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Use the same specified homogeneous Newtonian background and Jeans swindle, but describe stars by a mass-normalized galactic distribution function. The Collisionless Boltzmann equation is ${\partial }_{t}f+\mathbf{v}\cdot {\nabla }_{x}f-\nabla \varphi \cdot {\nabla }_{v}f=0$. Let $f=f_0(\mathbf{v})+f_1{e}^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ and similarly perturb the gravitational potential. To first order,Eliminating $f_1$ and the nonzero gravitational potential amplitude gives the collisionless Jeans dispersion relationFor growing modes $\mathrm{Im}\omega >0$ there is no real-velocity pole. Real and damped frequencies require a causal contour prescription or analytic continuation; one must not simply integrate through a pole without specifying it.

Take $b>0$ in the Cauchy velocity distribution. Its normalization is $\int f_0{d}^{3}v={\rho }_0$. Orient $\mathbf{k}$ along the $u=v_z$ axis and integrate over the perpendicular velocities:At the marginal mode $\omega =0$, $F^{\prime }(u)/u=-2{\rho }_{0}b/[\pi (u^2+b^2{)}^2]$, with its finite limiting value at zero. The provided Gamma function integral at power two givesConsequentlyTo check that this really separates growing modes, put $c=\omega /k$ in the upper half-plane. Residue integration, or integration by parts followed by the Cauchy resolvent, gives $\int F^{\prime }(u)/(u-c)du={\rho }_0/(c+ib{)}^2$. The upper-half-plane solution has $\omega =i(\sqrt{4\pi G{\rho }_0}-kb)$, which grows exactly for $0<k<k_J$. At the threshold it approaches the marginal mode continuously. Although $b$ plays the threshold role of a velocity scale, this distribution's second moment diverges: the radial integrand for $\int v^2{f}_0{d}^{3}v$ tends to a nonzero constant at large speed. It is therefore incorrect to identify $b^2$ with a finite velocity dispersion or to infer this result by substituting an rms speed into a fluid formula.

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

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 $z$.

The Mellin transform iswhere this integral converges. Rapid decay at infinity makes the integral over $[1,\infty )$ an entire function of $s$, but alone gives no control near zero. With the additional expansion, the integral initially converges absolutely for $\mathrm{Re}s>-{\sigma }_1$.

For $\mathrm{Re}s>1$, expand $(e^{\pi y}-1{)}^{-1}={\sum }_{r\ge 1}{e}^{-\pi ry}$. Absolute convergence justifies termwise integration, using the Gamma integral, and gives the scaled Bose integral

There is a missing hypothesis in the general continuation claim: one needs ${\sigma }_j\to +\infty$. The original PDF, like the TeX, only says that the sequence increases. Under the intended additional hypothesis, split the integral at one and subtract $r$ terms of the asymptotic expansion:Continuity of $g_r$ at zero makes it bounded on $[0,1]$. The last integral is holomorphic on $\mathrm{Re}s>-{\sigma }_{r+1}$: on every compact subset, its integrand and all its $s$ derivatives are dominated by an integrable power of $y$ times a power of $|\log y|$. The expressions for successive $r$ agree on their common initial domain and hence everywhere they overlap by the identity theorem. As ${\sigma }_{r+1}\to \infty$, these half-planes cover $\mathbb{C}$. Taking $r\ge j$ with ${\sigma }_{r+1}>{\sigma }_j$ isolates the term $c_j/(s+{\sigma }_j)$, while all other terms are holomorphic near $-{\sigma }_j$. Thus the corrected claim isThis is meromorphic continuation of a Mellin transform from an asymptotic expansion.

