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A phase-space distribution function describes the mass in a small cell of positions and velocities: $F(\mathbf{x},\mathbf{v},t)d^{3}x{d}^{3}v$ is that mass. Its velocity integral is the spatial mass density, and $F/\rho$ at a fixed position is a normalized velocity probability density. A number-normalized galactic distribution function differs by the stellar mass convention.

Take $a>0$ and $v_0>0$. For the cusped logarithmic spherical galaxy model, differentiate the gravitational potential and use the spherical Poisson equation for Newtonian gravity:ThusA circular orbit requires $v_c^2/r={\Phi }^{\prime }$. Its galaxy rotation curve is thereforeIt rises as $v_0\sqrt{r/a}$ at small radius and approaches $v_0$ from below at large radius.

The outer flat galaxy rotation curve is qualitatively useful for galaxies. However, the density has an inner $1/r$ cusp and an outer $1/r^2$ tail, while $M(r)\sim v_0^{2}r/G$ diverges. The model therefore requires an outer truncation for a finite galaxy, and does not describe a central constant-density core. Its gravitational potential cannot be normalized to zero at infinity.

For a star, the specific energy is $E=v^2/2+\Phi$ and the specific angular momentum is $\mathbf{L}=\mathbf{x}\times \mathbf{v}$. Since the gravitational potential is stationary and spherical,Thus both the specific energy and the entire angular-momentum vector are conserved.

For Jeans theorem, follow a stellar trajectory in phase space. The Collisionless Boltzmann equation is preciselyIn a steady system ${\partial }_{t}F=0$, so the value of the galactic distribution function is constant along every trajectory: it is itself an integral of motion. Conversely, if $I_1,\dots ,I_m$ are time-independent integrals of motion, then $F=h(I_1,\dots ,I_m)$ satisfies $dF/dt=\sum h_{,a}{\dot{I}}_a=0$. Locally, a complete set of invariant orbit labels expresses any steady galactic distribution function this way. This is the weak form of Jeans theorem; a stronger assertion about a small set of isolating integrals needs additional orbital regularity or mixing assumptions. In particular, the theorem does not say that every steady galaxy must have a galactic distribution function depending on only $E$ and $L$.

For the proposed constant-anisotropy distribution function, write $v_T=(v_{\theta }^2+v_{\varphi }^2{)}^{1/2}$ and $L=r{v}_T$. Introduce a polar angle $\psi$ in the tangential velocity plane. Then $d^{3}v=d{v}_r{v}_{T}d{v}_{T}d\psi$, so the apparent $1/L$ singularity cancels against the velocity Jacobian. This inverse-angular-momentum velocity integration makes the density finite for every $r>0$.

Let $t=e^{-\Phi /v_0^2}=a/(a+r)$, $c_1=1$, $c_2=2$, and ${\alpha }_k=k/(2v_0^2)$. Each exponential component contributesHere the two Gaussian integrals and the angular integral have product ${\pi }^2/{\alpha }_k$. Consequently,The galactic distribution function is nonnegative, gives the required density, and depends on conserved $E,L$; by Jeans theorem it is a self-consistent steady solution.

Differentiating the Gaussian integral with respect to ${\alpha }_k$ shows that each component has $⟨v_r^2{⟩}_k=1/(2{\alpha }_k)=v_0^2/k$, and the same value for $⟨v_T^2{⟩}_k$. Angular averaging gives $⟨v_{\theta }^2{⟩}_k=⟨v_{\varphi }^2{⟩}_k=v_0^2/(2k)$. Weighting these by ${\rho }_1,{\rho }_2$ yields the velocity dispersionsThe radial numerator includes $2r$, as in the original PDF. All mean velocities vanish by parity. The same parity, or the zero angular average of $\sin \psi \cos \psi$, gives $⟨v_r{v}_{\theta }⟩=⟨v_r{v}_{\varphi }⟩=⟨v_{\theta }{v}_{\varphi }⟩=0$.

