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A grey atmosphere model with mass absorption coefficient proportional to pressure. Under hydrostatic balance, $\kappa ={\kappa }_{s}p/p_s$ gives downward optical depth $t={\kappa }_s{p}^2/(2g{p}_s)$. For $T=T_s(p/p_s{)}^a$, a non-scattering optically thick column emits hemispheric radiant flux $F\sim \sigma T_s^4{t}_s^{-2a}\Gamma (2a+1)/(a+1)$. The angular factor follows from integrating ray attenuation over the upward hemisphere; a vertical-ray approximation omits it. The extension of the emitting-depth integral to infinity requires fixed $a$ as $t_s\to \infty$.

A probability density is a nonnegative function whose integral over an event gives its probability.

If $(X,Y)$ has a joint probability density $f$, the product has the displayed probability density function, interpreted almost everywhere. For fixed $x\ne 0$, substitute $z=xy$ in the integral defining $\mathbb{P}(XY\in A)$; the absolute Jacobian determinant is $|x|$. The Tonelli theorem then exchanges the nonnegative integrals. The line $x=0$ has two-dimensional Lebesgue measure zero, so its omission creates no probability mass. A density may be infinite at exceptional points without creating an atom.

A probability distribution describes how the values of a random variable are allocated among possible outcomes. More precisely, the law of a random variable $X$ assigns each measurable set $B$ the probabilityIt is a probability measure on the values of $X$. The underlying experiment can be more elaborate than the output: different experiments may produce the same probability distribution. The law of a random variable retains the probabilities needed to answer questions about that output. ^[ref].

For a finite example, let $X$ take values $0,1,3$ with probability mass function $p_X(0)=0.2$, $p_X(1)=0.5$, $p_X(3)=0.3$. Its probabilities sum to one. The event $0<X\le 1$ has probability $0.5$, whereas $0\le X\le 1$ has probability $0.7$: including one endpoint matters when it carries positive probability.

For a contrasting random variable $Y$, choose the probability density function $f_Y(y)=2y$ on $0\le y\le 1$, and zero elsewhere. It integrates to one. Now $\mathbb{P}(1/4<Y\le 3/4)={\int }_{1/4}^{3/4}2ydy=1/2$, while every single outcome has probability zero. The probability density function can have height $f_Y(0.8)=1.6$: probabilities come from integrals over outcome ranges.

The cumulative distribution function answers all threshold questions for a real random variable, regardless of whether there are jumps or a probability density function. A quantile function reverses the direction of the question: given a cumulative probability level, find a threshold. This connects descriptive summaries to inverse transform sampling, which constructs a chosen probability distribution from a uniform distribution.

For independent random variables $X,Y$ with the uniform distribution on $(0,b)$, use the probability density of a product of random variables and integrate $1/(b^{2}x)$ over $z/b<x<b$. The displayed probability density function is zero off $(0,b^2)$ and has integral one. More generally, for independent and identically distributed factors $U_j$ with this uniform distribution, $W_n={\prod }_{j=1}^n{U}_j$ satisfiesThe expectation of a product of independent random variables proves these identities directly.

An optically thin spherical source with volume emissivity $j(r)=j_0{r}^{-p}$ has surface brightnessSubstitute $z=Ru$ in the line-of-sight integral and evaluate ${\int }_{-\infty }^{\infty }(1+u^2{)}^{-p/2}du$ using the beta function. The condition $p>1$ is convergence at large distance. A core or finite outer boundary changes the projection near that boundary; the stated scaling applies between those scales.

Let ${\delta }_t=\sqrt{p_t}-\sqrt{p}=(t/2)g\sqrt{p}+r_t$, with $\Vert r_t{\Vert }_2=o(|t|)$. Since $p_t-p=2\sqrt{p}{\delta }_t+{\delta }_t^2$, the Cauchy-Schwarz inequality gives $\Vert p_t-p-tpg{\Vert }_1\le 2\Vert r_t{\Vert }_2+\Vert {\delta }_t{\Vert }_2^2=o(|t|)$. Thus differentiability in quadratic mean suffices to differentiate bounded density integrals, even when the density itself has no useful pointwise derivative.

