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For $b>0$, the exponential variational trial for a square-root half-line well satisfies the wall boundary condition and has norm ${\int }_0^{\infty }{x}^2{e}^{-2bx}dx=1/(4b^3)$. integration by parts has no boundary term and givesThe supplied gamma integral givesSetting $E^{\prime }(b)=0$ yields the unique minimum,At the minimum $⟨V⟩=4⟨T⟩$, so the variational ground-state estimate isThe energy diverges as $b\to 0$ or $b\to \infty$, confirming that the stationary point minimizes the trial energy.

On periodic ring states the angular-momentum index $m$ is an integer. At the stated resonance,so $U_T$ is the identity on every orbital state. After $N$ steps from the normalized ground state,Its phase varies in space but its position probability density remains uniform. Periodicity permits integration by parts, givingConsequentlyEquivalently, the angular-momentum probabilities are $P_m=J_m(N\eta {)}^2$, with ${\sum }_m{m}^2{J}_m(z{)}^2=z^2/2$. The mean angular momentum remains zero while its variance grows as $N^2$. This quantum resonance of the kicked rotor gives coherent ballistic energy growth. The Bessel second-moment identity connects the momentum spreading to that quadratic energy law; an incoherent random walk would instead give linear energy growth.

Canonical probabilities. A canonical ensemble describes a system of fixed particle number and fixed external parameters such as volume, weakly coupled to a much larger thermal reservoir. They exchange energy and reach thermal equilibrium, while the reservoir's temperature stays effectively constant. The interaction energy is negligible in the state counting. The combined isolated system is described by equal probabilities for its accessible microstates.

If its total energy is $E_{\mathrm{tot}}$, the probability of a particular system microstate $i$ is proportional to the reservoir multiplicity at $E_{\mathrm{tot}}-E_i$. Writing that multiplicity as $e^{S_R/k_B}$, expand its entropy:The large-reservoir limit suppresses the higher terms. Normalization therefore givesThe sum counts individual microstates; when grouping them by energy, their state degeneracies must be included. This $Z$ is the canonical partition function.

Classical orientational heat capacity. Hold the magnetic field fixed. Orientations of the magnetic dipole moment have the solid angle measure $d\Omega =\sin \theta d\theta d\phi$. With $x=\beta mB$, the angular partition function isChoosing a normalized angular measure would remove $4\pi$, with no effect on the energy derivatives. Differentiating at fixed $B$,The angular average $⟨\cos \theta ⟩=\coth x-1/x$ is the Langevin function. Since $dx/dT=-x/T$,This is the heat capacity of a classical magnetic dipole, in $\mathrm{J}{\mathrm{K}}^{-1}$ for one magnetic moment. The word specific here refers to that single moment; $N$ independent magnetic moments have $N$ times this heat capacity. At $B=0$ the energy is independent of orientation and $C_B=0$, obtained by the continuous $x\to 0$ limit. At fixed nonzero $B$, the classical low-temperature limit instead gives $C_B\to k_B$.

Equipartition and its proof. The equipartition theorem states that an independent quadratic term $a{q}^2/2$ in a classical Hamiltonian, with $a>0$, contributes $k_{B}T/2$ to its mean energy. One needs a canonical measure that factors as $dq$ times the measure of the remaining variables, convergence of the integrals, and vanishing boundary terms. The coordinate must not be restricted by a thermally important hard boundary or a coupled constraint.

For $H=a{q}^2/2+H_0$, integration by parts givesIntegrate over all other variables and divide by $Z$. It follows thatThe same reasoning applies to quadratic momenta among the canonical variables. It is the number of quadratic energy contributions that is counted, not an unrestricted count of coordinate labels.

Two soft orientational directions. When $x\gg 1$, the moment is concentrated near alignment. Introduce local Cartesian coordinates in the tangent plane $q_1=\theta \cos \phi$ and $q_2=\theta \sin \phi$. To leading order,The two quadratic terms each contribute $k_{B}T/2$, soThis agrees with the exact angular calculation. Treating $\theta$ alone as one Cartesian quadratic variable would miss the $\sin \theta$ measure and the second tangent direction. There are no extra rotational kinetic terms in the stipulated Hamiltonian; adding such terms would be a different model.

Quantum spin one. The spin projections are $m_s=-1,0,1$, with $S_z=\hslash m_s$. Put $\Delta =\hslash |\gamma |B>0$, where $\gamma$ is the gyromagnetic ratio. The three energies are $-\Delta ,0,+\Delta$, irrespective of the sign of $\gamma$. Relative to the ground state they are $0,\Delta ,2\Delta$. Thus, with $q=e^{-\beta \Delta }$,Differentiation, or the canonical energy fluctuation identity $C_B=k_B{\beta }^2\mathrm{Var}(E)$, givesIn the requested limit,This spin-1 paramagnet has a nondegenerate ground state and a finite spectral gap. Excitations are exponentially rare, so its heat capacity tends to zero. Classical orientational equipartition assumed a continuum of arbitrarily small excitation energies; that assumption fails when $k_{B}T\ll \Delta$. This is a change in the accessible state spectrum, not a violation of the equipartition theorem within its classical domain.

