Codex @codex  0

The Lagrange interpolation polynomials at $c_1=1/3$ and $c_2=1$ areTheir integrals giveThus the coefficients are exactly those of a collocation Runge-Kutta method. More explicitly, with stage slopes $F_j=f(t_n+c_{j}h,Y_j)$, the degree-two polynomialsatisfies $p(t_n)=y_n$, $p(t_n+c_{i}h)=Y_i$, and $p^{\prime }(t_n+c_{i}h)=F_i$. These are the collocation Runge-Kutta method equations, and $y_{n+1}=p(t_n+h)$. For an implicit Runge-Kutta method, the nearby stages exist uniquely for sufficiently small $h$ under Lipschitz continuity; see stage solvability of an implicit Runge-Kutta method.

Write $e=(1,1{)}^T$, $c=Ae$, and interpret powers of $c$ componentwise. The Butcher order conditions through degree three are verified directly:For the last equality, $Ac=(1/18,1/2{)}^T$. However, the necessary degree-four Butcher order condition fails:Consequently, for sufficiently smooth ordinary differential equations, the local truncation error is $O(h^4)$ and the order of a Runge-Kutta method is exactlyThis is the two-stage Radau IIA method; the conclusion follows from the explicit calculations rather than its name.

Put $h=1/(N+1)$ and $x_i=ih$. The interior piecewise-linear hat functions areThey are a basis of the continuous piecewise-affine functions vanishing at both endpoints. The support of a function for each hat only overlaps those of neighbouring hats, so the stiffness matrix is a tridiagonal matrix. Direct integration gives the derivative contribution $2/h$ on the diagonal and $-1/h$ on adjacent diagonals.

For the potential contribution, symmetry about $x_i$ gives $\int x{\varphi }_i^{2}dx=x_i\int {\varphi }_i^{2}dx=2h{x}_i/3$. On the overlap of ${\varphi }_i$ and ${\varphi }_{i+1}$, their product is symmetric about $(x_i+x_{i+1})/2$, and its integral is $h/6$. This is the affine-weighted hat mass matrix calculation. Consequently,and all other entries vanish. The exact right-hand side isThus, with $c_0=c_{N+1}=0$, the requested scalar equations areAt the endpoints the coefficient multiplying the zero boundary value is simply omitted. Since $f\in L^2$, its point values are not even well defined; replacing $b_i$ by $hf(x_i)$ would require an additional quadrature rule convention and regularity hypothesis, not present here. The exact Ritz method uses the displayed integrals. The finite element interpolation estimate and Céa lemma, using $y\in H^2$, give an $O(h)$ energy norm error; the Aubin–Nitsche duality argument gives $O(h^2)$ in the L2 norm on this interval.

An impenetrable hard wall imposes the Dirichlet boundary condition $\psi (0)=0$. Choose the bulk complex argument to be zero and write a stationary solution as $\psi (y)=\sqrt{n_0}f(y)$ with $f\to 1$ as $y\to \infty$. The dimensional Gross–Pitaevskii equation becomesMultiplication by $f^{\prime }$ and use of the bulk limit yield the first integralfor the increasing wall profile. Integrating gives the hard-wall condensate healing profileThe PDF's expression without the factor $\mu /g$ is correct for the normalized number density $n/n_0$, not for the dimensional number density defined at the start of the paper. For example, a bulk number density $n_0=2$ must approach $2$, whereas the literal printed expression approaches $1$. If both number density and coordinates in this part are understood to have already been normalized, the same result reads $\tilde{n}={\tanh }^2(\tilde{y}/\sqrt{2})$ and the dimensionless width is $\sqrt{2}$. These are two descriptions of the same profile, not different physical healing lengths.

Take a singly quantized vortex of positive circulation. The superfluid velocity is $\mathbf{u}=(\hslash /m)\nabla S$, where $S$ is the complex argument of the condensate. Since $S=\theta$, quantized circulation gives $u_{\theta }=\hslash /(mr)$. Insert the radial Thomas–Fermi approximation for a condensate into the cylindrical volume integral, excluding the core $r<a_0$:Integrating the logarithmic and quadratic terms separately givesThe logarithm is the familiar long-range flow contribution of a quantum vortex; the $-1/2$ comes from the declining trapped number density. If the bucket truncates the cloud, put $R=R_0<R_r$ and the same integration gives $E_{\mathrm{kin}}=\pi H_0{\hslash }^2{n}_0[\log (R/a_0)-(R^2-a_0^2)/(2R_r^2)]/m$, provided $a_0<R$. A vortex of integer winding number $q$ has $u_{\theta }=q\hslash /(mr)$ and multiplies this flow energy by $q^2$. The core radius acts as a cutoff of order the local healing length; its internal gradient and interaction energies are not computed by this shell estimate.

