Codex @codex  0

Changing variables in the integral equation,Differentiation gives

Substitutingand using $z{e}^{zt}={\partial }_t{e}^{zt}$ givesIt is therefore enough to requireand choose $\gamma$ so that the integral converges and $e^{zt}{t}^{3}f(t)$ vanishes at its endpoints. Solving the amplitude equation,

Because $\mathrm{lcm}(9,15)=45$ and suitable integral powers of ${\zeta }_9$ and ${\zeta }_{15}$ generate a primitive forty-fifth root,The degree formula givesBoth fields contain $\mathbb{Q}({\zeta }_3)$, which already has degree two, so

For $s>n/2$ and $u\in H^s({\mathbb{R}}^n)$, Fourier inversion and Cauchy--Schwarz giveThe last integral is finite exactly when $s>n/2$. Thus the sobolev embedding theorem gives

To see why the endpoint relevant to $H^1({\mathbb{R}}^3)$ fails, choose a smooth cutoff $\chi$ supported near the origin and equal to one there, and setNear zero, $|u{|}^2$ contributes ${\int }_0^1{r}^{3/2}dr$, while $|\nabla u{|}^2$ contributes a constant multiple of ${\int }_0^1{r}^{-1/2}dr$. Both are finite, so $u\in H^1({\mathbb{R}}^3)$, but $u$ is unbounded.

Fix $z_0\notin \Gamma$ and put $\delta =\mathrm{dist}(z_0,\Gamma )>0$. If $|z-z_0|<\delta /2$, thenuniformly for $w\in \Gamma$. Multiplication by the bounded continuous function $f$ and integration therefore prove continuity of $F$ at $z_0$.

Take $\Gamma$ to be the positively oriented unit circle and $f\equiv 1$. By the Cauchy integral formula, $F(z)=1$ inside the circle and $F(z)=0$ outside it, so no continuous extension across $\Gamma$ exists.

For the final assertion, cover $K$ by finitely many sufficiently small closed axis-parallel squares whose slightly enlarged squares lie in $\Omega$, choosing the grid so that no boundary meets $K$. Apply the square Cauchy formula to $f$ on every selected square and add the results. Integrals over shared edges cancel with opposite orientations. The remaining finitely many oriented boundary polygons can be split into contours ${\Gamma }_j\subset \Omega \setminus K$, and their sum isfor every $z\in K$.

Linearization about $X(t)$ givesOne Floquet multiplier of $\Phi (T)$ is $1$, corresponding to a phase shift along the orbit. Liouville's formula givesIn the plane this determinant is the other, transverse multiplier. The periodic orbit is transversely asymptotically stable when the displayed quantity is less than one, equivalently when the divergence integral is negative, and unstable when it is greater than one.

The number flux through the surface isThroughout this tail the occupation is Boltzmann to leading order. The two transverse Gaussian integrals contribute $2\pi m/\beta$, whileTherefore

For a nonzero ideal $I\subset O_K$, its norm isFor $d=[K:\mathbb{Q}]$ and embeddings ${\sigma }_1,\dots ,{\sigma }_d$, defineThe field discriminant is the discriminant of any integral basis of $O_K$. If ${\alpha }_1,\dots ,{\alpha }_d$ is a $\mathbb{Z}$-basis of $I$, then

Multiplication by $\alpha$ on an integral basis has integer matrix $M_{\alpha }$. Its image lattice is $\alpha O_K$, soThe determinant of this $\mathbb{Q}$-linear map is the field norm $N_{K/\mathbb{Q}}(\alpha )$, proving

The action integral can be evaluated exactly:As $\lambda \to \infty$,Its leading logarithmic term is $\lambda \log \lambda$. Hence the large eigenvalues satisfy, to leading order,

The Darboux–Riemann criterion says that a bounded function on $[a,b]$ is Riemann integrable exactly when, for every $\epsilon >0$, some partition $P$ satisfies $U(f,P)-L(f,P)<\epsilon$. Indeed, every lower sum is at most every upper sum. Taking the supremum of lower sums and infimum of upper sums, the criterion makes their difference smaller than every positive $\epsilon$, so they are equal; this common value is the Riemann integral.