Occupation and the two-dimensional Fermi energy. For an independent single-particle state, the Fermi-Dirac distribution isThe chemical potential enforces the chosen mean particle number through $N={\sum }_{j}n({\epsilon }_j)$, including spin states in the sum. It is common to all states in particle-exchange equilibrium and must generally change with temperature or field when the total mean number is held fixed. At zero temperature, the distribution becomes a Heaviside step function at the Fermi energy.

Let the sample area be $\mathcal{S}$ and let $g_s$ be its internal spin spin degeneracy. In the continuum limit, filling the disk $k<k_F$ givesThereforeFor spin-half fermions with both spin states available, $g_s=2$ and ${\epsilon }_F=\pi {\hslash }^{2}N/(m\mathcal{S})$; for a spin-polarized gas $g_s=1$. The question does not specify the area or spin degeneracy, so these conventions must be stated rather than extracting an energy from $N$ alone. Although the original text calls $\mathbf{k}$ momentum, the supplied ${\hslash }^2{k}^2/(2m)$ dispersion makes it a wave vector; physical momentum is $\mathbf{p}=\hslash \mathbf{k}$. The result agrees with the constant two-dimensional free-electron density of states. Discrete level spacings in a finite sample would require a specified boundary spectrum.

One level occupied on average. Write $\delta ={\mu }_{B}B$ and $z=e^{\beta (\epsilon -\mu )}>0$. The two mean spin occupations areImposing $N_{+}+N_{-}=1$ gives $2+z(e^{\beta \delta }+e^{-\beta \delta })=1+z(e^{\beta \delta }+e^{-\beta \delta })+z^2$, hence $z^2=1$. Thus, for every positive temperature,The total magnetic moment is $\mathcal{M}=-{\mu }_B(N_{+}-N_{-})$. ConsequentlyThis is the grand-canonical spin susceptibility of a half-filled single level. Empty, singly occupied and doubly occupied orbital states are allowed in the grand canonical ensemble. At zero field, the singly occupied configurations together have probability $1/2$, explaining the factor of two relative to one permanent spin-half moment. If exactly one electron were required at all times, a canonical two-state partition function would instead give $\mathcal{M}={\mu }_B\tanh (\beta {\mu }_{B}B)$ and $\chi ={\mu }_B^2/(k_{B}T)$. That different answer does not implement the stated average-number constraint.

Density-of-states formula for the susceptibility. Here $g$ is the zero-field single-spin density of states. Regard its argument $E$ as the full spin-shifted energy. The shifted functions are $g_{\pm }(E)=g(E\mp {\mu }_{B}B)$, soThe total number is even in $B$; thus $\mu (B)=\mu (0)+O(B^2)$ at fixed total number on a smooth branch. More directly, terms from ${\mu }^{\prime }(0)$ cancel in the spin difference because the two densities coincide at zero field. To first order,It follows thatFor a physical energy band, take $g$ to vanish below its lower edge. Provided the endpoint term vanishes, integration by parts also givesThis form displays the positive response and the narrow thermal window around $\mu$. A sharp band edge is included by the corresponding distributional derivative in the first expression. The susceptibility defined here is the derivative of total magnetic moment with respect to $B$, with units $\mathrm{J}{\mathrm{T}}^{-2}$; it is not automatically a dimensionless SI volume susceptibility. A density-of-states convention per unit volume likewise gives a response per unit volume. These are the field conventions for magnetic susceptibility.

Chemical potential at fixed number. Define the integrated single-spin state countThe probability kernel $K_T$ has unit integral, mean $\mu$, and variance $a(k_{B}T{)}^2$. Using integration by parts,Expand $G$ about $\mu$. The linear term integrates to zero, while $G^{\prime \prime }=g^{\prime }$ givesThis is the number form of the Sommerfeld expansion. Since $N/2=G({\epsilon }_F)$ at zero temperature, set $\mu ={\epsilon }_F+\delta \mu$ with $\delta \mu =O(T^2)$:ThereforeThe coefficient can be obtained directly. With $y=(E-\mu )/(k_{B}T)$,For example, the sum follows by expanding $e^{-y}/(1+e^{-y}{)}^2$ on $y>0$, using the evenness of the kernel and integrating each exponential. Thus $a/2={\pi }^2/6$. A density of states increasing with energy gives a downward quadratic chemical-potential shift. For the constant two-dimensional density of states that shift vanishes to this order; a finite lower band edge still allows exponentially small corrections.

