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The Euler spiral is the planar curve parametrized by the Fresnel integrals. In the optical convention,The tangent has unit length and argument $\pi u^2/2$, so its signed curvature is $\pi u$, proportional to signed arc length. In Fresnel diffraction, an interval integral is the directed chord joining its endpoint parameters on the Cornu spiral. Chord length gives the interval factor's modulus, while chord direction gives its complex argument.

In the unscaled convention, the finite Fresnel integrals are $C(\lambda )={\int }_0^{\lambda }\cos (t^2)dt$ and $S(\lambda )={\int }_0^{\lambda }\sin (t^2)dt$. Their Taylor series follow by termwise integration. At infinity both tend to $\sqrt{\pi /8}$, and their oscillatory tails have asymptotic expansions obtained by integration by parts. This convention differs from the convention with phase $\pi t^2/2$.

Completing the square in the integral over an intermediate position shows that two free-particle propagators compose to the propagator for the sum of their positive time intervals. For equal slices, the quadratic identity isThe Fresnel integral contributes a factor proportional to $\sqrt{k/(k+1)}$; multiplying successive factors gives $1/\sqrt{N}$. A consistent convergence regulator fixes the phases of these square roots.

At the geometric shadow boundary of a straight opaque half-plane, the transmitted Fresnel integral is exactly half the unobstructed complex amplitude for ideal point illumination. The Cornu spiral chord is from its origin to one limiting endpoint, rather than between both limiting endpoints. Therefore $U/U_0=1/2$ and $I/I_0=1/4$. Half the unobstructed intensity would confuse a field fraction with its square. Edge thickness, source extent and alignment can modify the ideal result.

For unit mass and $\hslash =1$, this prepoint kernel tends to the identity and generates the Time-dependent Schrödinger equation. The square-root branch follows the damped Fresnel integral. Its normalized moments are $1$, $0$, and $i\Delta t$ at orders zero, one and two; these fix both the normalization and the kinetic sign.

For the optical convention of Fresnel integrals, $C(w)={\int }_0^w\cos (\pi u^2/2)du$ and $S(w)={\int }_0^w\sin (\pi u^2/2)du$. Both functions are odd and tend to $1/2$ as $w\to +\infty$. If the unscaled convention uses phase $t^2$, multiply its argument by $\sqrt{\pi /2}$ and its value by $\sqrt{2/\pi }$ to obtain this convention. An aperture factor with phase $\pi x^2/(\lambda R)$ uses $u=x\sqrt{2/(\lambda R)}$. Stating the convention prevents missing factors of $\pi$ or $\sqrt{2}$ in diffraction calculations.

Fluid-loaded plate. Use complex amplitudes with time dependence $e^{-i\omega t}$ and take the velocity potential in the fluid above the elastic plate to be $\Phi$. Let ${\rho }_0$ and $c_0$ be the undisturbed mass density and speed of sound. The linearized Euler equations give $p=i{\rho }_0\omega \Phi$; the kinematic boundary condition is ${\Phi }_y(x,0)=-i\omega \eta (x)$. We model loading from one fluid half-space, with no fluid loading on the other face. Loading by identical fluids on both faces instead doubles the fluid term below.

Define the Fourier transform and its inverse byThe Sommerfeld radiation condition selects $\hat{\Phi }(k,y)=\hat{\Phi }(k,0)e^{-\gamma y}$, with ${\gamma }^2=k^2-k_0^2$ and $k_0=\omega /c_0$. Here and in the diffraction calculation below, the outgoing square root is specified by first taking $\mathrm{Im}\omega >0$. Draw a vertical cut upwards from $+k_0$ and another downwards from $-k_0$, and putThus ${\gamma }_u$ is analytic in the upper half-plane and ${\gamma }_l$ in the lower half-plane. In the real-frequency limit,The Fourier transform contour $C$ follows the real axis with the limiting-absorption pole prescriptions. In particular, an outgoing propagating component has positive vertical wavenumber.

