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Branches and cuts. For the square root of z squared minus one, a branch point is a point around which analytic continuation along an arbitrarily small closed loop can change the value of a chosen branch of a multivalued function. Near $z=1$, the square root behaves as $\sqrt{2}(z-1{)}^{1/2}$; near $z=-1$ it is a nonzero analytic factor times $(z+1{)}^{1/2}$. One circuit around either point reverses its sign. Thus the finite branch points are $\boxed{z=\pm 1}$. Infinity is not a branch point: in $w=1/z$ the expression is $w^{-1}\sqrt{1-w^2}$, which has two single-valued meromorphic choices near zero.

A single branch cut can join $-1$ to $1$; its shortest choice is the real segment $[-1,1]$. Any surviving loop then surrounds either both branch points or neither, so it produces no sign change. Alternatively, use two branch cuts, $(-\infty ,-1]$ and $[1,\infty )$, each joining a branch point to infinity. On the resulting slit region one can choose a consistent branch of a multivalued function.

For the finite cut, fix $f(z)\sim z$ at infinity, equivalently $f(x)>0$ for $x>1$. To give its actual values on both sides, putFor $y\ne 0$, the selected square root isHere $\mathrm{sgn}(0)=0$; when $x=0$, $U=0$. Squaring verifies the formula because $UV=|xy|$. Its real part has the sign of $x$, its imaginary part has the sign of $y$, and it matches the chosen value at infinity. In particular, the two boundary values on the branch cut areOutside the cut, $f(x)=\mathrm{sgn}(x)\sqrt{x^2-1}$. Reversing the global branch of a multivalued function reverses every displayed value.

Differential equation. Write $f(\zeta )=\sqrt{{\zeta }^2-1}$ with this finite cut, and orient $C$ counterclockwise with one circuit around it. Since $C$ is fixed and avoids the branch points, differentiation under the integral sign givesOn the contour $f^{\prime }=\zeta /f$. ThereforeThe last contour integral vanishes because the expression in brackets returns to its original value after circling both branch points. This proves $\boxed{z{u}^{\prime \prime }+u^{\prime }-zu=0}$, including at $z=0$; division by $z$ is unnecessary.

Value at zero. Near either endpoint ${\zeta }_0=\pm 1$, the integrand at zero is $O(|\zeta -{\zeta }_0{|}^{-1/2})$. A small arc of radius $\epsilon$ has length $O(\epsilon )$ and hence contributes $O(\sqrt{\epsilon })\to 0$. The upper lip runs from $1$ to $-1$, and the lower lip from $-1$ to $1$. ThusUsing $x=\sin t$ evaluates the final integral as $\pi$, and soThe same deformation gives $u(z)=2i{\int }_{-1}^1{e}^{zx}(1-x^2{)}^{-1/2}dx$. The sign of $u(0)$ reverses if either the global square-root choice or the contour orientation is reversed; the differential equation is unaffected.

The Laplace transform is $\overline{f}(p)={\int }_0^{\infty }{e}^{-pt}f(t)dt$ wherever it converges. For sufficiently large $\mathrm{Re}p$, suitable exponential bounds justify integration by parts and differentiation under the integral sign, givingFor a nonnegative integer $n$ and real $p>0$, substitute $u=pt$ in the gamma integral. The result extends holomorphically to $\mathrm{Re}p>0$, givingThis also follows recursively from the second identity, starting with $\mathcal{L}[1]=1/p$.

Put $G=\overline{g}$, $g_0=g(0^{+})$ and $g_1=g^{\prime }(0^{+})$. For a solution with finite initial values and the required transform bounds, the transformed equation isThe initial-value terms cancel, leaving $(p-p^2)G^{\prime }+(3-p)G=0$. On the real interval $p>1$, a nonzero solution satisfiesIntegration gives $G=C(p-1{)}^2/p^3=C(1/p-2/p^2+1/p^3)$. Inverting term by term yieldsTaking $C=1$ gives a nonzero solution, verified by direct substitution. It is $L_2(t)$, and the rational transform agrees with the general Laplace transform of a Laguerre polynomial. No claim about solutions with singular initial behavior is needed to produce this regular solution.