A quadrature rule approximates an integral by a finite weighted sum of sampled function values:The quadrature nodes $x_i$ specify where to sample, and the quadrature weights $w_i$ specify how much each sample contributes. The density $\rho (x)$ specifies the target integral, whereas the discrete quadrature weights belong to the approximation. Ordinary numerical integration has $\rho =1$. Gaussian quadrature and the midpoint quadrature rule choose interior quadrature nodes; the trapezoidal rule includes endpoints.^[ref]

For ordinary numerical integration, exactness on the constant function $1$ requires ${\sum }_i{w}_i=b-a$; exactness on $x$ additionally requires ${\sum }_i{w}_i{x}_i=(b^2-a^2)/2$. These two checks make the quadrature rule exact for every affine function. For example, both the midpoint quadrature rule and trapezoidal rule give $6$ for ${\int }_0^2(2x+1)dx$. Testing progressively higher polynomial degrees defines the degree of precision of a quadrature rule. Exactness describes a class of integrands, rather than promising exact results for every function.

Composite midpoint quadrature rules and composite trapezoidal rules apply a simple quadrature rule on every panel of a partition of an interval. This buys accuracy by shrinking the panels while keeping the elementary formula fixed. For continuous functions sampled exactly, their quadrature errors tend to zero as the largest panel width tends to zero. The familiar second-order bounds require stronger assumptions about derivatives. The signed approximation error used here is $Q(f)-I_{\rho }(f)$; some sources define quadrature error with the opposite sign, so the convention must accompany any formula.

Quantum escape of a current-biased Josephson junction is decay of a trapped collective quantum phase state through a barrier in the tilted Josephson washboard potential. This barrier lies in the potential energy of an electrical circuit coordinate. Transfer of Cooper pairs across the physical insulating layer already supplies the Josephson coupling energy; it does not by itself demonstrate this further collective escape. Single-electron superconducting tunneling is a third, distinct process.

At higher temperature $T$, thermal activation instead lifts the coordinate over the barrier. In the rare-escape regime its characteristic form is the Kramers escape ratewhere $k_B$ is the Boltzmann constant and the prefactor depends on damping and thermal equilibration. Cooling suppresses this route. For a quasistationary energy level at energy $E$, the leading conservative WKB approximation instead haswith the two turning points bracketing the forbidden barrier. This is an exponential estimate requiring a large action compared with $\hslash$, slow escape compared with intrawell motion, and weak environmental corrections. It is not a complete rate prediction near the barrier top or with substantial damping.

The familiar cubic-barrier exponent can be checked independently. Choose $V(x)=M_{\varphi }{\omega }_p^2{x}^2(1-x/a)/2$, with $a>0$. Its top is at $2a/3$ and $\Delta U=2M_{\varphi }{\omega }_p^2{a}^2/27$. At the well-bottom reference energy $E=0$,Substitution $u=x/a$ reduces the dimensionless integral to ${\int }_0^{1}u\sqrt{1-u}du=4/15$. This leading semiclassical action uses a cubic potential and a well-bottom reference; actual ground-state escape also needs the zero-point correction, prefactor and environment. In the chosen washboard example it gives $B\simeq 74.0$ after matching the exact barrier and curvature. This matching is an illustrative approximation, not an exact action integral for the cosine potential.

Escape need not yield a sustained voltage: the coordinate can become trapped again. The running-state readout assumes sufficient underdamping and bias to avoid such retrapping. The parallel model has quality factor $Q={\omega }_{p}RC$; large $Q$ makes energy loss per oscillation small. Low-temperature rate saturation alone cannot distinguish quantum tunnelling from unwanted heating or external noise.

A quasistatic process is an ideal limit in which a changing thermodynamic system passes through a sequence of states arbitrarily close to thermodynamic equilibrium. Its intermediate pressure, temperature and volume are therefore well defined and obey its equation of state. Driving must be slow compared with the system's own equilibration, rather than merely taking a long time on a clock. A reversible thermodynamic process additionally allows both system and surroundings to be restored without net change; friction can produce entropy even in a quasistatic process.^[ref]

For a computed example, take a fixed amount of substance $n$ of ideal gas in a sealed cylinder with a frictionless piston, in contact with a heat bath at absolute temperature $T$. Reduce the external load in infinitesimal steps so that the piston expands from $V_0$ to $2V_0$. In the reversible limit, the gas pressure exceeds the external pressure by an infinitesimal amount during expansion, and the heat transfer occurs across an infinitesimal temperature difference. Exact mechanical equilibrium would give no driving force; the limiting imbalance supplies motion while preserving equilibrium intermediate states.^[ref]

