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For $\mathrm{Re}s>1$, expand $(e^t-1{)}^{-1}={\sum }_{n\ge 1}{e}^{-nt}$. Absolute integrability, bounded near zero by a constant times $t^{\mathrm{Re}s-2}$, permits termwise integration. The Gamma function integral then givesSpecify the Hankel contour to run from the negative real axis below the cut towards zero, circle zero counterclockwise with fixed radius $0<r<2\pi$, and return above the cut. Take $-\pi <\arg t<\pi$. The circle excludes every nonzero pole $2\pi i\mathbb{Z}$ of the denominator. For $\mathrm{Re}s>1$ its radius can shrink to zero, and the lower and upper rays give respectively $-e^{-i\pi s}\Gamma (s){\zeta }_R(s)$ and $e^{i\pi s}\Gamma (s){\zeta }_R(s)$. By the Gamma reflection formula,away from integer $s$ in that half-plane, with limiting equality at the removable points.

With the circle radius kept fixed, the contour integral is an entire function of $s$: the rays decay exponentially and converge locally uniformly, and the circular part stays away from zero. Multiplication by $\Gamma (1-s)$ is analytic for $\mathrm{Re}s<1$ and meromorphic globally. The apparent singularities at positive integers $s\ge 2$ are removable because the integral vanishes there and it already agrees with the original analytic zeta function nearby. Only $s=1$ remains a simple pole. Thus this constructs the meromorphic continuation to the whole plane and an analytic continuation on $\mathbb{C}\setminus \{1\}$. Keeping the circular segment is essential when the integral is not locally integrable at zero.

At $s=-1$ the two rays cancel, since $t^{-2}$ is single-valued. The Laurent expansionshows that the residue of $t^{-2}/(e^{-t}-1)$ is $-1/12$. Since $\Gamma (2)=1$, the residue theorem gives

For nonnegative insect density assume $N_0>0$. Substitution of $N=e^{\alpha t}R$ gives $R_t=D{e}^{2\alpha t}(R^2{R}_x{)}_x$, so chooseThen ${\tau }^{\prime }=D{e}^{2\alpha t}>0$ and the porous medium equation is $R_{\tau }=(R^2{R}_x{)}_x$. Its total mass is conserved when the flux vanishes at infinity. For the mass-preserving similarity solution, direct differentiation givesOn the positive support, integrating once with zero boundary flux gives $F^2{F}^{\prime }=-\xi F/4$ and hence $F^2=({\xi }_0^2-{\xi }^2)/4$. The semicircle mass integral is $\pi {\xi }_0^2/4=N_0$, confirming the profile and its symmetric compact support.

The front is at $|x|={\xi }_0{\tau }^{1/4}$. When $\alpha <0$, $\tau (t)$ increases towards $D/(2|\alpha |)$, so the supremum of the distance reached isIt is approached as $t\to \infty$, rather than attained at a finite time. If $N_0=0$, the nonnegative solution is identically zero and the corresponding distance is zero.

Assume the holonomic constraints have linearly independent gradients, so their common level set is locally a smooth constraint manifold parametrized by $q_1,\dots ,q_n$. The multiplier form of the Euler-Lagrange equations isAt fixed time ${\sum }_A{\partial }_A{f}_{\alpha }{\partial }_i{x}_A=0$. The chain rule for the reduced Lagrangian givesConversely, if these reduced equations hold, the Cartesian Euler-Lagrange residual annihilates every tangent direction. Linear independence of the constraint gradients implies that it is a linear combination of those gradients, giving the required Lagrange multipliers. This proves the local equivalence, including time-dependent constraints; singular constraints would need separate analysis.

For the bead, $\dot{y}=f^{\prime }(x)\dot{x}$, soThe autonomous conservation of energy gives $E=\frac{1}{2}(1+f^{\prime 2}){\dot{x}}^2$. If $E>0$, the motion has a fixed sign $s=\mathrm{sgn}\dot{x}$ andThe integral is the signed arc length divided by the constant speed, and is locally invertible. The special solution $E=0$ is a stationary bead, for which writing time as a function of its unchanging position is not meaningful.