Let $B=|b|>0$. For $a>0$, integrate $e^{iaz}/(z^2+B^2)$ over the upper semicircle. Jordan's lemma removes the arc, and the residue theorem at $z=iB$ givesTaking the real part and using evenness in both $x$ and $a$ yieldsThe case $a=0$ follows directly from an inverse tangent primitive. The complex exponential is essential on the contour: the cosine alone has a growing exponential component.

For $a>0$, differentiation under the integral sign gives $-I_1^{\prime }(a)={\int }_0^{\infty }x\sin (ax)/(x^2+B^2)dx$. This derivative is a conditionally convergent improper integral; the Dirichlet test makes its tails uniform on compact subintervals of $a>0$, justifying the differentiation. Setting $a=1$ givesFinally let $J(a)={\int }_0^{\infty }\sin (ax)/[x(x^2+B^2)]dx$. Its differentiated integrand is absolutely integrable, so $J^{\prime }(a)=I_1(a)$ and $J(0)=0$. HenceIf $b=0$, $I_1$ and $I_3$ diverge at the origin, whereas $I_2={\int }_0^{\infty }\sin x/xdx=\pi /2$ is the corresponding Dirichlet integral.

The discrete base factors in the displayed construction inherit their variational parameters. Accordingly write $s_{\lambda }(\epsilon )$, not a parameter-independent $s(\epsilon )$, for the joint base probability mass function or probability density function. The continuous noise factors can be parameter independent while the discrete factors are $\lambda$ dependent. Calling the entire base law parameterless while retaining its nonzero score function would be inconsistent.

Let $t_{\lambda }$ push the base law forward to $q_{\lambda }$, and defineUse a fixed base measure $\nu$, with counting measure for discrete coordinates and Lebesgue measure for continuous coordinates. Assume locally parameter-independent support, differentiability and an integrable domination permitting differentiation under the integral sign. A valid reparameterization gradient givesApply the product rule before simplifying:This is the required combined score and pathwise gradient estimator. A sample from $s_{\lambda }$ and the integrand on the last line give an unbiased stochastic gradient under the stated regularity conditions. Its first term is a score-function gradient estimator contribution; the remaining terms differentiate the sampled objective along the continuous transformation. The same separation appears in the general stochastic computation graph construction.^[ref]

All derivatives inside the expectation hold $\epsilon$ fixed. They still include both explicit parameter dependence and dependence through $t_{\lambda }$. For a continuous coordinate $z=t_{\lambda }(\epsilon )$,The model joint density $p(z,x)$ is fixed as $\lambda$ changes, but its composition with $t_{\lambda }$ need not be constant. Discrete coordinates are not differentiated as real-valued paths; their changing probabilities enter through the score function. Moving supports or parameter-dependent branch boundaries require additional care and are not justified merely by formally interchanging a derivative and an integral.

Set $q=y-x$ and $z_t=x+tq$. The definition of pushforward measure gives the confinement energy $\mathcal{V}[{\mu }_t]=\int V(z_t)d\Pi$. The chain rule and differentiation under the integral sign, whenever justified, giveFor example, bounded Hessian matrix is sufficient: it gives $|\nabla V(z)|\le C(1+|z|)$, and the first two differentiated integrands are bounded by $C^{\prime }(1+|x{|}^2+|y{|}^2)$. The dominated convergence theorem applies because $\Pi$ has finite second absolute moments. Compactly supported endpoints also suffice because all interpolating points lie in one compact set.

The printed growth assumption alone does not ensure a classical second derivative. Here is a counterexample within the stated assumptions. On the line take the bounded nonnegative smooth functionand the probability measures $\mu ={\sum }_n{p}_n{\delta }_{n-1/2}$, $\nu ={\sum }_n{p}_n{\delta }_{n+1/2}$. Both have finite second absolute moments. Their translation transport plan is optimal: its cost is one, while every coupling has cost at least the square of its mean displacement, also one. At $t_0=1/2$, $z_{t_0}=n$, so $F(t)=\mathcal{V}[{\mu }_t]$ satisfies $F(t_0)=0$. Each summand is nonnegative, and the Fatou lemma givesThus the requested classical identity requires an additional integrability assumption on the Hessian matrix; quadratic growth controls the energy but not its second derivative.