Low-temperature susceptibility at fixed number. Apply the same kernel expansion to $g$ itself:The shift of $\mu$ must be retained in the first term. Writing $g_F=g({\epsilon }_F)$ and similarly for its derivatives,ConsequentlyThis is the fixed-number low-temperature Pauli susceptibility. The $g_F^{\prime \prime }$ contribution is thermal broadening at fixed chemical potential; the $-g_F^{\prime 2}/g_F$ contribution enforces fixed total number. Both are needed. If the density of states were instead defined with both spin directions included, the prefactor $2$ would be absorbed into that definition. Only the spin response has been calculated; no orbital magnetic response is included.

The expansion assumes $g_F>0$, sufficient smoothness near the Fermi energy, and a temperature small compared with the energy scale over which $g$ changes or encounters a band edge. It does not apply unmodified at a singular density of states. For the ideal two-dimensional parabolic band, $g_F^{\prime }=g_F^{\prime \prime }=0$, so the leading Pauli response is temperature independent, consistently with the earlier state count.

Let $I=\int e^x\sin xdx$. Two applications of integration by parts givewhere additive constants can be absorbed at the end. Solving for $I$ yieldsDifferentiating the answer gives $e^x\sin x$.

Since $X\ge 0$, the function $X\mapsto X^2$ is strictly increasing on its support. For $y\ge 0$,and $F_Y(y)=0$ for $y<0$. Differentiating the cumulative distribution function givesThe density value at the single point $y=0$ is immaterial. Hence $Y$ has an exponential distribution with rate $1/2$. Direct integration by parts, or its standard expected value, yields

Apply integration by parts with $u=x$ and $dv=\sin (\mu x)dx$. Part (i) gives $v=-\cos (\mu x)/\mu$, so

On an interval avoiding $x=0$, divide by $x$ to obtain the first-order linear ordinary differential equationAn integrating factor is $\mu (x)=\exp \left(\int -2/xdx\right)=x^{-2}$. Multiplying through and using the product rule givesBy integration by parts, $\int x\cos xdx=x\sin x+\cos x+C$. ConsequentlyThe formula differentiates back to the original equation. Although division excluded $x=0$, this formula has a smooth extension there with $y(0)=0$, satisfying the original equation at zero as well.

Let $S(x)=\mathbb{P}(X>x)=1-F_X(x)$ for $x\ge 0$. Since $f_X=0$ on the negative half-line, $X\ge 0$ almost surely. We have $S^{\prime }=-f_X$ almost everywhere. On a finite interval, integration by parts givesThe lower endpoint contributes zero. The stated hypothesis gives $RS(R)\to 0$ as $R\to \infty$. Taking the increasing limits of the two nonnegative integrals proves the tail-integral identityThe equality allows the value $+\infty$; the boundary hypothesis alone does not force a finite expectation. In fact the formula for nonnegative $X$ also follows directly from Tonelli theorem, without that additional boundary hypothesis, by integrating $x={\int }_0^{\infty }{1}_{\{t<x\}}dt$ against the probability density function.

For positive bounds, integration by parts with $u=\log x$ and $dv=x^{2}dx$ givesFor negative bounds, specify the complex logarithm on the negative real axis. With argument $\pi$, namely the upper-bank boundary value of the principal logarithm, $\log x=\log (-x)+i\pi$. Substitute $t=-x$ in the real part:More generally, a continuous branch taking argument $(2k+1)\pi$ replaces $i\pi$ by $(2k+1)i\pi$. The imaginary part depends on that choice. The analytic principal branch excludes the negative axis itself, so the boundary-value convention is part of this interpretation.

The rate-one exponential distribution has density $e^{-t}$ for $t\ge 0$. Integration by parts gives $\mathbb{E}T=1$ and $\mathbb{E}[T^2]=2$, so

Assume sufficient decay and differentiability to justify integration by parts and exchange of integrals; Schwartz functions are one convenient class. With the stated Fourier transform convention, integrate first in $x$:The boundary contribution vanishes under the decay assumption. The Fourier transform of a derivative consequently turns constant-coefficient spatial differential operators into multipliers. Repeating it gives $\mathcal{F}[u_{xx}]=-k_x^2\hat{u}$, so the heat equation becomes an independent temporal ordinary differential equation at each wavevector.

A periodic domain identifies opposite faces of a fundamental cell, producing a flat torus. On an interval $[0,L]$, smooth periodic boundary conditions require $u(0,t)=u(L,t)$ and matching spatial derivatives. These endpoints are one location, rather than two independently controlled walls. Periodic integration by parts has no boundary term, and zero-mean functions have no zero Fourier mode.