The kinematic boundary condition gives $\hat{\Phi }(0)=i\omega \hat{\eta }/\gamma$, so $\hat{p}(0)=-{\rho }_0{\omega }^2\hat{\eta }/\gamma$. Eliminating the fluid pressure from the bending balance gives the line-forced fluid-loaded bending plate kernelFor harmonic line force amplitude $F_0$ at the origin, the plate displacement and fluid velocity potential are thereforeAn arbitrary localized loading replaces $F_0$ by its Fourier transform. This representation both calculates the response and identifies its different wave components through poles and branch points.

For an observer $x=r\cos \theta$, $y=r\sin \theta$, with $0<\theta <\pi$, the method of steepest descent uses the saddle point $k_s=k_0\cos \theta$. At fixed angle away from grazing, the acoustic far field isThe $r^{-1/2}$ spreading is that of cylindrical acoustic waves. The phase velocity parallel to the plate is $\omega /|k|$: components with $|k|<k_0$ are supersonic in that sense and radiate, whereas those with $|k|>k_0$ are subsonic and exponentially decay normally to the plate. The latter have reactive fluid loading and zero mean normal acoustic energy flux.

There is also a real guided root $\kappa >k_0$. Indeed, $K(k_0+)=-\infty$, $K(+\infty )=+\infty$, andThe residue theorem then gives an outgoing displacement contribution $i{F}_0{e}^{i\kappa |x|}/K^{\prime }(\kappa )$. It transports energy along the plate while its fluid field decays as $e^{-\gamma (\kappa )y}$. Other zeros on continued sheets can describe leaky or resonant waves; branch point contributions describe near and grazing fields. A fixed-angle acoustic far field expansion is not uniform near the plate, where these components can matter.

For a propagating spectral component, write $\ell =(k_0^2-k^2{)}^{1/2}$. Its upward normal velocity is $i\ell \hat{\Phi }$ and its pressure is $i{\rho }_0\omega \hat{\Phi }$. Applying Parseval's identity to the time-averaged acoustic energy flux gives the upward power per unit length of the forcing line:Equivalently, the integrand is $|\hat{\eta }{|}^2/\ell$ with prefactor ${\rho }_0{\omega }^3/(4\pi )$. Integrating the radial acoustic intensity of the far field over a semicircle gives the same expression. The input power $\omega \mathrm{Im}(F_0^{\ast }\eta (0))/2$ also feeds guided waves along the plate, so it must not be identified with $P_y$ alone.

Rigid half-plane diffraction. Suppress the time factor and assume $0<{\theta }_0<\pi$. Write $a=k_0\cos {\theta }_0$, $b=k_0\sin {\theta }_0$, and denote the incident spatial field by $I=e^{-iax-iby}$. The Helmholtz equation, Sommerfeld radiation condition, and square-root branches are as above. Across the open positive axis the scattered field and its normal derivative are continuous. Across the rigid negative axis its normal derivative is also continuous, while its potential can jump. Let $J(x)$ be this jump, supported on $x<0$. The transformed scattered field isThe jump transform is analytic in the upper half-plane. The absence of an even scattered component follows from derivative continuity and the outgoing condition. Its normal derivative on either face is $-\gamma {\hat{J}}_u/2$.

On the negative axis this derivative must cancel $-ib{e}^{-iax}$. Since the Half-range Fourier transform of $H(-x)e^{-iax}$ is $i/(k+a)$, the Wiener-Hopf equation has the formwhere $G_l$ is the transform of an unknown function supported on $x>0$ and analytic in the lower half-plane. The incident pole $k=-a$ lies below the Fourier contour. Divide by ${\gamma }_l$ and perform pole subtraction in a Wiener-Hopf equation:The first term is upper analytic and the second lower analytic, since the latter's apparent pole cancels. Equality of the two continued sides gives an entire function. The finite-energy edge condition, with $J(x)=O(\sqrt{-x})$, and the decay of the transforms at infinity force that function to vanish by Liouville's theorem. ConsequentlyThis is the Wiener-Hopf solution of rigid half-plane diffraction with the present transform and time conventions.

Deforming to a steepest descent contour gives a residue from the incident pole precisely when $x+|y|\cot {\theta }_0<0$. At that pole $\gamma (-a)=-ib$; its residue producesThus the total geometrical optics field is $I$ plus a reflected wave in the upper reflection region; in the lower shadow region the scattered residue cancels $I$, and outside that shadow the direct incident wave remains. This gives both the amplitude and the spatial support of the geometrical optics fields rather than merely naming a reflected wave.