For a confinement potential $V$ that is a $\lambda$-strongly convex function, the derivative formula, where valid, yieldsThe underlying displacement convexity conclusion is valid under the printed assumptions without that extra differentiation hypothesis. Apply convexity of $V(x)-\lambda |x{|}^2/2$ pointwise and integrate:The same argument on every subinterval makes $t\mapsto \mathcal{V}[{\mu }_t]-\lambda D^2{t}^2/2$ convex, so the second derivative lower bound also holds in the sense of distributions.

Use an independent copy $(x^{\prime },y^{\prime })$ of $(x,y)$ under the product measure $\Pi \otimes \Pi$. Write $q=y-x$, $q^{\prime }=y^{\prime }-x^{\prime }$, andSince the pushforward measure of $\Pi \otimes \Pi$ under $(z_t,z_t^{\prime })$ is ${\mu }_t\otimes {\mu }_t$, the interaction energy isTwo applications of the chain rule giveThe factor $1/2$ is retained. This classical identity requires differentiation under the integral sign to be justified, just as in part (b). Bounded Hessian matrix of $W$ is sufficient, since $|q-q^{\prime }{|}^2\le 2|q{|}^2+2|q^{\prime }{|}^2$ is integrable; compactly supported endpoints are another sufficient case. The printed quadratic-growth condition alone does not provide this domination.

The failure can occur for the interaction energy too. Take the even interaction potential $W(x)=1-\cos (2\pi x^3)$ and replace the endpoint probability measures in part (b) by $\frac{1}{2}{\delta }_0+\frac{1}{2}\mu$ and $\frac{1}{2}{\delta }_0+\frac{1}{2}\nu$. Keep the origin fixed and translate every other atom by one. This coupling is increasing on the line, hence is optimal: its assigned values extend to a nondecreasing transport map, which is the gradient of a convex function. Differences between translated atoms remain integers, so their mutual interaction vanishes. Only interactions with the origin contribute, giving $\mathcal{W}[{\mu }_t]=F(t)/4$, with $F$ from part (b). It therefore also lacks a finite classical second derivative at $t=1/2$.

A finite first moment suffices for the velocity, but may not suffice for the printed potential. For example, $W(z)=|z{|}^2/2$ has all the stated properties. If $\mu$ is the law of $(S,0)$ with density $2s^{-3}$ on $s\ge 1$, then $\mathbb{E}S=2$ but $\mathbb{E}{S}^2=\infty$, and $(W\ast \mu )(x)=\infty$ for every $x$. Thus the literal definition of $V$ needs a finite second absolute moment, or another hypothesis ensuring integrability of $W$.

For the specified class ${\mathcal{P}}_1$, define the velocity fields directly by $b_{\mu }=K\ast \mu$ and $b_{\hat{\mu }}=K\ast \hat{\mu }$. Equivalently, use the renormalized interaction potentialThe bounded Hessian matrix and $\nabla W(0)=0$ implyThus $V_{\mathrm{ren}}$ is finite, is twice continuously differentiable in space, and has $\nabla V_{\mathrm{ren}}=\nabla W\ast \mu$, by dominated differentiation under the integral sign. Its rotated negative gradient is exactly $b_{\mu }$. Whenever $W\ast \mu$ is finite, the two potentials differ only by an $x$-independent constant and give the same transport equation.

We check the hypotheses for characteristic representation of a measure-valued continuity equation. For every $x,y$,For any coupling $\gamma \in \Gamma ({\mu }_t,{\hat{\mu }}_t)$, the Lipschitz condition on $K$ also givesTaking the infimum over couplings proves the uniform boundApplying the same estimate at two times of one curve proves time continuity of $b_{\mu }$, uniformly in $x$. Also $|m_1({\mu }_t)-m_1({\mu }_s)|\le d_1({\mu }_t,{\mu }_s)$, so $m_1({\mu }_t)$ is bounded on compact time intervals. Spatial partial derivatives of $b_{\mu }$ are continuous and bounded by $L$, using dominated convergence in space and weak convergence of probability measures in time for the bounded continuous kernel $DK$. All corresponding facts hold for $\hat{\mu }$.