The velocity-anisotropy parameter isThe constant-anisotropy distribution function therefore favors radial motion at every radius: the radial variance is twice either individual tangential variance. It still has a substantial population with nonzero angular momentum, rather than exclusively radial orbits. The $L^{-1}$ factor enhances low-angular-momentum stars. The radial velocity dispersion approaches $3v_0^2/4$ at the centre and $v_0^2$ far out; each individual tangential variance is half as large. There is no mean rotation, despite the well-defined galaxy rotation curve for hypothetical circular orbits.

The stationary points of the exponent are $t=1/3$ and $t=1$, because ${\varphi }^{\prime }(t)=(3t-1)(t-1)$. The first is a local maximum with $\varphi (1/3)=4/27$ and ${\varphi }^{\prime \prime }(1/3)=-2$; the second is a local minimum. Alsoso the endpoint overtakes the interior maximum only when $b>4/3$.

For fixed $0<b<1/3$, the upper endpoint is the unique maximum. Linearizing the exponent there gives the Laplace endpoint estimate $e^{\lambda \varphi (b)}/(\lambda {\varphi }^{\prime }(b))$. At $b=1/3$, only half of the local Gaussian integral is included. For $1/3<b\le 4/3$, the whole Gaussian near $1/3$ contributes. At $b=4/3$ an endpoint has the same exponential height, but its prefactor is ${\lambda }^{-1}$, smaller than the interior ${\lambda }^{-1/2}$ prefactor. For $b>4/3$, the endpoint has strictly greater height and dominates.

Consequently all fixed-$b$ regimes areThe estimate follows from Laplace's method; the tail toward negative infinity is exponentially harmless since the cubic tends to negative infinity.

PutIn free space, the parabolic wave equation is $E_x=i{E}_{zz}/(2k)$. Under the Fourier transform convention $\hat{E}(p)=\int E(z)e^{-ipz}dz$, it becomesThe Gaussian integral is legitimate because $\mathrm{Re}A=1/D^2>0$. Invert the transform after multiplying by $e^{-i{p}^{2}x/(2k)}$. A second Gaussian integral, or equivalently one-dimensional transverse Fresnel propagation, givesChoose the square-root branch continuously from $q(0{)}^{-1/2}=1$; for real $x\ge 0$ there is no zero of $q$. This gives the correct incident field at $x=0$, and direct differentiation verifies the free parabolic wave equation.

For clarity, the squared envelope magnitude isThe Gaussian beam with one transverse coordinate remains Gaussian, with one-transverse-coordinate amplitude factor $q^{-1/2}$, not the $q^{-1}$ of a beam with two transverse coordinates. The negative initial quadratic phase produces focusing for $F>0$; diffraction prevents a singularity at $x=F$. These expressions describe the paraxial approximation to free propagation, rather than an exact unrestricted Helmholtz equation beam.

Set $a=2m{L}^2/({\hslash }^2\beta )$ and $C=\sqrt{m/(2\pi {\hslash }^2\beta )}$. For the reflected images, the intervals $rL-q$, $0\le q\le L$, tile the real line once. A Gaussian integral givesConsequently $Z=CL{\sum }_r{e}^{-a{r}^2}-1/2$. The Poisson summation formula applied to the Gaussian yieldsThe zero dual mode cancels the reflected contribution, while the positive and negative modes pair:This proves equality with the energy eigenstate calculation. The image expansion converges rapidly at small $\beta$; the spectral expansion converges rapidly at large $\beta$.

For the normalized standard Gaussian density,The two terms cancel, so the Ornstein-Uhlenbeck Fokker-Planck equation has stationary density for a Fokker-Planck equationEquivalently its Fokker-Planck probability current $-\nabla \gamma -v\gamma$ vanishes identically. The normalization follows from the Gaussian integral in each coordinate.

