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Equality of constant terms is preserved by componentwise addition, subtraction, and multiplication, and $(1,1)$ belongs to the set, so $R$ is a subring. The projection is surjective andThe nonzero elements $(x,0)$ and $(0,y)$ have product zero, so $R$ is not an integral domain and hence not a PID. In fact $R\cong \mathbb{C}[x,y]/(xy)$, which also shows that it is Noetherian.

A euclidean domain is an integral domain with a function $d$ on nonzero elements such that for $a,b\ne 0$ there are $q,r$ with $a=bq+r$ and either $r=0$ or $d(r)<d(b)$.

Write the rational root in lowest terms as $a/b$. The factor $bx-a$ is primitive and divides $f$ over $\mathbb{Q}[x]$; Gauss's lemma makes it divide in $\mathbb{Z}[x]$. Comparison of leading coefficients makes $b$ divide the leading coefficient one, so $b=\pm 1$ and the root is integral.

For a closed surface, $\chi =V-E+F$ for any finite cell decomposition. Gauss–Bonnet for a geodesic polygon saysand for a closed smooth surface it says ${\int }_{S}KdA=2\pi \chi (S)$.

For a large radius, the graph surface bounded over that circle is flat near its boundary. The boundary geodesic-curvature integral is $2\pi$, so Gauss–Bonnet on the disc gives $\int K=0$. Since $K\ge 0$, continuity forces $K\equiv 0$.

For the torus,Thus $T_{+}$ has $\cos u>0$ and $T_{-}$ has $\cos u<0$. Cap the two boundary circles of the outer half $T_{+}$ by flat discs. The result is a sphere, and the caps contribute no Gaussian curvature, so Gauss–Bonnet gives ${\int }_{T_{+}}K=4\pi$. The full torus has Euler characteristic zero, hence total curvature zero and

Sincethe divergence theorem proves Green's second identity.

If $x_0=(x_0,y_0,z_0)$ and $x_0^{\ast }=(x_0,y_0,-z_0)$, the image solution isIt vanishes on the plane and has the required unit delta source.

Green's identity yields the Poisson-kernel integralOn the axis this becomes

If the monic orthogonal polynomial $p_n$ had fewer than $n$ distinct sign-changing zeros in $(-1,1)$, multiply the factors corresponding to those zeros to form a polynomial $q$ of degree below $n$. Then $p_{n}q$ has one sign and is not identically zero, so its positive-weight integral cannot vanish, contradicting orthogonality. Thus all $n$ zeros are distinct and lie in the open interval.

The Euler–Lagrange equations areFor $\omega \ne 0$ their general solution iswhile for $\omega =0$, $x=c_{1}t+c_2$.

The action has continuous translation symmetry in $x$; it is also invariant up to endpoint terms under adding homogeneous Jacobi solutions $at+b$ to $x$ and $a\cos t+b\sin t$ to $y$. For $\omega =0$ it additionally has continuous time-translation symmetry; for nonzero $\omega$ only the corresponding discrete period remains.

A complete set of four independent first integrals isIn particular $({\dot{y}}^2+y^2)/2=(C_3^2+C_4^2)/2$ is conserved; when $\omega =0$, the usual total energy is conserved as well.

The Heaviside factor restricts the integral to $t\ge \alpha$. With $u=t-\alpha$,

In a parallel-plate rheometer, a point at radius $r$ on the upper plate moves at speed $\Omega r$. The thin-gap approximation therefore gives the local shear rateDefine the rim valueAn annulus of radius $r$ and width $dr$ has area $2\pi rdr$; its tangential force is $\tau (r)2\pi rdr$, and its moment arm is $r$. The measured torque is consequently

For a generalized Newtonian fluid, $\tau =\eta (\dot{\gamma })\dot{\gamma }$. Changing the integration variable from $r$ to $\dot{\gamma }$ givesThus the requested kernel isMultiplying by ${\dot{\gamma }}_R^3$ and taking a derivative with respect to ${\dot{\gamma }}_R$ turns the upper-limit contribution of the integral into the rim viscosity:ThereforeA measured torque curve and its slope hence determine $\eta$ over the range of imposed rim shear rates.

The experimental law impliesSubstitution givesThe graph is a decreasing rectangular hyperbola with a positive high-rate plateau $2/(\pi \alpha R^3)$ and a $1/\dot{\gamma }$ divergence at the origin. Equivalently,so the inferred material is a Bingham plastic with yield stress $3T_c/(2\pi R^3)$.

Use the Hankel contour about the negative real axis: it starts at $-\infty$ below the cut, circles the origin counterclockwise, and returns to $-\infty$ above the cut. Take $-\pi <\arg t<\pi$. For $\mathrm{Re}s>0$, the small circle vanishes and the two banks giveThe gamma reflection formula then gives

Near zero,Subtracting any desired number of these terms makes the local contour integral converge in successively larger left half-planes. The resulting apparent singularities are removable after multiplication by $\Gamma (1-s)$, so the Hankel formula analytically continues $\beta$ to all $s\in \mathbb{C}$.

The nonzero poles areTheir paired residues sum toThe paired series converges for $\mathrm{Re}s<1$ by the alternating-series test, and absolutely for $\mathrm{Re}s<0$. ThereforeUsing the stated contour closure, the Hankel integral is $-2\pi i$ times this sum. Substitution into part (a) givesNow $1/\Gamma (1-s)=\Gamma (s)\sin (\pi s)/\pi$ and $\sin (\pi s)=2\sin (\pi s/2)\cos (\pi s/2)$, soIt was derived on a nonempty half-plane. Both sides have analytic continuations, with the apparent gamma poles cancelled by the sine factor or the trivial zeros from part (b); the identity theorem therefore extends it to every $s\in \mathbb{C}$.

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 Schrödinger 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.