The characteristic representation of a measure-valued continuity equation says that a continuous velocity field, continuously differentiable and globally Lipschitz in space and with locally uniform linear growth, has a global characteristic flow map $X_{\mu }(t;s,a)$ solvingFor each $f\in {\mathcal{P}}_1$, its pushforward measure is the unique weak solution in $C([0,\infty ),{\mathcal{P}}_1)$ of the associated continuity equation. The hypotheses have all been verified above, and the fields have zero divergence, so these are also the two required transport equations. Explicitly,For completeness, the Gronwall inequality givesso these pushforward measures have finite first absolute moments. Integrating $|X_{\mu }(t;0,a)-X_{\mu }(s;0,a)|$ against $f$ proves continuity in the Wasserstein distance. The chain rule proves the weak identity. For uniqueness at a fixed terminal time $t$, use the backward test function $\phi (s,x)=\psi (X_{\mu }(t;s,x))$, where $\psi$ is continuously differentiable with compact support. It satisfies ${\partial }_s\phi +b_{\mu }\cdot \nabla \phi =0$ and has compact spatial support uniformly for $0\le s\le t$, by the backward linear-growth bound. The weak identity with temporal cutoffs therefore gives $\int \psi d{\rho }_t=\int \psi (X_{\mu }(t;0,a))df(a)$, identifying every weak solution with the displayed pushforward measure.

Finally compare the two characteristic curves with the same initial point $a$. Put ${\Delta }_t=X_{\mu }(t;0,a)-X_{\hat{\mu }}(t;0,a)$. The spatial Lipschitz condition and the boxed velocity estimate givefor almost every time; the integral form gives the same conclusion at zeros of $\Delta$. The Gronwall inequality impliesThe pushforward measure $(X_{\mu }(t;0,\cdot ),X_{\hat{\mu }}(t;0,\cdot ){)}_{\#}f$ is a coupling of ${\rho }_t,{\hat{\rho }}_t$. Integrating the last bound against the probability measure $f$ provesThe same constant $L$ depends only on the uniform bound for the Hessian matrix of $W$, and is independent of the prescribed curves and their common initial data.

On any bounded parameter interval $|t|\le T$, the derivative of the integrand has the integrable majorantThe dominated convergence theorem therefore justifies differentiation under the integral sign at every real $t$, givingUse integration by parts in this absolutely convergent integral, with $x{e}^{-x^2}dx=-\frac{1}{2}d(e^{-x^2})$. Both boundary terms vanish, since the sine is zero at $x=0$ and the Gaussian tends to zero at infinity. HenceIt follows that $f^{\prime }=-4t^{3}f$, or equivalently $(e^{t^4}f(t){)}^{\prime }=0$. This identity holds on all of $\mathbb{R}$, including zero. With the supplied Gaussian integral value,The fourth power comes from the printed phase $2x{t}^2$; the parameter has been differentiated with that phase throughout.

For every integer $j\ge 0$, the $j$th derivative with respect to $t$ of $\cos (t\cos x)$ is $(\cos x{)}^j$ times a sine or cosine with a sign. Its absolute value is at most $1$ on the finite interval $[0,\pi /2]$. This is an integrable dominating function independent of $t$. Repeated differentiation under the integral sign is therefore justified for every order, giving $J_0\in C^{\infty }(\mathbb{R})$.

A sufficient differentiation under the integral sign theorem is the following local domination criterion. Suppose $u(t,\cdot )$ is measurable for each $t$, $u(t_0,\cdot )$ is integrable at one point of a neighborhood $J$ of $t_0$, and for almost every $x$ the function $u(\cdot ,x)$ is differentiable on $J$. Assume an integrable function $g_J$ satisfies $|{\partial }_{t}u(t,x)|\le g_J(x)$ for $t\in J$ and almost every $x$. Then $I$ is differentiable on the interior of $J$ andThe mean value theorem bounds the difference quotients by $g_J$, and the dominated convergence theorem passes their pointwise limit through the integral. Applying this criterion locally throughout $(a,b)$ proves differentiability there. One may impose the stronger condition that every $u(t,\cdot )$ is integrable from the outset.

A useful differentiation under the integral sign theorem is this: if $f(\cdot ,y)$ is integrable, $f$ is differentiable in $y$ almost everywhere, and on every compact parameter interval $|{\partial }_{y}f(x,y)|$ is bounded by an integrable function of $x$, then $F^{\prime }(y)={\int }_I{\partial }_{y}f(x,y)dx$, under the usual pointwise continuity hypotheses. It follows by applying dominated convergence to difference quotients. The theorem can be iterated when a corresponding bound holds for the second derivative.