The moment-generating function of a random variable $Z$ is $m_Z(\theta )=\mathbb{E}[e^{\theta Z}]$, at the real values of $\theta$ for which that expectation is finite. For a standard normal distribution variable $X$, combining the exponentials in its probability density function yieldsThe last equality follows by scaling the Gaussian integral. The variables $X_i^2$ are independent random variables, so the expectation of their product factorizes. ThereforeThis is the moment-generating function of a chi-squared distribution with $n$ degrees of freedom. For positive $n$ the defining integral diverges at and above $\theta =1/2$.

Use the Fourier transform $\tilde{f}(k)={\int }_{\mathbb{R}}f(x)e^{-ikx}dx$. For a rapidly decaying smooth function, integration by parts gives $\widetilde{f^{\prime }}=ik\tilde{f}$, and differentiation under the integral gives $\widetilde{xf}=id\tilde{f}/dk$. Both are justified here by Gaussian decay. Since $f^{\prime }=-2a^{2}xf$,The Gaussian integral supplied in the hint gives $\tilde{f}(0)=\sqrt{\pi }/a$. Solving this first-order equation yieldsThe calculation states the transform normalization and the two differentiation properties explicitly.

The constant Fourier coefficient is $A_0(y,s)={\int }_0^{1}G(x+iy,s)dx$. The $m=0$ terms give $2\zeta (2s)y^s$. For fixed $m\ne 0$, unfolding the sum over $n$ givesIndeed, the substitution $u=mx+n$ introduces a factor $1/|m|$, while the translated intervals of length $|m|$ cover the real line exactly $|m|$ times, cancelling that factor. Now scale $u=|m|yt$. The beta function integral yieldswhich can be checked by inserting $(1+t^2{)}^{-s}=\Gamma (s{)}^{-1}{\int }_0^{\infty }{v}^{s-1}{e}^{-v(1+t^2)}dv$ and evaluating the inner Gaussian integral. Thus the fixed-$m$ contribution is $|m{|}^{1-2s}{y}^{1-s}\sqrt{\pi }\Gamma (s-1/2)/\Gamma (s)$. Summing over positive and negative $m$ proves the constant term of a nonholomorphic Eisenstein seriesWith the printed normalization $\xi (u)={\pi }^{-u/2}\Gamma (u/2)\zeta (u)$ of the completed Riemann zeta function, this is exactlyBoth the unfolding and the original series calculation are justified for $\mathrm{Re}s>1$. Beyond this region the identities are understood meromorphically: $G(\tau ,s)$ is the Epstein zeta function of the unit-covolume lattice $(\mathbb{Z}\tau +\mathbb{Z})/\sqrt{y}$, so question 4 supplies its continuation. The symbol $\xi$ here has the completed-zeta normalization displayed above, which has poles at zero and one; it does not include the extra polynomial factor sometimes used to define an entire xi function.

Use the Euclidean path integral with weight $e^{-S}$ and Fourier transform convention $\varphi (x)=\int d^{d}p(2\pi {)}^{-d}{e}^{ip\cdot x}\varphi (p)$. The momentum-space Feynman rules are: an internal scalar propagator $1/(p^2+m^2)$; a quartic vertex $-\lambda$ together with $(2\pi {)}^d{\delta }^{(d)}(\sum p_i)$ for incoming momenta; and an integral $\int d^{d}k/(2\pi {)}^d$ over each independent loop momentum. Multiply by the Feynman-diagram symmetry factor; external propagators are retained for a full correlation function and removed for an amputated connected correlation function.

To enumerate the requested one-particle-irreducible Feynman diagrams, let $V$ be the number of quartic vertices and $I$ the number of internal edges. Four external legs imply $4V=2I+4$, while one loop order implies $I-V+1=1$. Thus $V=I=2$. Both internal edges must connect the two vertices: an alternative with a tadpole and one connecting edge would disconnect on cutting that edge. The only graphs are therefore the three pairings of labelled external legs, each with Feynman-diagram symmetry factor $1/2$.

The corresponding channel momenta are $q_s=p_1+p_2$, $q_t=p_1+p_3$, and $q_u=p_1+p_4$, with all external momenta incoming. This exhausts the connected one-loop four-point one-particle-irreducible Feynman diagrams.