Choose a branch of $t=1-z$ with $|\arg t|<\pi$ and write $d=c-a-b\notin \mathbb{Z}$. The Gauss hypergeometric equation isChanging to $t$ turns this into the same equation with third parameter $a+b+1-c=1-d$. Its solution regular at $t=0$ is $y_1=F(a,b,1-d;t)$. Substitution of $y=t^{d}v$ gives the Gauss hypergeometric equation for $v$ with parameters $c-a,c-b,1+d$. Thus $y_2=t^{d}F(c-a,c-b,1+d;t)$ is also a solution. Their distinct nonintegrally separated characteristic exponents at a regular singular point $0,d$ imply linear independence near $z=1$, soThis also follows from the exponent transformation in the Papperitz symbol.

Initially assume $\Re c>\Re b>0$, so the Euler integral for the hypergeometric function applies. If $\Re d>0$, taking $z\to 1$ gives an integral evaluated by the beta function and determinesIndeed $y_1\to 1$ and $y_2\to 0$ there. To find $B$, instead work in the nonempty parameter region $\Re d<0$, put $z=1-ϵ$ and scale the integration variable by $1-s=ϵu$. The singular contribution isThe defining integral of the beta function evaluates this integral to $\Gamma (c-b)\Gamma (-d)/\Gamma (a)$. Since ${ϵ}^{-d}{y}_1\to 0$ and ${ϵ}^{-d}{y}_2\to 1$, it determines $B$. Analytic continuation in the parameters then yields the hypergeometric connection formulaThe usual power-series definition requires $c\notin \{0,-1,-2,\dots \}$; exceptional parameters need a separately specified analytic limit. Reciprocal gamma functions make vanishing coefficients at denominator poles meaningful. The branch must remain fixed when continuing in $z$.

Let $\mathbf{r}$ be position relative to the center of mass, and let the new origin have vector $\mathbf{c}$ relative to it. The inertia tensor about this origin isExpanding and using $\int \mathbf{r}dm=0$ proves the tensor form of the parallel axis theorem:For a uniform cube of side $L$ and mass $M$, centered coordinate integrals give $I_0=(M{L}^2/6)I$. The displacement to a vertex is $(L/2,L/2,L/2)$ up to signs. Using axes along its incident edges givesThe principal moments of inertia are $\boxed{M{L}^2/6,11M{L}^2/12,11M{L}^2/12}$. The first principal axis is the body diagonal through the vertex; every direction perpendicular to it has the repeated moment.

By the argument principle, the number of zeros minus poles, counted with multiplicity, in the fundamental parallelogram is $(2\pi i{)}^{-1}\oint f^{\prime }/fdz$. The logarithmic derivative is periodic with the same lattice, so the integrals over opposite sides cancel. Thus the numbers of zeros and poles are equal.

The Weierstrass elliptic function gives a degree-two map $\wp :\mathbb{C}/\Lambda \to {\mathbb{P}}^1$ whose fibers identify $z$ and $-z$. An even elliptic function is constant on these fibers, so it descends to a single-valued function $Q$ on the sphere. This is meromorphic even at a branch value: near a half-period the involution is $t\mapsto -t$, the Laurent series of $f$ has only even powers, and $\wp -{\wp }_0$ has a double zero and provides a local coordinate in $t^2$. At the image of $0$, $\wp \sim t^{-2}$ gives the same conclusion at infinity. A meromorphic function on the Riemann sphere is a rational function, so $\boxed{f=Q(\wp )}$.

A meromorphic antiderivative need not be elliptic. Take $f=\wp$. Its double poles have zero residue, so integrating on the punctured plane gives a single-valued primitive, since every closed-contour integral is a sum of zero residues. The primitive extends meromorphically over each pole, with local principal part $-1/z$. If $F$ were elliptic, it would have precisely one pole modulo the lattice, with residue $-1$, contradicting the zero sum of residues in a fundamental parallelogram. This is a nonconstant counterexample satisfying the stated premise; the preceding descent proves that even elliptic functions are rational in the Weierstrass function.

