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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.

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 $s\in \mathbb{C}$.