For the Euclidean massive loop integral, first take $m^2>0$ and an integer $n>d/2$, where the integral converges. Schwinger parameterization givesThe supplied Gaussian integral then yieldsIntegrating over $s$, and using the Gamma function recurrence to obtain $\Gamma (n)=(n-1)!$, provesOutside its initial convergence range, the right-hand side defines the meromorphic continuation used in dimensional regularization; the divergent ordinary integral is not being assigned a convergent value.

For the scalar bubble integral, the Feynman parameter identity $1/(AB)={\int }_0^{1}dx[xA+(1-x)B{]}^{-2}$ followed by a translation of loop momentum givesWith $d=4-ϵ$, the Gamma function satisfies $\Gamma (ϵ/2)=2/ϵ+O(1)$, and the parameter integral tends to one. Hence the ultraviolet pole is independent of external momentum:One can also see why the same pole occurs without introducing a Feynman parameter: at large $k$, the difference between this integrand and $(k^2+m^2{)}^{-2}$ is ultraviolet integrable near four dimensions, so both have the same dimensional regularization pole. The positive mass avoids an infrared ambiguity in this argument.

In the quantum effective action convention, the tree-level four-point vertex is $+\lambda$, the negative of the amputated connected correlation function tree vertex. The three bubble corrections giveThus the pole counterterm is $\delta \lambda =3{\lambda }^2/(16{\pi }^2ϵ)$. The modified minimal subtraction scheme also subtracts the conventional finite $\log 4\pi -{\gamma }_E$ combination, or equivalently absorbs it into the subtraction-scale convention; that does not change the one-loop renormalization-group beta function. Using that scale convention, writeHere $\lambda$ is dimensionless, $[{\lambda }_B]=ϵ$, and the conventional constant scale factor is implicit. The one-loop tadpole is momentum independent, so wave-function renormalization does not contribute at this order. Differentiating at fixed ${\lambda }_B$, with $a=3/(16{\pi }^2)$, gives $0=ϵ(\lambda +a{\lambda }^2/ϵ)+\beta (1+2a\lambda /ϵ)+O({\lambda }^3)$. ThereforeFor the stable quartic scalar field theory, with $\lambda >0$, the one-loop quartic scalar beta function is positive: the running coupling increases toward the ultraviolet and decreases toward the infrared. The theory is not asymptotically free; extrapolating the one-loop flow gives a Landau pole.

Rescale $W\mapsto tW$ with $t>0$. After Berezin integration, the unnormalized holomorphic function insertion isIts derivative is an ordinary total derivative:The last equality again follows from polynomial growth and exponential decay. Thus the insertion is independent of $t$, and supersymmetric localization allows evaluation as $t\to \infty$ at the zeros of $P$, as in localization of a zero-dimensional polynomial model.

For a simple zero $z_a$, $P(z)=P^{\prime }(z_a)(z-z_a)+O((z-z_a{)}^2)$. The local Gaussian integral cancels the factor $|P^{\prime }(z_a){|}^2$ and contributes $\pi f(z_a)$. For a zero with multiplicity of a root $m_a$, write $P(z)=a_a(z-z_a{)}^{m_a}+\cdots$. Its leading radial integral isThe insertion tends to $f(z_a)$ there; contributions away from the zeros vanish. Therefore the full answer, including degenerate critical points, isThis is an unnormalized expectation. For simple zeros all $m_a=1$; division by $Z=\pi r$ would give the normalized average, but is not part of the requested insertion. Taking $f=1$ recovers the partition function, and taking $f=Pg$ recovers its vanishing supersymmetric Ward identity.