On a compact interval $0<a\le y\le b$, the original integrand is bounded by $b/(1+x^4)$, using $|\sin (x^{2}y)|\le b{x}^2$. Its first and second parameter derivatives arewith integrable bounds $(1+x^4{)}^{-1}$ and $x^2/(1+x^4)$. Thus $F$ is twice differentiable. Subtracting the integrals givesThe final integral is finite: its integrand tends to one at zero and is absolutely bounded by $t^{-2}$ at infinity. Consequentlywhich establishes the required parameter-independent constant without needing an explicit Fresnel integral evaluation.

Write $F_j(p)=\mathcal{L}{f}_j(p)$, using the square-root branch positive on real $p>0$. Integrating the differential equation and using boundedness at zero givesThe integration constant is zero since $x{f}_1(x)\to 0$. Continuity of $f_0$ then shows $f_1(x)=O(x)$, so $f_1$ extends continuously with $f_1(0)=0$.

The Laplace transform of a derivative and differentiation under the integral sign giveIntegrating and using $F_1(p)\to 0$ as real $p\to \infty$ fixes the constant:The initial value theorem for Laplace transforms gives $f_0(0)={\lim }_{p\to \infty }p{F}_0(p)=1$. It is tempting to apply the derivative rule immediately to $f_0$, but its stated regularity is only continuity. To justify that step, define $h(x)=1-{\int }_0^x{f}_1(t)dt$. ThenUniqueness of the Laplace transform gives $h=f_0$, and hence $f_0$ is continuously differentiable and

With $H>0$, setThen the first integral gives $1/\sqrt{1+y^{\prime 2}}=s$. On the descending and ascending halves of the physical track choose respectively $0<\varphi <\pi$ and $\pi <\varphi <2\pi$, so that $y^{\prime }=\cot (\varphi /2)$. Solving for $y$ givesThe unsquared equation requires $1-bs>0$. At $s=1$ this says $b<1$; it is not permissible to retain other values merely because the displayed formula for $y$ has a squared denominator.

For completeness, the Euler-Lagrange equation also gives $y^{\prime \prime }=(A^{\prime }(y)/A(y))(1+y^{\prime 2})<0$. Thus $y^{\prime }$ decreases through zero exactly once, so the chosen parameter runs through the whole single arch without an extra branch. Differentiating the formula for $y$ and dividing by $y^{\prime }=\cot (\varphi /2)$ gives, continuously also at $\varphi =\pi$,Using $x(0)=0$ yieldsSince $\sqrt{1+y^{\prime 2}}=1/\sin (\varphi /2)$, the arc length is obtained by cancelling one factor of $\sin (\varphi /2)$ in $dx/d\varphi$. The conditions $x(2\pi )=1$ and $S=L$ becomeThese determine a unique parameter pair for each $L>1$. To see uniqueness, regard $L$ as the expected value of $1/s$ under a probability density function proportional to $s^2/(1-bs{)}^3$ on $0<u<2\pi$. Differentiation under the integral sign gives the covariance identityThe first argument decreases with $s$ and the second increases. The strict sign follows by applying $(f(s)-f(t))(g(s)-g(t))<0$ to two independent unequal values and averaging; the integrable endpoint singularity of $1/s$ causes no problem. As $b\uparrow 1$, the weight concentrates at $s=1$, so $L\downarrow 1$. As $b=-k\to -\infty$, endpoint estimates give numerator $\sim 2/k^2$ and denominator $\sim 4\log k/k^3$, so $L\sim k/(2\log k)\to \infty$. Continuity then covers every $L>1$. At $b=0$ one recovers the cycloid and $L=4/\pi$.

Smoothness qualification. Near $\varphi =0$,and the other endpoint has $y^{\prime }\to -\infty$. These are smooth graphs on $0<x<1$, not $C^1$ graphs on the closed interval. Indeed a finite endpoint slope would make the conserved $H$ diverge as $y\to 0$. The travel time remains finite becauseThus the printed phrase “smooth solution” must be understood on the interior, with improper endpoint limits. The variational calculation supplies necessary stationary arches; it does not by itself prove global minimality in every larger class of tracks.