To show why the correction matters, set ${\sigma }_j=1-1/j$, $c_j=2^{-j}$, andwhere $\eta$ is continuous, equals one for $0<y\le 1$ and vanishes for $y\ge 2$. On $[0,1]$, the remainder after division by $y^{{\sigma }_{r+1}}$ is a uniformly convergent series of nonnegative powers of $y$, so it extends continuously to zero with value $c_{r+1}$. Away from zero, define $g_r$ by the required remainder quotient; it is continuous on the rest of $[0,\infty )$ as well. Thus all the printed hypotheses hold. Its Mellin transform equalsThe series continues meromorphically on $\mathrm{Re}s>-1$ and has genuine poles at $-1+1/j$, accumulating at $-1$. A meromorphic function on $\mathbb{C}$ cannot have such an accumulation of poles. This is the accumulating asymptotic exponents obstruct Mellin continuation counterexample.

For the Gamma function, the Gamma function recurrence givesAt $s=-j$ the numerator is $\Gamma (1)=1$ and the product of the nonzero denominator factors is $(-1{)}^{j}j!$. Therefore the residues of the Gamma function are

The generating function of the Bernoulli numbers gives the convergent expansion near zeroThe subtraction proof above applies to this expansion, including its zero coefficients, which give no pole. At $s=1-n$, for $n\ge 1$, the residue of its transform is $B_n{\pi }^{n-1}/n!$, while the residue of ${\pi }^{-s}\Gamma (s)$ is ${\pi }^{n-1}(-1{)}^{n-1}/(n-1)!$. Their quotient defines a holomorphic continuation of the Riemann zeta function at this point, including when the first residue is zero. Dividing gives the Bernoulli formula for zeta values at nonpositive integers:In particular the generating-series convention is $B_1=-1/2$, so the formula includes $\zeta (0)=-1/2$.

Put $\Delta B_h=B_{t+h}-B_t$ and $R_h={\int }_t^{t+h}(H_s-H_t)d{B}_s$. The constant-in-time coefficient $H_t$ on $(t,t+h]$ is previsible, so the ratio minus $H_t$ is $R_h/\Delta B_h$. Since $|H|\le C$, continuity and the dominated convergence theorem implyThe Itô isometry gives $\mathbb{E}{R}_h^2\le h{\eta }_h$. For $\epsilon ,\delta >0$, use the Chebyshev inequality and the normal distribution of $\Delta B_h/\sqrt{h}$ to obtainFirst let $h\downarrow 0$, then $\delta \downarrow 0$; the normal distribution has no atom at zero. This provesNo independence between $R_h$ and the denominator is needed.

The first estimate suggested by the hint is valid: Brownian scaling and the Gamma function give $\mathbb{E}|\Delta B_h{|}^{-1/2}=h^{-1/4}{2}^{-1/4}\Gamma (1/4)/\sqrt{\pi }$. The second hint requires care about the placement of its expectation: with $V_h={\int }_t^{t+h}(H_s-H_t{)}^{2}ds$, the interpretation $\mathbb{E}|R_h{|}^{1/2}\le (\mathbb{E}{V}_h{)}^{1/4}$ is correct by the Jensen inequality and the Itô isometry, whereas the interpretation $\mathbb{E}|R_h{|}^{1/2}\le \mathbb{E}(V_h^{1/4})$ is false with constant $1$. For the latter interpretation, the Burkholder-Davis-Gundy inequality supplies a constant $C_{1/2}$ instead. One can see the obstruction with a Brownian motion stopped on exiting $(-1,1)$: if $\tau$ is that Brownian exit time, then $|B_{\tau }|=1$, $\mathbb{E}\tau =1$, and strict Jensen inequality gives $\mathbb{E}{\tau }^{1/4}<1=\mathbb{E}|B_{\tau }{|}^{1/2}$.

This obstruction also survives the continuity condition on $H$. On a long finite interval approximate $1_{\{s<\tau \}}$ by a continuous adapted process equal to $1$ until $\tau$ and decreasing linearly to $0$ during the next $ϵ$ units of time, and take its initial value to be $0$ by inserting a short continuous ramp before starting the exit experiment. The Itô isometry controls these approximations, and letting the interval length tend to infinity recovers the strict inequality above; hence some bounded continuous adapted example also violates the latter interpretation with constant $1$. The direct proof avoids relying on this ambiguous hint.