Use the time convention $e^{-i\omega t}$. The free-space Helmholtz equation, with $\psi =e^{ikx}E$, gives the exact reduced equationFor a forward plane wave at angle $\alpha$, $E_x=O(k{\alpha }^{2}E)$ and $E_{zz}=O(k^2{\alpha }^{2}E)$, whereas $E_{xx}=O(k^2{\alpha }^{4}E)$. Thus the paraxial approximation neglects $E_{xx}$ relative to $2ik{E}_x$. The reduced field satisfies the parabolic wave equationWith the Fourier transform convention used below and its inverse, this becomes an ordinary differential equation for each transverse wavenumber:Consequently the initial-value solution isEquivalently, evaluating the oscillatory Gaussian integral gives the one-dimensional transverse Fresnel propagation formulaThe square-root branch has $\sqrt{1/i}=e^{-i\pi /4}$. The formula is an oscillatory integral for general data, or the Fresnel propagator acting on square-integrable functions. It approaches the initial field as $x\downarrow 0$. A plane wave has transverse wavenumber $\nu =k\sin \alpha$; the reduced axial phase $-{\nu }^{2}x/(2k)$ agrees with $k(\cos \alpha -1)x$ through order ${\alpha }^2$. This also checks the sign. The paraxial approximation applies to $|\nu |\ll k$.

Set $d=1/\sqrt{2}$. At the positive real endpoint the contour is real and crosses the saddle point at zero. The Gaussian integral givesBoth omitted tails are exponentially small. At the negative real endpoint the contour is oriented leftward and does not cross the saddle point:The endpoint contribution in steepest descent, or one integration by parts, gives ${\int }_d^{\infty }{e}^{-\lambda x^2}dx\sim e^{-\lambda d^2}/(2\lambda d)$. The upper-end correction is $O(e^{-\lambda }/\lambda )$ and is much smaller. ThereforeThe sign comes from contour orientation, not from a different Gaussian convention.

The sudden approximation keeps the wavefunction unchanged at the instant of switching: a finite step of potential creates no delta-function impulse in the Schrodinger equation. Just before switching the old ground state differs from the printed real wavefunction only by an overall phase, which does not affect probabilities.

Let ${\tilde{\psi }}_0(x)=(\pi \hslash {)}^{-1/4}{e}^{-(x-d{)}^2/(2\hslash )}$, $d=qE$, be the new ground state. Its overlap with the initial state is, up to that phase,Since $x^2+(x-d{)}^2=2(x-d/2{)}^2+d^2/2$, the Gaussian integral gives $A_0=e^{-d^2/(4\hslash )}$. The Born rule and the ground-state overlap after sudden oscillator displacement thus giveThis is an overlap probability, not a claim that the original wavefunction instantaneously relaxes to the new ground state. It becomes one when $qE=0$.

Use a real orthonormal basis of Fourier modes on the time circle:The eigenvalue of $L_{\omega }$ on the constant mode is ${\omega }^2$, and each sine/cosine pair has eigenvalue ${\omega }^2+{\nu }_r^2$. Retaining these modes giveswhere $C_N$ can depend on $N,\beta$ but is independent of $\omega$. Each real Gaussian integral contributes $\sqrt{2\pi /{\lambda }_r}$. Consequently the Fourier-cutoff oscillator functional determinant isA spectral cutoff between successive eigenvalue pairs selects precisely this finite subspace. In comparing two frequencies, the displayed regularization keeps the same number of modes $N$; holding an absolute spectral threshold fixed need not give the same $N$ at finite cutoff.

Use a genus-zero Polyakov path integral, target signature $(-,+,\dots ,+)$ and all four momenta incoming. Define the Mandelstam variables by $s=-(p_1+p_2{)}^2$, $t=-(p_1+p_3{)}^2$, $u=-(p_1+p_4{)}^2$. The string mass-shell condition is $p_i^2=4/{\alpha }^{\prime }$ and momentum conservation is ${\sum }_i{p}_i=0$, soThe reduced amplitude omits the overall momentum-conservation Dirac delta distribution and any conventional overall scattering-matrix phase. Interpret each tachyon vertex operator as normal ordered; self-contractions must not be included.