For $x=r\cos \theta$, $y=r\sin \theta$, $-\pi <\theta <\pi$, the saddle contribution uses ${\gamma }_l(-a)=-i\sqrt{2k_0}\cos ({\theta }_0/2)$ and ${\gamma }_u(k_s)=\sqrt{2k_0}\cos (\theta /2)$. The diffracted wave isThe total field is $I+{\varphi }_{\mathrm{GO}}+{\varphi }_d$ to this order. This far-field expression applies away from the shadow/reflection boundaries $|\theta |=\pi -{\theta }_0$. There the pole and saddle coalesce, and a uniform Fresnel integral approximation replaces the divergent separate terms by a smooth transition. The exact contour integral remains well defined.

With the prescribed magnetic-only Lorentz force $m\ddot{\mathbf{r}}=q\dot{\mathbf{r}}\times \mathbf{B}$, the component equations areDifferentiate the squared planar speed:Thus $V=\sqrt{{\dot{x}}^2+{\dot{y}}^2}$ is constant, since the Lorentz force is perpendicular to the velocity.

For $V>0$, introduce the velocity angle $\theta$ by $\dot{x}=V\cos \theta$, $\dot{y}=V\sin \theta$. The two planar equations imply $\dot{\theta }=-q{B}_z(t)/m$. Therefore the velocity rotation in a time-dependent axial magnetic field has phaseEquivalently, the complex velocity satisfies $\frac{d}{dt}(\dot{x}+i\dot{y})=-iq{B}_z(t)(\dot{x}+i\dot{y})/m$, giving the same integrated phase. Integrating the velocities and $\ddot{z}=0$ provesThe constants $x_0,y_0,z_0$ are the initial position coordinates, $v_z$ is the constant velocity parallel to the field, $V\ge 0$ is the initial planar speed, and $\varphi$ is the initial planar-velocity direction modulo $2\pi$. Together they specify the initial position and velocity. If $V=0$, the planar position stays fixed and the angle $\varphi$ is immaterial.

For $B_z(t)=\beta t$, the inner integral gives $\theta (t)=-a{t}^2+\varphi$, with $a=q\beta /(2m)>0$. Under the specified initial data,Rescale $u=\sqrt{a}s$. The given Fresnel integrals then giveHence the limiting position isThe negative $y$ coordinate follows from clockwise rotation of planar velocity for positive charge and positive $B_z$. The position converges through cancellation of increasingly rapid oscillations; its convergence does not mean the constant planar speed tends to zero.

Integrate the entire function $e^{i{z}^2}$ around the positively oriented sector boundary: the real segment $0\to R$, the arc $R{e}^{i\theta }$ for $0\le \theta \le \pi /4$, and the radial segment back to zero. Cauchy integral theorem givesThe last term follows from $z=r{e}^{i\pi /4}$ and $i{z}^2=-r^2$; its minus sign is the return orientation.

On the arc, $|e^{i{z}^2}|=e^{-R^2\sin (2\theta )}$. The supplied sine bound gives $\sin (2\theta )\ge 4\theta /\pi$ throughout this sector, soThe Gaussian integral then yields ${\int }_0^{\infty }{e}^{i{x}^2}dx=e^{i\pi /4}\sqrt{\pi }/2$. Taking real and imaginary parts proves both Fresnel integrals:The contour estimate proves existence of the improper oscillatory integrals as well as their values; absolute convergence is neither asserted nor required.

For a smooth real phase with no stationary point, integration by parts using $e^{ixf}=(ix{f}^{\prime }{)}^{-1}(e^{ixf}{)}^{\prime }$ gives an $O(x^{-1})$ contribution, provided $f^{\prime }$ stays away from zero and the differentiated amplitude is integrable. Near an interior nondegenerate stationary point $u_0$, write $f(u)=f(u_0)+f^{\prime \prime }(u_0)(u-u_0{)}^2/2+O((u-u_0{)}^3)$ and scale $u-u_0=x^{-1/2}s$. The Fresnel integral then gives the stationary phase contributionOne sums these contributions over the stationary points and adds endpoint terms where needed. Mere differentiability is insufficient to guarantee this particular nondegenerate formula; it requires the stated local smoothness and nonzero second derivative. Degenerate or endpoint stationary points need their own scaling.

