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Using $\nabla |x-x\prime {|}^{-1}=-\nabla \prime |x-x\prime {|}^{-1}$, the divergence is a line integral of a total derivative, zero for a closed wire or suitable vanishing endpoints. Taking the curl and moving it under the integral gives the stated Biot–Savart formula.

Bilinearity and symmetry follow from the integral. Also $⟨p,p⟩={\int }_0^{1}p(t{)}^{2}dt\ge 0$, and equality forces the continuous polynomial $p$ to vanish identically.

Fermat’s principle makes $\int ds/c$ stationary. Writing the ray as $x=x(y)$ gives a Lagrangian independent of $y$, and its first integral is $\sin \theta /c=K$. At the origin $K=(1/\sqrt{2})/(1/\sqrt{3})=\sqrt{3/2}$. At $x=1$, $c=1/\sqrt{2}$, so $\sin \theta =\sqrt{3}/2$ and the forward angle is $\theta =\pi /3$.

Away from $x=y$ the Newton kernel is harmonic, while integrating its normal derivative over a small sphere gives one, proving ${\nabla }^2\hat{G}=\delta$. Green’s second identity is ${\int }_V(u{\nabla }^{2}v-v{\nabla }^{2}u)={\int }_{\partial V}(u{\partial }_{n}v-v{\partial }_{n}u)$. For the half-space, images give $G(x,y)=-(4\pi |x-y|{)}^{-1}+(4\pi |x-y^{\ast }|{)}^{-1}$. Substitution in the boundary formula yields the Poisson kernel $z/[2\pi ((x_1-y_1{)}^2+(x_2-y_2{)}^2+z^2{)}^{3/2}]$; polar integration against the Gaussian gives exactly the stated one-dimensional integral.

Integrate $(1+z^p{)}^{-1}$ around the wedge $0\le \arg z\le 2\pi /p$. The radial integrals differ by the factor $e^{2\pi i/p}$, the arc vanishes, and the wedge contains the simple pole $e^{i\pi /p}$ with residue $-[p{e}^{-i\pi /p}{]}^{-1}$. Solving the resulting identity gives $I=(\pi /p)\csc (\pi /p)$.

Twice integrating by parts, with both endpoint variations fixed, gives $g_x-d(g_{\dot{x}})/dt+d^2(g_{\ddot{x}})/d{t}^2=0$. In the given integral the mixed $\dot{x}\ddot{x}$ term is a boundary term, and the equation is $(D^2-4{)}^{2}x=0$. Decay removes the growing solutions, so $x=(A+Bt)e^{-2t}$. The initial data give $A=1,B=3$, hence $x=(1+3t)e^{-2t}$.

Vertical hydrostatic balance gives $p=\rho g(h-z)$; depth-independent horizontal momentum and depth-integrated incompressibility yield the two shallow-water equations. Curling momentum and combining with mass conservation gives $D[(\zeta +f)/h]/Dt=0$. Dot momentum with $\rho hu$ and use continuity to obtain $E_t+\nabla \cdot J=0$ with $E=\rho h|u{|}^2/2+\rho g{h}^2/2$ and $J=u(E+\rho g{h}^2/2)$. It is kinetic plus gravitational potential energy. A no-normal-flow boundary kills the flux and conserves its integral; a localized nonuniform height has excess positive energy relative to the uniform rest state with the same mass, so it cannot decay to that state without dissipation.

Noetherian means every ideal is finitely generated, equivalently every ascending ideal chain stabilises. Quotients preserve this; for example $\mathbb{Z}[\sqrt{-5}]$ is Noetherian as a quotient of $\mathbb{Z}[X]$ but not a UFD. A UFD need not be Noetherian (a polynomial ring in infinitely many variables is a counterexample). Stabilisation of kernels proves every surjective endomorphism of a Noetherian ring injective. The shift $k[x_1,x_2,\dots ]\to k[x_1,x_2,\dots ]$, $x_1\mapsto 0$, $x_{i+1}\mapsto x_i$, is surjective noninjective; $x\mapsto x^2$ on $k[x]$ shows injective need not mean surjective. Finally $f_n(X)=\left(\genfrac{}{}{0ex}{}{X}{n}\right)$ is integer-valued. Finite differences prove uniquely that every integer-valued polynomial is an integral linear combination of the $f_n$, so they are a $\mathbb{Z}$-basis. Int$(\mathbb{Z})$ is not Noetherian: the denominators in $f_p$ yield an ideal chain requiring new prime denominators.

Write the power-law size distribution as $n(D)=k{D}^{-\alpha }$. The total geometric cross-section of the optically thin debris disk isBecause blackbody grains at radius $r$ intercept the fraction ${\sigma }_{\mathrm{tot}}/(4\pi r^2)$ of the stellar luminosity, the fractional luminosity of a debris disk givesAnother integral then yieldsfor $D_{\min }\ll D\ll D_{\max }$.

The trace/norm integrality conditions give the stated congruences for
$\theta =(a+bi+c\sqrt{m}+di\sqrt{m})/2$. Solving them yields these integral bases:The three quadratic subfields have fundamental discriminants $-4,D_m,D_{-m}$. By discriminant of a biquadratic field,in both residue classes.

Hamilton's equations are ${\dot{x}}_i=p_i$ and ${\dot{p}}_i=-{\partial }_{i}V$. A first integral $F$ satisfies $\{F,H\}=0$. See integrable Hamiltonian system: in three degrees of freedom, integrability requires three independent commuting first integrals.

Insert the incident plane wave for $\psi$ in the integral. At large $r$, $|\mathbf{r}-{\mathbf{r}}^{\prime }|\simeq r-\hat{\mathbf{r}}\cdot {\mathbf{r}}^{\prime }$, so the outgoing amplitude isThus Born approximation gives the stated differential cross section.

The ordinary integral fails because the simple real poles at $-2$ and $3$ produce nonintegrable logarithmic singularities. Partial fractions giveClosing above for $\omega >0$ and below for $\omega <0$, with symmetric semicircular indentations, yieldsTaking real and imaginary parts,All integrals here run over $\mathbb{R}$.

Given a nonzero fractional $I$, scale it to an integral ideal $aI$. By (i), $(aI)J=(r)$ for some nonzero ideal $J$. Then $I^{-1}=a{r}^{-1}J$ is fractional and $I{I}^{-1}=R$.

Since $\beta$ is integral, $R[\beta ]$ is generated over $R$ by finitely many powers, hence is fractional. It is invertible by (c). Because it is a ring, $I^2=I$; multiplying by $I^{-1}$ gives $I=R$. Thus every $\beta \in {\mathcal{O}}_K$ lies in $R$, so $R={\mathcal{O}}_K$.

The nontrivial Floquet multiplier is the factor by which a transverse perturbation changes after one period. Liouville's formula makes its logarithm the integral of the divergence. Here $\nabla \cdot f=\mu (1-3a{x}^2-y^2)$, yielding the stated formula.

For $k_{B}T\ll \hslash {\omega }_D$, extend the integral to infinity:Thus $X=\zeta (3)/\pi$.

Repeated integration by parts in the cosine-coefficient integral shows $|c_k|\le C_r{k}^{-r}$ for every $r$ when $f$ is smooth (endpoint terms vanish after using the even periodic extension). Parseval then bounds the squared tail by $C_r{\sum }_{k>N}{k}^{-2r}$, which is $O(N^{-2r+1})$. Choosing $r$ arbitrarily large proves spectral convergence.