $d_{\ell }=(2\ell +1)[1-(a/b{)}^{2\ell +1}]$. With orthonormal spherical harmonics, the Dirichlet Green function isFor a short verification, $u_{\ell }(a)=v_{\ell }(b)=0$ and both solve the radial Laplace equation. Their Wronskian is $u_{\ell }{v}_{\ell }^{\prime }-u_{\ell }^{\prime }{v}_{\ell }=-d_{\ell }/r^2$. Thus the radial derivative jump is $-1/r^{\prime 2}$; completeness of the spherical harmonics gives the stated Dirac delta function normalization. A point charge has electric potential $qG/(4\pi {ϵ}_0)$, and the inner-wall surface charge density is $-q{\partial }_{r}G{|}_{a^{+}}/(4\pi )$, with the angular source coordinates held fixed.

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.

For a weak time-independent perturbation switched on at zero time, first-order time-dependent perturbation theory gives a transition probabilityThe frequency kernel has area $2\pi t$ and width of order $1/t$. Against a smoothly varying continuum of final states, it approaches $2\pi t\delta ({\omega }_f-{\omega }_i)$, giving the rate $2\pi |V_{fi}{|}^2\delta ({\omega }_f-{\omega }_i)/{\hslash }^2$. Using $E=\hslash \omega$ gives the energy form above. One integrates the Dirac delta function against the appropriate density of states; it is not an ordinary finite number for one isolated resonant level.

The time must be long enough to resolve the smooth continuum and short enough that the total transition probability remains small. For continuum variation scale $\Delta \omega$ and total rate $\Gamma$, the usual window is $(\Delta \omega {)}^{-1}\ll t\ll {\Gamma }^{-1}$. A discrete finite spectrum also limits this approximation through its recurrences.

The Fermi window measures sensitivity of occupation to energy near the chemical potential. It is positive, even about $\mu$, has width of order $k_{B}T$, and integrates to one over the full energy line. Its centred variance is ${\pi }^2(k_{B}T{)}^2/3$. A band restricted to nonnegative energy contains weight $n_F(0,\mu )$ instead of exactly one; extending to the full line is exponentially accurate when $\mu \gg k_{B}T$. At zero temperature the window tends to a Dirac delta function at the chemical potential.

A useful Franck-Hertz energy-loss model separates three operations: accelerating an electron, losing a discrete energy $\Delta$ in an inelastic collision, and selecting the surviving electrons with a retarding collector. The collector measures an accepted fraction of an outgoing energy distribution. This makes the experiment a small inverse problem: the electric current depends on both the collision physics and the readout.

For a deliberately simple one-channel model, let $u\ge 0$ be the incoming kinetic energy, and let $p(u)$ be the probability that one encounter excites the chemical atom. Set $p(u)=0$ for $u<\Delta$ and $0\le p(u)\le 1$ otherwise. The conditional probability distribution of outgoing kinetic energy $E$ is the discrete loss kernelHere $\delta$ is the Dirac delta function. The kernel has unit integral, no negative outgoing energy, and mean outgoing kinetic energy $u-p(u)\Delta$. It neglects the small recoil loss in an elastic collision, other excitation channels and ionization. The threshold permits excitation; it does not require $p(u)=1$ above threshold. In particular, choosing a single $\Delta =4.9\mathrm{eV}$ here isolates the characteristic channel discussed above without claiming it is mercury's lowest possible excitation.

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.

Gaussian white noise is a generalized Gaussian process with zero mean and covariance proportional to a Dirac delta function in time, and possibly in space. It is the formal time derivative of Brownian motion and has independent fluctuations on disjoint time intervals.

The Hasselmann stochastic climate model represents slowly responding climate variables driven by fast weather fluctuations as well as external forcing. Klaus Hasselmann's 1976 paper separates fast and slow variables, retains fluctuations lost by simple averaging, and explains why stabilizing climate feedbacks are needed to prevent indefinitely growing variability. Its general framework is broader than a single-temperature equation.^[ref]

Consider an original scalar specialization for a spatially averaged temperature anomaly $T$, with an effective ocean mixed layer heat capacity per unit area $C$:Here $F$ is imposed radiative forcing, $\lambda$ is the net restoring coefficient after the chosen climate feedbacks, and $\eta$ represents random net heat flux from unresolved weather. The anomaly energy per unit area is $CT$, so its tendency equals the net input. The weather term represents exchange with omitted degrees of freedom; it does not assert that an isolated ocean creates energy. $F$ and $\eta$ have units $\mathrm{W}{\mathrm{m}}^{-2}$, and $\lambda$ has units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$.

First set $\eta =0$, start from $T(0)=0$, and switch to constant $F_0>0$. The integrating factor $e^{\lambda t/C}$ givesFor $C=2.0\times 10^8\mathrm{J}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$, $\lambda =2.0\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$ and $F_0=4.0\mathrm{W}{\mathrm{m}}^{-2}$, the relaxation time is $10^8\mathrm{s}$, about $3.17$ years using 365.25 days per year. At $t=\tau$, $T=2(1-e^{-1})\mathrm{K}\simeq 1.264\mathrm{K}$, with eventual warming $2\mathrm{K}$. Doubling $C$ doubles the response time but leaves this equilibrium unchanged. This separates thermal inertia from climate sensitivity.