For $1\le k\le n$, the nearest neighbour distance ${\rho }_{(k)}(x)$ is the $k$th order statistic of $D_i=|X_i-x|$. Each $D_i$ has continuous distribution function $p_x(r)$ on $[0,\infty )$. The probability integral transform makes $U_i=p_x(D_i)$ independent and identically distributed random variables with the uniform distribution on $(0,1)$. Monotonicity of $p_x$ gives $P=U_{(k)}$, a uniform order statistic. Strict monotonicity is not needed at this stage.

For $0<s<1$, counting observations below $s$ givesDifferentiating and cancelling consecutive terms gives the Beta distribution:The Gamma function identity $\Gamma (j)=(j-1)!$ identifies this with the stated normalization. Ratios of Beta function integrals yield the momentsThis is the probability content of a nearest neighbour ball identity, specialized to intervals on the real line.

Now write $f_0=f(x)>0$. The Lipschitz bound givesThusStrict positivity of the probability density function makes $p_x$ continuous and strictly increasing from zero to one, so its inverse function is defined for every $s\in (0,1)$.

For the requested conditional calculation, suppose $f_0^2\ge L$. At $r_0=s/f_0$ the preceding bound givesTherefore $r=p_x^{-1}(s)\le s/f_0$, and substituting in the approximation proves the inverse probability content bound for a Lipschitz density:The same argument works, without $f_0^2\ge L$, on the meaningful local range $0<s<1$ with $s\le f_0^2/L$.

Since the nearest neighbour density estimator obeys $f(x)/{\hat{f}}_{(k)}(x)=2(n+1)f_0{p}_x^{-1}(P)/k$, its conditional bound follows from the Beta distribution moments:Hence, under the printed location condition,The conditional inverse function bound also bounds $p_x^{-1}(P)$, so the expected value in this calculation is finite under those formal hypotheses.

There is, however, a genuine flaw in the original PDF: the printed condition $f(x)\ge \sqrt{L}$ has no admissible points when $f$ is strictly positive on all of $\mathbb{R}$. A probability density function on all of $\mathbb{R}$ cannot have Lipschitz constant zero, so $L>0$. Its Lipschitz bound forcesIntegrating this triangular lower envelope gives the Lipschitz density height boundThe inequality is strict because the probability density function has positive mass outside the finite interval. Thus $f(x)<\sqrt{L}$ everywhere. The last two conditional conclusions above are valid implications but vacuous as printed. Dropping strict positivity gives only $f(x)\le \sqrt{L}$, with equality forcing the entire probability density function to be the normalized triangular envelope. The local inverse bound does not justify replacing the printed condition by its reverse in the final expected value bound, since $P$ ranges over all of $(0,1)$.

Start from the product of two Gamma function integrals:Use the sum-and-ratio substitution for gamma integralswhose Jacobian is $r$. Absolute convergence for $\mathrm{Re}p,\mathrm{Re}q>0$ justifies the change of variables and separates the integral:Therefore

At $a=1$, the endpoint is degenerate: $S(t)=\sinh t-t=t^3/6+O(t^5)$. The contributing width is therefore $t=O({\nu }^{-1/3})$, rather than the width in part (i). With $v=\nu t^3/6$, the formula for a degenerate endpoint in Laplace's method givesThe coefficient is a Gamma function value obtained by the Gamma integral.

The functional equation isEquivalently,

For a proof, letThe Poisson summation formula applied to a Gaussian function gives the theta transformationThe standard Gamma function integral and termwise integration initially give, for $\Re s>1$,Split the integral at one, substitute $u\mapsto 1/u$ in the lower half, and use the theta transformation. The result isThe integral is an entire function of $s$ because $\Theta (u)-1$ decays exponentially. The right side is visibly invariant under $s\mapsto 1-s$, proving both the analytic continuation and the functional equation of the Riemann zeta function. This is the Mellin representation of the completed Riemann zeta function.