Along this reversible isothermal expansion, the ideal gas law givesHere $R$ is the molar gas constant; specifying $nRT$ fixes an energy scale, without choosing $n$ and $T$ separately. The initial and final equilibrium states are $(V_0,p_0,T)$ and $(2V_0,p_0/2,T)$. Count work delivered to the surroundings as positive. The reversible work is the shaded area beneath the pressure-volume curve:The integral of $1/V$ supplies the natural logarithm. Since internal energy of an ideal gas depends only on temperature, $\Delta U=0$; the first law of thermodynamics, $\Delta U=Q_{\mathrm{in}}-W_{\mathrm{out}}$, gives $Q_{\mathrm{in},\mathrm{rev}}=0.693147\mathrm{kJ}$ to the displayed precision. The heat bath replaces the energy delivered as work.^[ref]

Now start from the same initial state, but abruptly lower the external pressure to the constant value $p_{\mathrm{ext}}=p_0/2=0.5\mathrm{MPa}$. There is a finite initial pressure imbalance, so this is an irreversible expansion rather than a quasistatic process. Assume the piston begins and ends at rest, has no net change in stored energy, and has no external friction; relaxation occurs within the gas and its heat bath. After motion has settled and the heat bath has restored temperature $T$, mechanical equilibrium and the ideal gas law give $V_f=nRT/p_{\mathrm{ext}}=2V_0$. The work delivered to the constant load isThis work follows directly from load force times piston displacement, using $dV=Adx$ for piston area $A$. Intermediate gas states need not have a single equilibrium pressure or temperature. In particular, neither $p(V)=nRT/V$ at the bath temperature nor $p(V)=p_{\mathrm{ext}}$ describes the actual gas during this irreversible motion. The rectangle below plots the known external load; its upper edge is not an equilibrium gas path.

The entropy accounting makes the difference precise. Because entropy is a state function, both expansions have the same gas entropy change, computed along the reversible isothermal expansion:The thermal reservoir stays at $T$ and loses heat $Q_{\mathrm{in}}=W_{\mathrm{out}}$, so $\Delta S_{\mathrm{bath}}=-W_{\mathrm{out}}/T$. Assume the ideal load stores work without an entropy change. The combined entropy production is thereforeNo numerical entropy unit is implied by $nRT=1\mathrm{kJ}$ alone: $nR$ or $T$ must also be specified. The lost work opportunity is $T\Delta S_{\mathrm{total}}=0.193147\mathrm{kJ}$ for this comparison. This calculation uses equilibrium endpoints and the reservoir's temperature, without assigning equilibrium intermediate states to the irreversible gas. ^[ref]

For another boundary choice, let the same ideal gas expand into an initially evacuated compartment of volume $V_0$ inside a rigid, insulated vessel of total volume $2V_0$. Removing the partition causes free expansion: no load is moved and no heat crosses the outer boundary, so $W_{\mathrm{out}}=Q_{\mathrm{in}}=0$ and $\Delta U=0$. After equilibration the temperature is again $T$, the pressure is $p_0/2$, and the gas entropy has increased by $nR\ln 2$. The outer vessel has fixed volume; the gas's accessible volume doubles. Restoring the initial gas state requires compression and removal of heat, changing the surroundings. Equal endpoints do not make free expansion reversible. ^[ref]

A slowly moving piston with dry friction supplies a different caution: its gas can remain arbitrarily close to thermodynamic equilibrium, yet rubbing still converts work into heat and produces entropy. If the frictional force has a nonzero slow-motion limit, its dissipated energy over a fixed travel does not vanish with speed. Reversing the motion reverses friction, so infinitesimally changing the external load cannot retrace the same operation and restore the surroundings. The heat capacity calculations additionally exclude friction and other dissipative work.^[ref]

The usefulness of a quasistatic process is that it replaces detailed motion by a readable sequence of equilibrium states. An equation of state and an integral then turn a pressure-volume drawing into an energy budget. The beautiful distinction is that internal energy and entropy depend on the endpoints, while heat and work record how the change was performed. Reversible benchmarks expose how much work an irreversible device sacrifices, while entropy production explains why simply making it slower need not recover that opportunity.

Assign an arbitrary finite value at the origin. Polar integration and the substitution rule $u=r^2$ reduce the joint L1 norm exactly to $\pi {\int }_0^{\infty }|\sin u|u^{-\alpha }du$. Near zero, $|\sin u|\le u$ gives $u^{1-\alpha }$; at infinity, $|\sin u|\le 1$ gives $u^{-\alpha }$. These are integrable when $\alpha <2$ and $\alpha >1$, respectively. Their explicit integrals give the displayed bound. This is a planar version of sine power kernel integrability, with the radial area element determining the exponents.