The free-boson worldsheet propagator is $⟨X^a(z)X^b(w)⟩=-({\alpha }^{\prime }/2){\eta }^{ab}\log |z-w{|}^2$. Wick theorem then evaluates the normal-ordered exponential correlator as the Koba-Nielsen factorThe worldsheet zero mode supplies momentum conservation. Four vertices give $g_s^4$ and the sphere contributes $g_s^{-2}$ through the string genus expansion, leaving $g_s^2$.

The sphere's residual Möbius transformations fix three insertion points. In sphere gauge fixing for four string vertices, take $(z_1,z_2,z_3,z_4)=(z,0,1,\infty )$. The worldsheet ghost fields, equivalently the Faddeev-Popov determinant for this residual group, supply $|z_{23}{z}_{24}{z}_{34}{|}^2$. At finite $z_4$, this determinant grows as $|z_4{|}^4$, while the matter correlator decays as $|z_4{|}^{-4}$ by momentum conservation and the tachyon mass shell; their product has a finite limit. Equivalently the weight-$(1,1)$ matter operator at infinity is normalized with $|z_4{|}^4$, while the ghost pair uses the inverse factor. After stripping the fixed-position normalization, onlyremains. Here choose the standard complex-coordinate measure $d^{2}z=idz\wedge d\bar{z}=2dxdy$. With $dxdy$ instead, the reduced normalization must acquire a factor two to give the same requested amplitude. Absolute vertex/sphere normalization is a convention; this choice fixes it consistently with the displayed $2\pi$ prefactor.

Set $a=-1-{\alpha }^{\prime }s/4$, $b=-1-{\alpha }^{\prime }t/4$, $c=-1-{\alpha }^{\prime }u/4$. Then $a+b+c=1$ and the integrand is $|z{|}^{2a-2}|1-z{|}^{2b-2}$. To evaluate the complex beta integral first work where $\mathrm{Re}a,\mathrm{Re}b,\mathrm{Re}c>0$. This ensures convergence near $0$, $1$ and infinity. Set $v=1-a$ and $w=1-b$. Schwinger parameterization givesCompleting the square, the $dxdy$ Gaussian integral is $\pi (\lambda +\rho {)}^{-1}\exp [-\lambda \rho /(\lambda +\rho )]$. Change variables to $q=\lambda +\rho$ and $x=\lambda /q$; their Jacobian determinant is $q$. Integrating $q$ gives $\Gamma (v+w-1)[x(1-x){]}^{-(v+w-1)}$. The remaining integral is the ordinary beta function $B(1-w,1-v)=B(b,a)$. ThusRestoring $d^{2}z=2dxdy$ and substituting the Mandelstam variables proves the Virasoro–Shapiro amplitude:For physical scattering the original position integral generally fails to converge. The formula defines the amplitude by analytic continuation from the convergence domain, with the desired scattering boundary value at real poles; the convergent integral should not be claimed valid for every physical momentum.

The gamma function is a meromorphic function, with simple poles at nonpositive integers and no zeros, while its reciprocal $1/\Gamma$ is an entire function. Consequently it is a meromorphic function of the independent invariants and a crossing-symmetric scattering amplitude: permuting $s,t,u$ leaves it unchanged. Its generic channel poles areand the same tower in $t$ and $u$. They represent the exchanged bosonic string mass spectrum: the tachyon, massless states including the graviton, and an infinite sequence of massive closed-string levels. There are no threshold branch cuts at this tree order.

A precise check of factorization is the Virasoro–Shapiro amplitude pole residue. Near $a=-n$, set $c=1+n-b$ in the nonsingular factor. The gamma function recurrence givesCombining this with $\Gamma (a)\sim (-1{)}^n/[n!(a+n)]$ showsThe residue is a degree-$2n$ polynomial in $t$, consistent with exchange up to spin $2n$ and scattering-amplitude factorization. At exceptional kinematics reciprocal Gamma zeros can remove apparent poles. In particular, when both $a$ and $b$ approach nonpositive integers, the zero from $1/\Gamma (a+b)$ cancels the putative double pole, leaving channel simple-pole terms. Thus one must not count products of numerator poles without using $a+b+c=1$.