For real $s>0$, collapse the Hankel contour onto the negative axis and use the jump in the argument of $t^{s-1}$. The reflection formula for the gamma function yieldsso $K(z,s,t)=z{t}^{s-1}/[\Gamma (s)(e^t-z)]$. At positive integer $s$, the original Hankel expression is interpreted by analytic continuation: the pole of $\Gamma (1-s)$ cancels the corresponding zero of the contour integral.

When $z$ crosses the slit at $x>1$, the pole of the real-integral kernel at $t=\log x$ crosses the integration path. The two boundary values differ by $2\pi i$ times its residue. Since $d(e^t-z)/dt=x$ there, that residue is $(\log x{)}^{s-1}/\Gamma (s)$, giving the jump

Substitution ofand integration by parts giveThus takeand require the endpoint term to vanish. Infinite contour ends must lie in sectors where $\mathrm{Re}(t^4)>0$, namely sectors centred on the positive and negative real and imaginary axes, each of angular width $\pi /4$.

Let the common final segment be the positive real ray from $0$ to $+\infty$, and choose the initial rays from $-\infty$, $+i\infty$, and $-i\infty$ to $0$. Call the resulting contours ${\gamma }_1,{\gamma }_2,{\gamma }_3$. PutSince $w_i^{(r)}(0)={\int }_{{\gamma }_i}{t}^r{e}^{-t^4/4}dt$, for $r=0,1,2$ the three rows areConsequentlyUsing the Gamma function reflection identity,The initial-data vectors are therefore linearly independent, so the three contour integrals are linearly independent solutions.

Taking reciprocals in the finite product defining the gamma function givesLet $H_n={\sum }_{k=1}^n{k}^{-1}$. Sinceand $H_n-\log n\to \gamma$, where $\gamma$ is the Euler--Mascheroni constant, passage to the limit gives the Weierstrass product for the reciprocal gamma function

The logarithmic derivative of this identity isSince $k^{-1}-(z+k{)}^{-1}=z/[k(z+k)]$, the digamma function therefore satisfies

For real $z>0$, termwise differentiation gives the trigamma functionThus $\psi$ is strictly increasing on the positive real axis. At the two positive integers needed here, telescoping givesThe intermediate value theorem supplies a zero in $(1,2)$, and strict increase makes it unique. This is the positive zero of the digamma function.

Although $x\mapsto |x{|}^{-1}$ is not integrable on all of ${\mathbb{R}}^3$, it defines a tempered distribution. For $x\ne 0$, the Gamma function integral givesThe three-dimensional form of the Fourier transform of a Gaussian isTherefore, up to a nonzero constant depending on the Fourier-transform convention,For $\xi \ne 0$, the substitution $u=|\xi {|}^2/(4t)$ turns the last integral intoThus, as asserted by the Fourier transform of inverse distance in three dimensions,as a tempered distribution, where $C^{\prime }\ne 0$.

For $\Re s>1$, absolute convergence permits separation into odd and even terms:Using the defining integral of the Gamma function and the substitution $u=nt$ givesTermwise integration is justified by absolute convergence when $\Re s>1$, and the geometric sum isConsequently the Dirichlet eta function satisfies

Near zero the integrand is $O(t^{\Re s-1})$, while at infinity it decays exponentially. The integral therefore defines a holomorphic function for $\Re s>0$. Thusprovides the desired continuation wherever the displayed denominator is nonzero. At a nonreal zero of $1-2^{1-s}$, use insteadwith an integer $k$ for which $1-k^{1-s}\ne 0$; the bounded partial sums of $a_n$ give convergence for $\Re s>0$. This shows that those apparent singularities are removable and yields the Analytic continuation of the Riemann zeta function to the right half-plane.