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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.

Approximate $h$ uniformly by finite sums $\sum f_j(x)g_j(y)$; their integral operators have finite-dimensional range and converge in operator norm, so $T_h$ is compact. If $h(x,y)=0$ for $x\le y$, iterated integration over an ordered simplex gives $\Vert T_h^n\Vert \le \Vert h{\Vert }_{\infty }^n/n!$. Thus $r(T_h)=0$, and compact spectral theory gives $\sigma (T_h)=\{0\}$.

Not for an arbitrary signed $L^1$ kernel. One can place very tall, very narrow spikes of $g$ at separated locations, with summable masses and total integral one, and matching still narrower spikes of an $L^1$ function accumulating at a Lebesgue point. Along a selected subsequence of dilations, a kernel spike lands on the matching function spike and contributes a fixed amount, while the total widths make the point a Lebesgue point. Thus compact support (or a suitable integrable radial majorant) cannot simply be dropped.

On disjoint events $A_n$ with $P(A_n)=c/n^2$, set $X_n=n{1}_{A_n}$. Then $X_n\to 0$ almost surely and $E{X}_n^2=c$, but any common dominator has integral at least ${\sum }_{n}nP(A_n)=c\sum 1/n=\infty$. Thus no integrable dominating variable exists.

With phase $\varphi (z)=(1+i)z-z^2/2$, the original ray is a contour integral. At the endpoint $z=-1$, the steepest direction solves $\mathrm{Im}(\varphi (z)-\varphi (-1))=0$ with decreasing real part; locally it is the direction for which ${\varphi }^{\prime }(-1)(z+1)$ is negative real.

Stirling and replacing the magnetization sum by an integral gives the stated form with $P(m)=-\beta (Bm+3J{m}^2+2K{m}^3)$, up to subexponential constants.

The second Adams operation $\chi \mapsto [g\mapsto \chi (g^2)]$ lies in the integral character ring, so $\tilde{\chi }=\sum n_i{\chi }_i$ with integers $n_i$. Norm one forces exactly one coefficient to be $\pm 1$; evaluation at the identity makes the sign positive. Squaring is bijective on conjugacy classes as well, so distinct irreducibles remain distinct. Thus the ${\tilde{\chi }}_i$ permute the irreducible characters.

A Poisson point process of intensity $\lambda (x)$ has independent counts on disjoint Borel sets, with $\Pi (A)\sim \mathrm{Poisson}({\int }_A\lambda (x)dx)$ whenever the integral is finite.

Multiplying two asymptotic eigenfunctions by $q$ reduces the integral, after the phase coordinate $s={\int }_a^x\sqrt{q}$, to a constant multiple of ${\int }_0^L\sin (n\pi s/L)\sin (m\pi s/L)ds=0$.

The identity theorem says analytic functions agreeing on a set with an interior accumulation point agree throughout a connected domain. Analytic continuation is an analytic function on a larger connected domain agreeing on a nonempty overlap. Factor the denominator as $(t+i)(t-z)$. For $\Im z<0$, contour closure below giveswith a removable value at $-i$; this entire expression is the continuation. For $\Im z>0$, the displayed real-axis integral encloses only $-i$ and equals $2\pi i{e}^{-5}/(z+i)$, so it is not that continuation. Finally $1/n\to 0$ only at a boundary point of $D$, so the zeros do not determine $G$; for example $G(z)=\sin (\pi /z)$ is nonzero and has them all.

Use $E=m{c}^2+p^2/(2m)+\cdots$ and replace the Fermi denominator by its Boltzmann form. The Gaussian integral gives

The continuous function $|h|$ attains a finite maximum $M$ on $[a,b]$ by the extreme value theorem. Additivity and the triangle inequality for the Riemann integral imply, for any $x,y\in [a,b]$,For reversed endpoints, reverse the orientation of the integral before taking its absolute value. Given $\epsilon >0$, choose $\delta =\epsilon /(M+1)$. Then $|x-y|<\delta$ implies $|H(x)-H(y)|<\epsilon$. Thus $H$ is continuous on $[a,b]$, including one-sided continuity at the endpoints; it is even Lipschitz continuous. If $a=b$, the function is defined on one point and the conclusion is immediate.

The Fourier cosine series on $[0,L]$ isIt is the Fourier series of the even extension to $[-L,L]$, repeated with period $2L$. The Fourier sine series isIt comes from the odd extension and its $2L$-periodic extension. These Fourier coefficient formulas follow by Fourier orthogonality, since every nonconstant sine or cosine has squared integral $L/2$ on $[0,L]$, while the constant function has squared integral $L$.

Set $u=x/L$ and $q=n\pi$. The required integrals follow fromConsequently the Fourier coefficients areThus the Fourier cosine series isand its sum is $e^{x/L}$ everywhere on $[0,L]$. The Fourier sine series isand its sum is $e^{x/L}$ on $(0,L)$ and zero at the two endpoints. The open and filled circles in the sketches distinguish these endpoint limits from the actual values of the sum.

For $\alpha =3$, let $F_n$ denote the time factor normalized by $F_n(0)=0$, $F_n^{\prime }(0)=1$. By superposition the Fourier sine series solution isAt $t=0$, its initial velocity is ${\sum }_{n\ge 1}{b}_n\sin (nx)={\sin }^{2}x$. Multiply by $\sin (mx)$, integrate on $[0,\pi ]$, and use the permitted interchange and Fourier orthogonality. This yields the requested integral expressionUnder $x\mapsto \pi -x$, ${\sin }^{2}x$ is unchanged and $\sin (n(\pi -x))=(-1{)}^{n+1}\sin (nx)$. Hence every even-index Fourier coefficient is zero, including $b_2$.

For odd $n$, use ${\sin }^{2}x=(1-\cos 2x)/2$ and the product-to-sum formula:ThusIn particular $b_1=8/(3\pi )$ and $b_3=-8/(15\pi )$. An explicit version of the solution of a differential equation isOnly the first, overdamped normal mode, the third, critically damped normal mode, and the odd underdamped normal modes of index at least five occur. No odd-index Fourier coefficient vanishes.