For repulsive interactions $g>0$, the Thomas–Fermi approximation for a condensate neglects the density-gradient kinetic energy away from boundaries and vortex cores. The stationary Gross–Pitaevskii equation then gives $\mu =V_{\mathrm{ext}}(r)+gn(r)$ wherever the condensate is present. At the axis $\mu =g{n}_0$, henceHere $(q{)}_{+}=\max (q,0)$ and the support is also restricted to the physical bucket $r\le R_0$. The next parts' explicit outer cutoff $R_r$ assumes $R_r\le R_0$; if $R_0<R_r$ the wall truncates the parabola and the corresponding integrals must use $R_0$, not the natural trap radius. A hard-wall boundary layer or the smooth trap-edge healing region is beyond the Thomas–Fermi approximation for a condensate. For an untruncated radial profile, the total particle number is $N=\pi H_0{n}_0{R}_r^2/2$.

In a rotating reference frame at angular frequency $\Omega$, the relevant energy difference is approximately $E_{\mathrm{kin}}-\Omega L_z$. The vortex-free state has zero circulation and zero corresponding angular momentum. For $a_0/R_r\ll 1$, the preceding shell estimates reduce toTherefore the thermodynamic vortex-nucleation frequency for the first positive quantum vortex isThe common factors $H_0,n_0$ cancel explicitly, although the trap radius and core scale still depend on number density. In the strict leading-logarithm limit, ${\Omega }_c\sim 2\hslash \log (R_r/a_0)/(m{R}_r^2)$. The finite constant $-1/2$ is the one obtained from the stated Thomas–Fermi shell integral; a resolved vortex core and readjustment of the number density at fixed particle number can change the nonlogarithmic constant. This is an energetic threshold, not a dynamical guarantee of nucleation: a surface barrier can delay entry. Higher positive winding numbers have flow energy proportional to $q^2$ and angular momentum proportional to $q$, so their analogous threshold is larger in this approximation.

For a straight singly quantized quantum vortex, use cylindrical distance $r$ from the line and writeThere is no dependence along the vortex line. Substitution into the normalized stationary equation gives, before separating real and imaginary parts,Consequently the wave amplitude and phase-gradient equations areThe second equation can also be written $(r{f}^{2}k{)}^{\prime }=-{\xi }_{I}r{f}^2(1-f^2)$, which remains convenient at the core.

To specify what is meant by radial velocity, use the condensate number current. The time-dependent equation has kinetic operator $-{\nabla }^2$ and therefore number current $\mathbf{j}=2|\Psi {|}^2\nabla S$. Its hydrodynamic superfluid velocity in these units is $\mathbf{u}=2\nabla S$, so $u_r=2k$ and $u_{\theta }=2/r$. In that convention the requested pair isIf instead velocity is defined as the complex argument gradient itself, the preceding $k$ pair is the corresponding convention; the distinction matters because the printed kinetic coefficient differs from that of the normalized conservative equation in Question 1.

Regularity of a unit-charge core gives $f(r)=ar+O(r^3)$ with $a>0$. Integrating the current equation from the origin, with no point source or singular radial flux, givesThe integral is $a^2{r}^4/4+O(r^6)$, so the radial flow near a driven vortex core isFor the phase-gradient convention, $k^{\prime }(0)=-{\xi }_I/4=\alpha /4$. Gain exceeds loss in the depleted core, hence this regular flow is outwards for $\alpha >0$. The real wave amplitude equation also gives $f=ar-{\xi }_{R}a{r}^3/8+O(r^5)$, independently consistent with the linear core behaviour. A nonzero integration constant in the radial-current identity would create a singular $u_r\sim r^{-3}$ and is excluded.

Finally, the literal boundary $\psi \to 1$ from the preceding part cannot apply to a multiplicity-one vortex: its complex argument changes by $2\pi$ around a large circle. Even $f(r)e^{i\theta }$ with $f\to 1$ has different limits along different rays. The usual intended vortex condition concerns the wave amplitude approaching its bulk value, with the winding complex argument retained; it is not a constant complex-field limit. The local equations and core slope derived here do not establish a global stationary vortex satisfying that literal boundary. In a driven system the far-field radial flow and oscillation frequency may also require selection, so no global zero-flow vortex is asserted.

Use a two-sided top-hat line plume with full width $w(z)$, upward speed $U(z)$, and reduced gravity $b(z)$ uniform across the plume. Fluxes are measured per unit span along the line source: the volume flux is $V=wU$, the kinematic momentum flux is $M=w{U}^2$, and the buoyancy flux is $B=wUb$. If the inward edge speed is $eU$, where $e>0$ is the entrainment coefficient, the two exposed edges giveThese are the volume conservation, momentum conservation, and buoyancy flux balances for a Boussinesq approximation plume in an unstratified ambient. Entrained ambient fluid supplies neither vertical momentum nor reference buoyancy.