On the small clockwise contour $z=ϵe^{i\theta }$, $\theta$ decreases from $\pi$ to $-\pi$. Uniformly on that arc, $e^{z-\sqrt{z}}=1+O(\sqrt{ϵ})$, and henceThe large arcs also vanish. To make this precise for the displayed geometry, take them on the left semicircle $z=\gamma +R{e}^{i\theta }$, $\pi /2\le \theta \le 3\pi /2$, split at the cut. On the principal holomorphic branch, $\mathrm{Re}\sqrt{z}\ge 0$, $|z|\ge R-\gamma$, and $\mathrm{Re}z=\gamma +R\cos \theta$. The estimation lemma givesFor the last bound, symmetry and $\sin u\ge 2u/\pi$ on $0\le u\le \pi /2$ imply that the angular integral is at most $\pi /R$. The exact choice of large arcs does not affect the limiting result. The vertical improper integral converges absolutely because $e^{-\mathrm{Re}\sqrt{\gamma +iy}}/|\gamma +iy|$ decays as $O(e^{-c\sqrt{|y|}}/|y|)$.

Taking limits in the closed contour integral now yields $2\pi iJ+2iK-2\pi i=0$, where $K={\int }_0^{\infty }{e}^{-x}\sin (\sqrt{x})dx/x$. ThusAs an optional explicit evaluation, put $x=t^2$ and define $K(a)=2{\int }_0^{\infty }{e}^{-t^2}\sin (at)dt/t$. Differentiation under the integral sign and the Gaussian integral give $K^{\prime }(a)=\sqrt{\pi }{e}^{-a^2/4}$ and $K(0)=0$. Therefore $K=K(1)=\pi \mathrm{erf}(1/2)$ and $J=\mathrm{erfc}(1/2)\simeq 0.479500$ in terms of the error function and complementary error function.

For the independent statistical sample, write the likelihood function and log-likelihood function asThe expected Fisher information in the whole statistical sample isUnder the regularity conditions that permit differentiation under the integral sign, the score function has expected value zero andFor an interior statistical parameter, assuming identifiability and $0<I_1(\theta )<\infty$, the regular asymptotic normality of a maximum likelihood estimator statesThus the large-sample normal approximation is ${\hat{\theta }}_n\dot{\sim }N(\theta ,I_n(\theta {)}^{-1})$. The inverse information is the leading sampling variance approximation; the displayed convergence specifies its precise asymptotic meaning.

For a contour integral with factor $e^{{\zeta }^4/4+z\zeta }$, the quartic term dominates the linear term on rays at large radius. Writing $\zeta =r{e}^{i\theta }$ gives real part $r^4\cos (4\theta )/4$ for the leading exponent. There are four decay sectors, centered at $\theta =\pi /4+j\pi /2$ and each of angular width $\pi /4$. On a closed subsector the integrand and all its $z$ derivatives decay faster than any exponential in $r$, locally uniformly in $z$. Contours joining two such sectors therefore allow differentiation under the integral sign and have vanishing infinite endpoint terms. Distinct geometric contours still need an independence proof; their initial moments often give a Vandermonde matrix calculation.

For a twice continuously differentiable interaction potential with $\Vert D^{2}W\Vert \le L$ and a probability measure with finite first absolute moment, this subtraction produces a finite potential even if $W\ast \mu$ is infinite. Taylor's formula and linear growth of the gradient giveDominated differentiation under the integral sign gives $\nabla V_{\mathrm{ren}}=(\nabla W)\ast \mu$ and $D^2{V}_{\mathrm{ren}}=(D^{2}W)\ast \mu$. If the unrenormalized convolution is finite, the two potentials differ only by a constant, so they generate the same velocity field when that field depends on their gradient.

A reparameterization gradient differentiates an expected value by representing the simulated random variable as a differentiable transformation of parameter-independent noise. For $X_{\eta }=g_{\eta }(\epsilon )$, the identity ${\nabla }_{\eta }\mathbb{E}[F_{\eta }(X_{\eta })]=\mathbb{E}[{\nabla }_{\eta }{F}_{\eta }(g_{\eta }(\epsilon ))]$ requires justified differentiation under the integral sign. For a normal distribution, use $X=m+e^{\rho }\epsilon$ with $\epsilon$ drawn from the standard normal distribution.