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If $\beta \in {\mathcal{O}}_L$, all of its $K$-conjugates are algebraic integers, so their sum and product give

Conversely, if these two quantities lie in ${\mathcal{O}}_K$, then $\beta$ is a root ofa monic polynomial whose coefficients are algebraic integers. Part (a), or transitivity of integral dependence, implies $\beta \in {\mathcal{O}}_L$.

Let $\beta =a+b\alpha \in {\mathcal{O}}_L$. The nontrivial $K$-automorphism sends $\alpha$ to $-\alpha$, soAlso $\alpha$ is integral, hence $\alpha \beta$ is integral, andBut $4+\sqrt{17}$ has norm $-1$, so it is a unit of ${\mathcal{O}}_K$. Multiplying by its inverse proves $2b\in {\mathcal{O}}_K$.

PutAway from $x=x^{\prime }$, this satisfies $(d^2/d{x}^2+k^2)G_k=0$. Its first derivative has jump $2ik$, so, distributionally,Applying $d^2/d{x}^2+k^2$ to the proposed integral equation therefore givesSince $k^2=2mE/{\hslash }^2$, this is precisely the stated Schrodinger equation. The term $e^{ikx}$ is the incident wave and the choice $e^{ik|x-x^{\prime }|}$ gives outgoing waves on both sides of each source point.

SetThe integral equation becomesEvaluating it at the three delta functions givesThus

For $x>a$, all three Green-function terms are proportional to $e^{ikx}$. Hence

Now define $\gamma =ik{\hslash }^2/(\lambda m)$, so that $q=-1/\gamma$. If $M$ denotes the matrix multiplying $\mathbf{\Psi }$, direct evaluation givesThe displayed algebraic equation is therefore exactly the condition that the finite-dimensional scattering system become singular. Its solutions are poles of the analytically continued scattering amplitude.

To see the imaginary-axis poles directly, write $k=i\kappa$ with $\kappa >0$. When $a\to 0$, the equation reduces toand its nonzero root is $\gamma =-3$, orThis is the bound state of the coincident potential $-3\lambda \delta (x)$. When $a\to \infty$, the three centres decouple and the roots approachwith exponentially small splitting. A pole at $k=i\kappa$ has energy $E=-{\hslash }^2{\kappa }^2/(2m)$ and an exponentially decaying wavefunction, so these upper imaginary-axis singularities represent bound states.

For discrete states $s$ of energy $E_s$, the canonical partition function isFor a classical continuous system the sum is replaced by the appropriately normalized phase-space integral, including the Gibbs factor $1/N!$ for identical particles.

It packages the equilibrium probabilities $P_s=e^{-\beta E_s}/Z$ and hence all canonical thermodynamics. In particular,and $S=k_B(\log Z+\beta ⟨E⟩)$.

Take $x$ upward, let $y$ measure distance from the wall, and work in the cylinder frame. The wall at $y=0$ moves upward with speed $V$. Choose the sign of $\Omega$ so that the near-cylinder surface at $y=h$ moves upward with speed $a\Omega$. Lubrication theory givesThusThe vertical volume flux per unit cylinder length is independent of $x$ and equalsThereforeBecause $p(+\infty )=p(-\infty )$, its integral over $x$ vanishes. Withthe quoted integrals giveIt follows that

The partial-fraction decomposition isThe residues at $i$ and $-i$ are respectively $1/(2i)$ and $-1/(2i)$. If two paths from $0$ to $z$ have winding-number differences $n_{+}$ and $n_{-}$ about those poles, their integrals differ byLoops can wind around either pole independently, so the value changes by arbitrary integer multiples of $\pi$ and $F$ is multivalued.

Along the straight real segment,Consequently the complete set of values is

Let $\Gamma$ be any closed curve in $\mathbb{C}\setminus B$. Since the arc $B$ joins $-i$ to $i$ without meeting $\Gamma$, those two points lie in the same component of $\mathbb{C}\setminus \Gamma$. Their winding numbers about $\Gamma$ are therefore equal. The residue theorem givesThus every closed-path period vanishes. Integrals from $0$ to $z$ along any two paths avoiding $B$ agree, so $F_B$ is single-valued. Notice that this argument uses cancellation of the two residues; simple connectedness of the slit domain is not required.

False without continuity. The function equal to $1$ at one point and $0$ elsewhere is nonnegative, Riemann integrable with integral $0$, but is not identically zero.

Differentiation under the integral gives $I\prime \prime ={\int }_0^1{y}^4{e}^{xy}dy$. Integrating the derivative of $e^{xy}(x{y}^3-3y^2)$ and then once more yields $x^2{\int }_0^1{y}^4{e}^{xy}dy-12{\int }_0^1{y}^2{e}^{xy}dy=e^x(x-4)$.

The transformed inhomogeneous equation is $z\prime \prime +z=x^2$. Taking $z=x^2-2$ gives the particular integral $y_p=1-2/x^2$.

Substitution gives $({\dot{X}}^2-c^2)\varphi \prime \prime -(\ddot{X}+\gamma \dot{X})\varphi \prime +{\omega }^2\varphi =0$. Multiply by $\varphi \prime$ and integrate over $\mathbb{R}$; decay kills the first and last integrals, leaving $(\ddot{X}+\gamma \dot{X})\int (\varphi \prime {)}^2=0$. Thus a nonconstant profile requires $X=X_{\infty }+C{e}^{-\gamma t}$ and approaches a fixed position.

The clockwise boundary integral at $z=0$ is
$-{\int }_0^{2\pi }{a}^4({\sin }^4\varphi +{\cos }^4\varphi )d\varphi =-3\pi a^4/2$. Direct integration of the curl over the side and top gives the same value (the $h$-dependent terms cancel), verifying Stokes.

This isotropic rank-three integral is $0$ by oddness under $x\mapsto -x$.

Divergence theorem says ${\int }_V\nabla \cdot FdV={\int }_{\partial V}F\cdot ndS$ for the outward normal. Here $\nabla \cdot (x{e}^{-r^2})=(3-2r^2)e^{-r^2}$, so radial integration gives $4\pi a^3{e}^{-a^2}$. On the sphere $F\cdot n=a{e}^{-a^2}$, whose integral is the same $4\pi a^3{e}^{-a^2}$.

On $C$, $dx=a{e}_{\varphi }d\varphi$, so $I={\int }_0^{2\pi }d\varphi =2\pi$. Locally $f=\varphi$, but no single-valued global potential exists on the punctured domain; this is why irrotationality does not force this closed integral to vanish.

The series for $S$ converges uniformly by the M-test. Therefore termwise integration is valid and every term integrates to $\sin (n^3\pi )/n^7=0$, so the integral is zero.

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.