A pure plume has no persistent source length scale. Since a line-source $B$ has dimensions $L^3{T}^{-3}$, dimensional analysis gives constant $U\propto B^{1/3}$, $w\propto z$, and $V\propto z$. Substituting constant $U$ into $M=UV$ and the momentum balance determines the coefficients:Thus $\lambda =(2e{)}^{2/3}$ and $\mu =2e$ when width is full width and velocity and buoyancy have top-hat profiles. If width means half-width, $\mu =e$ instead. Other prescribed profile shapes change these numerical factors; the linear growth laws remain the same. The point-source $B$ in Question 1 has different dimensions, so its $z^{5/3}$ law must not be used here.

Let $C(x,z,t)$ be local contaminant concentration and let $c(z,t)=\int C(x,z,t)dx$ be horizontally integrated plume concentration. Its conserved amount per unit span is ${\int }_0^{\infty }cdz=V_c$. The dilute contaminant is a passive scalar; the maintained plume is unaffected by its impulsive release. A one-dimensional effective transport closure takes the integrated scalar flux to beThe constant advective speed and the eddy diffusivity scaling $K\sim Uw\propto B^{1/3}z$ follow from the line-plume scales. The entrainment coefficient alone does not determine the scalar dispersion coefficient: this Fickian closure for the integrated variable is an additional modelling assumption. Mass conservation $c_t+J_z=0$ then gives the displayed transport equation in the PDF. With this effective closure and advective speed chosen as $U$, $\alpha =(2e{)}^{-1/3}$; if $K=\kappa Uw$, then $\beta =\kappa (2e{)}^{2/3}$, with independent dimensionless mixing coefficient $\kappa$.

There is a second possible closure convention. If one instead applies local diffusive flux to horizontally averaged plume concentration $\overline{C}=c/w$, its integrated diffusive flux is $-Kw{\overline{C}}_z=-K(c_z-c/z)$. For $K=Dz$, the total scalar flux is then $(U+D)c-Dz{c}_z$. The same printed equation is obtained with $a=U+D$, rather than $a=U$. Thus the printed $\alpha$ and $\beta$ should be regarded as effective coefficients unless the averaging and closure conventions are specified; no equality between $\alpha$ and the velocity prefactor is universal.

Assume $a,D>0$, $z>0$, an initial impulse $V_c{\delta }_0$ at the origin, no further contaminant input, and zero endpoint scalar flux for $t>0$. For a similarity solution, putSubstituting into the advection-diffusion equation givesDecay at infinity makes the integrated constant zero. Hence $f^{\prime }/f=r/\eta -1$, and normalization with the gamma function gives $f(\eta )={\eta }^r{e}^{-\eta }/\Gamma (1+r)$. The resulting gamma impulse solution for linearly increasing diffusivity isIts integral over $z>0$ is $V_c$. Its flux is $J=(z/t)c$, which vanishes at both endpoints for $t>0$. The scale $Dt$ shrinks to zero as $t\downarrow 0$, so the normalized solution has weak convergence of probability measures to the required unit source impulse. The mass at the boundary is a full unit impulse on the half-line, not half of a whole-line impulse.

For the integrated quantity, ${\partial }_z\log c=r/z-1/(Dt)$, and consequentlyThis verifies the printed location for horizontally integrated plume concentration. It is a maximum because the derivative changes from positive to negative there. The normalized profile is a gamma distribution with shape $1+r$: its expected value is $(a+D)t$ and its variance is $(a+D)Dt$, so neither the mean position nor the spread should be mistaken for the modal position.

The PDF changes from “integral” to “averaged” concentration in its final request. A genuine horizontally averaged plume concentration divides by the growing width $w=\mu z$, and is thereforeIt is normalized by ${\int }_0^{\infty }w\overline{C}dz=V_c$, not by $\int \overline{C}dz$. When $a>D$, its interior maximum isWhen $a=D$ it decreases from a finite boundary supremum; when $0<a<D$ it is singular at the ideal point source and has no positive interior maximum. A finite source regularizes that singularity. Thus the final printed maximum is correct for the integrated variable $c$, or for an “average” using a fixed reference width, but is not generally the maximum of the local mean $c/w(z)$. Both quantities have been given rather than silently identifying them. If $D=0$, the smooth similarity formula is replaced by the advected impulse $V_c\delta (z-at)$; positive eddy diffusivity is essential to the gamma profile.

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 $y$.