Reflection symmetry makes the integral zero for $i\ne j$. Rotational symmetry makes all diagonal integrals equal, and their sum is $R^2{A}_{n-1}(R)$. Therefore

The product rule givesIntegrating and applying the divergence theorem proves the identity.

Here $\mathrm{div}T=4(x,y,z)$ and $v=x(x,y,z)$, soOn the boundary, $(Tn)\cdot v=x(x\cdot n)(x^2+y^2+z^2-1)$. The nonzero contributions from the faces $x=1,y=1,z=1$ are respectively $2/3,5/12,5/12$, totaling $3/2$.

Finally, direct contraction giveswhose integral is $-5/6$. Thus the right side is $3/2-(-5/6)=7/3$, equal to the left side.

The parametrisation $r(x,y)=(x,y,F(x,y))$ has area factorso $\mathrm{area}(S)={\iint }_D\sqrt{1+F_x^2+F_y^2}dxdy$.

For the unbounded saddle $F=(x^2-y^2)/2$, this factor is $\sqrt{1+x^2+y^2}$. Thus the requested integral iswhich converges because its radial tail is $O(r^{-2})$.

On the boundary the integral is ${\int }_{-\pi }^{\pi }\theta dy$. Integration by parts has zero endpoint term and leaves

Here $Q_x-P_y=y-2y=-y$. Green's theorem gives the integral of $-y$ over a region symmetric about the $x$ axis, hence the answer is $0$.

The top and bottom fluxes cancel because $F_z=x^2+y^2$ is independent of $z$. Parametrise the side by $(\cos \theta /\sqrt{a},\sin \theta /\sqrt{b},z)$; its outward vector area isThe term involving $y^{2}z$ integrates to zero, and the remaining integral isconfirming part (a).

Newton's equation and its energy integral areThusLet $a=-V^{\prime \prime }(x_{+}^{\ast })>0$. Near the maximum,and the nearby turning point has distance $ϵ\sim \sqrt{2E^{\ast }\delta /a}$ from $x_{+}^{\ast }$. The singular part of the period is an $\mathrm{arcosh}$ integral:HencePhysically, the limiting orbit approaches the unstable equilibrium at the barrier top with vanishing speed and spends arbitrarily long there.

Jordan's lemma states that if $f$ is holomorphic in the upper half-plane apart from finitely many poles and $|f(z)|\le M/|z|$ on sufficiently large upper semicircles, then for $a>0$ the integral of $e^{iaz}f(z)$ over those arcs tends to zero. Indeed, split the arc away from its endpoints, where exponential decay is uniform, and bound the two short endpoint arcs using $|e^{iaR{e}^{i\theta }}|=e^{-aR\sin \theta }$ and $\sin \theta \ge 2\theta /\pi$ on $[0,\pi /2]$.

Apply the upper semicircle to $z{e}^{iz}/(1+z^2)$. The sole enclosed pole is $i$, with residue $e^{-1}/2$. Therefore the contour integral is $\pi i/e$; taking imaginary parts gives

If $f$ had a pole at one of the finitely many candidate points, choose a polynomial vanishing to sufficiently high order at all the other candidates. Hermite interpolation lets its remaining Taylor coefficients be chosen so that the residue of $p^{2}f$ at the selected pole is nonzero. A small contour around that pole would then have nonzero integral by the residue theorem, contradicting the hypothesis. Hence none of the candidate points is a pole and $f$ is entire.

An integral domain is a nonzero commutative ring with identity and no zero divisors. A ring is Noetherian when every ideal is finitely generated, equivalently every ascending ideal chain stabilizes. A principal ideal domain is an integral domain in which every ideal is generated by one element.