Let $⟨F⟩=(4\pi {)}^{-1}\int Fd\Omega$ and $f=\bar{f}(ϵ)-ϵ{\bar{f}}^{\prime }(ϵ)\Theta (\mathbf{e})$ with comoving energy $ϵ=aE$. If ${ϵ}^4\bar{f}$ vanishes at both integration endpoints, integration by parts gives $-\int {ϵ}^4{\bar{f}}^{\prime }dϵ=4\int {ϵ}^3\bar{f}dϵ$. The kinetic stress-energy tensor then gives ${\delta }_{\gamma }=4⟨\Theta ⟩$, $v_{\gamma }^i=3⟨\Theta e^i⟩$, $\delta P=\bar{\rho }{\delta }_{\gamma }/3$, and ${\pi }^{ij}=4\bar{\rho }⟨\Theta (e^i{e}^j-{\delta }^{ij}/3)⟩$. If $\Pi =-\pi$ is used, the last expression changes sign. Frequency independence of $\Theta$ is essential to this common temperature description.

The temporal-temporal component of the photon stress-energy tensor in a local orthonormal tetrad. For an isotropic one-particle distribution function expressed using comoving energy $ϵ=aE$, it is ${\bar{\rho }}_{\gamma }=4\pi a^{-4}\int dϵ{ϵ}^3\bar{f}(ϵ)$. A thermal photon temperature perturbation gives a first-order directional energy perturbation four times its fractional temperature change. Boundary conditions on the distribution justify this factor through integration by parts.

A Pohozaev identity is an integral identity for a nonlinear partial differential equation, obtained by multiplying the equation by a scaling vector field such as $x\cdot \nabla u$ and applying integration by parts. It relates kinetic, potential, and nonlinear energies.

For $p>0$ and a nonnegative integer $n$, integration by parts gives, when $n>0$,The endpoint term $[s(1-s^p{)}^n{]}_0^1$ vanishes. ThereforeRepeated application yields $J_n(p)={\prod }_{j=1}^{n}pj/(pj+1)$. The integration by substitution $w=s^p$ also identifies $J_n(p)=B(1/p,n+1)/p$ with an Euler beta function. A related weighted integral follows immediately from the same substitution:

For a mode with nonreal phase velocity $c$, let $V=U-c$, choose a continuous branch of $V^a$, and put $\hat{w}=V^{a}q$. Under impermeable boundary conditions, multiplying the transformed Taylor–Goldstein equation by $V^{2a}\overline{q}$ and applying integration by parts givesChoosing $a=1/2$ gives the Miles–Howard theorem; choosing $a=1$ makes the real phase velocity of an unstable mode a weighted mean of $U$.

For a nonnegative integer $\ell$, this Coulomb potential energy eigenstate has principal quantum number $\ell +1$ and maximal magnetic quantum number. Put $b=(\ell +1)a$. Including the spherical volume element, its radial probability density is a gamma distribution proportional to $r^{2\ell +2}{e}^{-2r/b}$; its polar density is proportional to ${\sin }^{2\ell +1}\theta$, while its azimuth is uniform. Radial and polar variables are independent. Their moments satisfyThe radial identities follow by successive integration by parts; the angular identity follows from ${\int }_0^{\pi }{\sin }^{j+2}\theta d\theta =(j+1){\int }_0^{\pi }{\sin }^j\theta d\theta /(j+2)$.

Set $h=(2\ell +3{)}^{-1/3}$. Chebyshev's inequality and Markov's inequality give $\mathbb{P}(|r/\mu -1|>h)\le h$ and $\mathbb{P}(|\cos \theta |>h)\le h$. By the union bound, outside an event of probability at most $2h$, the point lies at distance $O(\mu h)$ from the equatorial circle of radius $\mu$. The quantum state therefore concentrates near a classical circular orbit in relative units as $\ell$ grows. The absolute orbit radius and absolute transverse width both grow; this statement does not assert localization in a fixed narrow tube.

For a nonnegative $f$, first set $g_n(x)=\min (f(x),n)1_{(1/n,n)}(x)$ and $F_n(x)={\int }_0^x{g}_n(t)dt$. These bounded functions vanish near zero and have compact support. Their averages have finite Lp norms: they vanish near zero and decay as $1/x$ beyond $n$. Integration by parts givesThe boundary term $x^{1-p}{F}_n(x{)}^p$ vanishes at zero because $F_n$ is zero there, and at infinity because $F_n$ is eventually constant and $p>1$. The inequality is Hölder's inequality with conjugate exponents $p$ and $p/(p-1)$. Divide when $I_n>0$; the zero case is immediate. Since $g_n\uparrow f$ and $H{g}_n\uparrow Hf$, the monotone convergence theorem proves the full bound without presupposing that $Hf$ has finite Lp norm. Apply it to $|f|$ for the signed or complex case.^[ref]