In geometrical optics, suppose radiation has a nonnegative radiance $L(\mathbf{x},\mathbf{s})$ whose directional contributions add as power. Here $\mathbf{s}$ is the actual propagation unit vector, $d\Omega$ is a solid angle element, and the collecting oriented surface $A$ has chosen unit normal $\mathbf{n}(\mathbf{x})$. Assume the following integrals exist. The outward and inward radiant fluxes areBoth quantities are nonnegative; their difference is the signed net radiant flux. The dot product supplies the cosine of the angle with the unit normal, hence the projected area. For incoming radiation the outward unit normal gives a negative dot product; the minus sign makes the collected power positive. Measuring the incoming angle from a source-facing unit normal instead gives the familiar positive $\cos \theta$. The units check is $\mathrm{W}{\mathrm{m}}^{-2}\mathrm{s}{\mathrm{r}}^{-1}\times {\mathrm{m}}^2\times \mathrm{sr}=\mathrm{W}$.

Equivalently, the incoming surface integral is ${\Phi }_{-}={\int }_{A}EdA$ with local irradiance $E$; the emitted part of ${\Phi }_{+}$ is ${\int }_{A}MdA$ with radiant exitance $M$. For a small flat patch and a narrow accepted cone, nearly constant radiance and angle give $\Phi \simeq LA\cos \theta \Delta \Omega$. This approximation requires uniformity across the patch and cone. Direction-independent outgoing radiance over a whole hemisphere gives the different exact result ${\Phi }_{+}=\pi LA$, by Lambert's cosine law and the spherical coordinates calculation in radiant exitance. Replacing the cosine-weighted integral by the hemisphere's $2\pi$ solid angle would double the answer.^[ref]

These radiance formulas presume a description in additive directional power; they do not reconstruct arbitrary coherent wave interference from separate ray contributions. For general electromagnetic fields, the signed instantaneous power crossing an oriented surface is ${\int }_A\mathbf{S}\cdot \mathbf{n}dA$, using the Poynting vector $\mathbf{S}$; optical measurements commonly average over many wave periods. A zero net radiant flux can conceal equal nonzero incoming and outgoing radiant fluxes.

A radiation pattern specifies emitted power per solid angle as a function of direction. For a slowly moving point electric charge with acceleration $\mathbf{a}$, $dP/d\Omega =q^2|\mathbf{n}\times \mathbf{a}{|}^2/(16{\pi }^2{ϵ}_0{c}^3)$. Its integral over directions is the Larmor formula; its fore-aft symmetry implies zero net radiated three-momentum in the instantaneous rest inertial frame. Directional detector power includes the distinction between source emission time and arrival time.

A bounded continuous factor $k_1$ and an integrable factor $k_2$ give a bounded velocity-integral gain on bounded continuous functions, with bound $\Vert Kf{\Vert }_{\infty }\le \Vert k_1{\Vert }_{\infty }\Vert k_2{\Vert }_1\Vert f{\Vert }_{\infty }$. The dominated convergence theorem proves joint continuity of the velocity integral. Combined with damped transport, the factorial bound for a Volterra iterate proves existence on arbitrary finite time slabs.

For $t>0$, the natural logarithm satisfiesTo prove it, put $h(t)=\ln (1+t)-2t/(2+t)$. Then $h(0)=0$ and the derivative is $h^{\prime }(t)=t^2/[(1+t)(2+t{)}^2]>0$. More precisely,This positive integral controls errors in symmetric approximations to the logarithm. After writing $x=1+t$, the bound is $\ln x>2(x-1)/(x+1)$ for $x>1$; the inequality reverses for $0<x<1$ by replacing $x$ with $1/x$.

The reachable set ${\mathcal{R}}_T(x_0)$ collects all endpoints attainable from $x_0$ at exactly $T$ by admissible control inputs. If arrival any time up to $T$ is allowed, use ${\mathcal{R}}_{\le T}(x_0)={\bigcup }_{0\le t\le T}{\mathcal{R}}_t(x_0)$, with ${\mathcal{R}}_0(x_0)=\{x_0\}$. These definitions differ: a general dynamical system may drift away after early arrival, so waiting need not preserve the target. The reachable set is always relative to a declared input class and any constraints on the entire path.