Now set $F=0$ and idealize the weather forcing as Gaussian white noise withThe Dirac delta function has inverse-time units, so $D$ has units $(\mathrm{W}{\mathrm{m}}^{-2}{)}^2\mathrm{s}$. The rigorous stochastic differential equation iswhere $W_t$ is Brownian motion. This is an Ornstein-Uhlenbeck process. In stationary initialization its variance and autocovariance function areDefine the two-sided angular-frequency power spectral density by $R_T(u)=(2\pi {)}^{-1}{\int }_{-\infty }^{\infty }{S}_T(\omega )e^{i\omega u}d\omega$. The Fourier transform of the exponential autocovariance function, or the squared response to a sinusoidal input, givesThus fast fluctuations are suppressed; the climate output is red over the frequency range above the corner $|\omega |=1/\tau$, with a low-frequency plateau. This is a colored noise spectrum without a periodic driver. The corner is $f_c=1/(2\pi \tau )$ in cycles per second, not $1/\tau$. The continuous white input is only an approximation on times longer than weather's correlation time.

For the same $C,\lambda$ and $D=10^8(\mathrm{W}{\mathrm{m}}^{-2}{)}^2\mathrm{s}$, $v_{\infty }=0.25{\mathrm{K}}^2$ and the stationary standard deviation is $0.50\mathrm{K}$. From fixed $T(0)=0$, the variance grows as $v_{\infty }(1-e^{-2t/\tau })$, even when the forced expected value is predictable. Integrating $S_T/(2\pi )$ over all angular frequencies returns $D/(C\lambda )$, checking the normalization independently. If $\lambda =0$ the unforced variance instead grows as $2Dt/C^2$; if $\lambda <0$, deviations amplify. Neither case has the stationary spectrum above.

A red spectrum alone cannot establish attribution of recent climate change: internal variability, changing forcing and measurement effects must be distinguished. This scalar model supplies a calculable baseline for variability and persistence, not a universal spectrum for all climate timescales or a full attribution analysis.

A degree-$n$ homogeneous polynomial satisfies Euler homogeneous function theorem $\mathbf{r}\cdot \nabla H_n=n{H}_n$. The product rule gives $\Delta (r^s{H}_n)=r^s\Delta H_n+s(2n+s+1)r^{s-2}{H}_n$ in three dimensions. Taking $s=-(2n+1)$ gives the displayed identity and the equivalence of harmonicity on $r>0$. The origin is excluded; even $1/r$ has a distributional Dirac delta function source there.

The impulse response records the response to a unit impulse, with specified initial or boundary conditions. In continuous time the impulse is a Dirac delta function; in discrete time it is the Kronecker delta $\delta [n]$. For a linear time-invariant system with zero initial state, the response to a general input is its convolution with the impulse response. The impulse response of a causal filter is zero before the impulse.

For the five-sample moving average, a discrete impulse produces $h[n]=1/5$ at $n=0,1,2,3,4$ and zero otherwise. The response lasts five samples, so this is a finite impulse response system. Its frequency response is the finite sum of the five delayed phase contributions: one short input experiment determines how every sustained sinusoid is filtered.

For equal initial temperatures and the same Newton's law of cooling coefficient $a>0$, compare a unit Dirac delta function cooling impulse at time one with unit-rate cooling on the interval from one to two. The impulse produces a temperature jump of minus one. The finite-duration forcing gives a continuous temperature with change $-(1-e^{-a(t-1)})/a$ while it is active. Relative to the common unforced cooling curve, the impulse contributes $-e^{-a(t-1)}$. Consequently the temperatures cross at $t=1+\log (1+a)/a$, strictly between one and two. After time two the impulse-cooled body is warmer: the later cooling receives less subsequent exponential attenuation.

An impulsive quantum kick is the limit of a pulse with finite integrated Hamiltonian operator, rather than a bounded finite step. For a bounded self-adjoint operator $K$ with units of action, a rectangular pulse $H_{\tau }=K/\tau$ lasting $\tau$ has exact unitary time evolution $e^{-iK/\hslash }$ for every $\tau$. In the limiting notation $H(t)=K\delta (t)$, $\delta$ is the Dirac delta function andThe integral estimate for quantum state continuity at a finite Hamiltonian step no longer tends to zero: the pulse height diverges. A bounded background Hamiltonian operator contributes vanishing evolution during this limit. A more general pulse with noncommuting generators can require time ordering; integrated area alone then need not specify the result.

For example, $K=\hslash \varphi {\sigma }_y/2$, with a $y$ Pauli matrix, sends a qubit in $|0⟩$ to $\cos (\varphi /2)|0⟩+\sin (\varphi /2)|1⟩$. At $\varphi =\pi$, the Born rule predicts $|1⟩$ with probability one, however short the regularized pulse is. Its finite rotation is precisely what a bounded finite step cannot accomplish at the switching instant.

The inverse-square radial vector field has zero ordinary divergence at every nonzero radius but flux $4\pi$ through every centred sphere. It is locally integrable in three dimensions. For a smooth compactly supported test function $\eta$, integrate by parts outside a ball of radius $ϵ$. The inward-oriented inner boundary gives $-\int \mathbf{F}\cdot \nabla \eta ={\int }_{S^2}\eta (ϵ\hat{\mathbf{r}})d\Omega$, which tends to $4\pi \eta (0)$. This proves the displayed distributional divergence, with the three-dimensional Dirac delta function representing the point source.

For an axisymmetric thin disc supplied at $R_0$ with the local Keplerian rotation specific angular momentum, the displayed source augments the Keplerian viscous diffusion equation. The radial velocity remains $v_R=-3(\Sigma R^{1/2}{)}^{-1}{\partial }_R(\nu \Sigma R^{1/2})$: subtracting local specific angular momentum times conservation of mass from conservation of angular momentum cancels the source term. The normalization of the Dirac delta function makes the integrated supplied mass per unit time equal to $\dot{M}(t)$.