For $f\in L^1({\mathbb{R}}^d)$, use the Fourier transform conventionThe integral is absolutely convergent for every $\xi$, and $|\hat{f}(\xi )|\le \Vert f{\Vert }_1$.

First ${\int }_{\mathbb{R}}{e}^{-x^2/2}dx=\sqrt{2\pi }$: square the positive integral and use polar coordinates to obtain $2\pi {\int }_0^{\infty }r{e}^{-r^2/2}dr=2\pi$. Scaling therefore normalizes the one-dimensional Gaussian function $g_t$.

Let $I(\xi )={\int }_{\mathbb{R}}(2\pi t{)}^{-1/2}{e}^{-x^2/(2t)}{e}^{-i\xi x}dx$. Differentiation under the integral is justified by the integrable majorant $|x|g_t(x)$. Since $g_t^{\prime }(x)=-x{g}_t(x)/t$, integration by parts, with zero endpoint terms, gives $I^{\prime }(\xi )=-t\xi I(\xi )$. With $I(0)=1$, this yields $I(\xi )=e^{-t{\xi }^2/2}$. Multiplication of the one-dimensional integrals givesApplying the same computation with reciprocal scale $1/t$ evaluates its inverse integral:For $f\in L^1$, Fubini's theorem gives $\widehat{f\ast g_t}=\hat{f}{e}^{-t|\xi {|}^2/2}$, which is integrable because $\hat{f}$ is bounded. Its inverse integral isAbsolute integrability of $|f(y)|e^{-t|\xi {|}^2/2}$ justifies this second application of Fubini's theorem. Thus Fourier inversion holds for every Gaussian convolution, in fact pointwise for this continuous representative.

Define $g$ by the inverse integral. Its absolute convergence gives $|g(x)|\le (2\pi {)}^{-d}\Vert \hat{f}{\Vert }_1$. If $x_n\to x$, the exponential factors converge pointwise and their differences are bounded by $2|\hat{f}|$. The dominated convergence theorem proves $g(x_n)\to g(x)$. Therefore $g$ is bounded and continuous and equals $f$ almost everywhere.

Assume the zero set $A=\{s\in [0,t]:g(s)=0\}$ is measurable, as is ensured by measurability of $g$. This is an implicit necessary hypothesis: for an arbitrary real function the printed integral and event need not be measurable. Put $a={\int }_0^t{1}_A(s)ds$. Conditional on $N_t=n$, the uniform order statistics description makes the probability of no jump in $A$ equal to $(1-a/t{)}^n$. Averaging over the Poisson distribution givesConsequentlyIn particular a measure-zero zero set is almost surely missed, despite possibly containing infinitely many points.

Keep $\nu$ fixed as $x\to +\infty$. On both vertical lines $\sin (\pm \pi +iy)=-i\sinh y$. The right side is traversed upward and the left downward, so their combined contribution isSet $v=\sinh y$. The integral becomes ${\int }_0^{\infty }{e}^{-xv}a(v)dv$, where $a(v)=e^{-\nu \mathrm{arsinh}v}/\sqrt{1+v^2}$, $a(0)=1$ and $a^{\prime }(0)=-\nu$. Successive integration by parts, with exponentially vanishing terms at infinity, givesHence the leading vertical contribution is $\boxed{V(x)=-\sin (\pi \nu )/(\pi x)+O(x^{-2})}$. It is exactly zero for integer $\nu$. These estimates hold for fixed real $\nu$; they are not uniform when the order grows with $x$.

Replace the sum by the integral against the density of states and set $x=ϵ/(kT)$, $y=\mu /T$:ThereforeFor the given $a>0$, the integral converges at zero and infinity for every finite $y$.