For the double integrator starting from $(0,0)$, now require measurable functions as control inputs with $|u(t)|\le 1$ almost everywhere. Write $(p,v)$ for its endpoint at $T>0$. Two integrals determine the endpoint:Our mathematical example has the exact reachable setTo see the upper boundary, fix $v$: put $u=1$ first and $u=-1$ last, switching at $s=(T+v)/2$. This maximizes the weighted integral for position because early commands have the largest weight $T-t$; the required net velocity fixes $s$. Formally, $(T-t)-(T-s)=s-t$ has the same sign as this maximizing command, so changing to any other admissible command with the same integral cannot increase $p$. Reversing the order gives the lower boundary. Pointwise convex combinations of these two commands preserve $v$ and attain every intermediate $p$, so the formula describes the whole region, not merely a bound. The extreme commands are bang-bang control.

In particular, the bounded double integrator can move from rest to position $d$ and rest again exactly when $|d|\le T^2/4$. Its minimum duration is therefore $2\sqrt{|d|}$ in these normalized units. With freely chosen time every rest-to-rest distance is attainable; with a fixed deadline many are excluded. Changing the amplitude bound to $|u|\le a$ with $a>0$ scales this condition to $|d|\le a{T}^2/4$. This shows why specifying the controls matters even when the Kalman rank criterion gives full rank of a matrix.

In closed quantum control, a density operator reachable set lies inside its unitary orbit of a density operator, since coherent evolution preserves the spectrum. That orbit need not be fully reachable under the available control inputs.

Real analysis studies limits, continuity, differentiation, integration, and convergence of sequences and functions taking values in the real numbers. Its central question is how finite approximations determine an exact answer: what does an arbitrarily small error mean, and which operations respect that error? A derivative is a limit of quotients, while a Riemann integral is obtained from increasingly fine Riemann sums. Both need conditions that make the answer independent of the approximations chosen.

Two organizing ideas are completeness of the real numbers, which ensures that suitable approximations have a destination, and uniform convergence, which controls approximation error over an entire domain. Pointwise convergence supplies separate guarantees at separate inputs; uniform convergence supplies one guarantee for all inputs. This distinction explains why replacing an approximating function by its limit can preserve continuity or integrals under appropriate hypotheses, yet fail without them.

For historical context, Cauchy's 1821 Cours d'analyse explicitly discusses convergence of series. Its Note VII treats double series and compares summing by rows and by columns under stated convergence assumptions: controlling an infinite process was already a concrete problem. This is historical context for questions about limits, rather than an attribution of the modern definitions below. ^[ref].

With inverse-time convention $e^{i\omega t}$, the half-line Klein-Gordon equation ${\sigma }_{tt}-{\sigma }_{rr}+{\mu }^2\sigma =0$ has spatial normal modes $A{e}^{ikr}+B{e}^{-ikr}$, where $k=\sqrt{{\omega }^2-{\mu }^2}>0$ above threshold and $k=i\sqrt{{\mu }^2-{\omega }^2}$ below it. For the branch even in $\omega$, reality above threshold requires $A(\omega )=B(-\omega {)}^{\ast }$. Below threshold boundedness eliminates $B$, and reality requires $A(\omega )=A(-\omega {)}^{\ast }$. At threshold the spatial differential equation instead has the basis $1,r$; boundedness retains only the constant mode. A spectral integral with ordinary integrable functions as amplitudes ignores the isolated threshold points, while distribution amplitudes can represent the surviving monochromatic constant-spatial modes.

Rectangular duct flow is steady flow of a Newtonian fluid down a straight rectangular pipe under a constant axial pressure gradient $-G$. For stationary walls and unidirectional flow, incompressibility eliminates axial dependence and the Navier-Stokes equation becomes a two-dimensional Poisson equation with zero Dirichlet boundary conditions. Subtracting a plane Poiseuille flow profile leaves a harmonic function that repairs the side-wall values using separation of variables. Uniqueness follows from Green's first identity: the difference of two solutions has zero boundary values and zero integral of the squared gradient, and hence vanishes.

The rectangular function has value one on $|t|<1/2$ and zero on $|t|>1/2$. Endpoint values are convention-dependent and do not affect its ordinary integrals or Fourier transform. Shifts, amplitude changes and time rescaling produce a rectangular pulse of arbitrary height, center and duration. With $\hat{f}(k)=\int f(t)e^{-ikt}dt$, for $a>0$ the width-$2a$ pulse $f(t)=1_{|t|<a}$ has Fourier transform $2\sin (ak)/k$, with value $2a$ at zero by a limit. Its convolution with itself is a rescaled triangular function, $(2a-|t|)$ on $|t|<2a$ and zero outside, because the convolution measures the overlap length of two intervals.