For the gamma integral, rotate the positive real contour toward the positive imaginary axis, using the logarithm in the first quadrant. The limiting imaginary-axis integral is understood in the Abel sense, with a damping factor removed after integration; it is not an ordinarily convergent integral at infinity. With $u=iv$ and then $v=ns$ this givesThe rotation can first be made to an angle below $\pi /2$, where the exponential decays, and then continued to the boundary. There is one stationary point at $s=1$, with phase $-1$ and second derivative $-1$. Thus its contribution is $e^{-in-i\pi /4}\sqrt{2\pi /n}$. To check that the endpoints do not compete, use a cutoff around one. On the remaining region integrate by parts with $1/(1/s-1)$: it vanishes at zero, its derivative is integrable away from one, and Abel damping controls infinity, giving $O(n^{-1})$. This is smaller than the saddle contribution.

Multiplication by the prefactor therefore yieldsIn particular the modulus behaves as $\sqrt{2\pi n}{e}^{-\pi n/2}$; retaining the contour-rotation factor is essential to this decay and to the phase $\pi /4$.

Use polar coordinates with $-\pi <\theta <\pi$; the two faces of the screen are $\theta =\pm \pi$. Take $0<{\theta }_0<\pi$, so the incident wave approaches from the upper half-plane, and retain the specified time dependence $e^{i\omega t}$. Outgoing cylindrical waves then have radial factor $e^{-i{k}_{0}r}$. Define the normalized Fresnel integralThe exact Neumann Sommerfeld half-plane diffraction solution for the total spatial field isOne way to verify the Helmholtz equation is to use parabolic coordinates $\xi =\sqrt{2r}\cos (\alpha /2)$, $\eta =\sqrt{2r}\sin (\alpha /2)$. Each term has the form $e^{i{k}_0({\xi }^2-{\eta }^2)/2}\mathcal{F}(\pm \sqrt{k_0}\xi )$. The operator is $({\xi }^2+{\eta }^2{)}^{-1}({\partial }_{\xi }^2+{\partial }_{\eta }^2)$, and ${\mathcal{F}}^{\prime \prime }(s)+2is{\mathcal{F}}^{\prime }(s)=0$ makes its Helmholtz residual zero. At $\theta =\pm \pi$, the two Fresnel arguments coincide, while the derivatives of both the arguments and plane-wave phases cancel in pairs. Hence ${\partial }_{\theta }\Phi =0$, which is the required normal boundary condition.

Away from the Fresnel transition regions, integration by parts gives the Fresnel tail expansionwhere $H$ is the Heaviside step function. The same expansion follows from the complementary-error-function asymptotics in NIST DLMF. Because $s_{\pm }^2=k_{0}r[1+\cos (\theta \pm {\theta }_0)]$, the tails in both terms have outgoing phase $e^{-i{k}_{0}r}$. The scattered far field is consequentlyThe first term cancels the incident geometrical optics field in its shadow, the second is its reflected plane wave and the last is the diffracted edge wave. The expansion is nonuniform at $\theta ={\theta }_0-\pi$ and $\theta =\pi -{\theta }_0$; the exact Fresnel expression resolves those regions. Incidence from below follows by reflecting $\theta$ and ${\theta }_0$.

Use position eigenstates normalized by $⟨q|q^{\prime }⟩=\delta (q-q^{\prime })$, and the free Hamiltonian operator $\hat{H}={\hat{p}}^2/(2m)$. The time-sliced path integral is the matrix element of the time-evolution operator:Insert momentum eigenstates, with $⟨q|p⟩=(2\pi \hslash {)}^{-1/2}{e}^{ipq/\hslash }$. The free-particle propagator becomesComplete the square in the exponent:The Fresnel integral, defined by the damping prescription $T\to T-i0$, gives, for $T>0$,The square-root branch is $i^{-1/2}=e^{-i\pi /4}$; it is fixed by the regulated Gaussian integral and the short-time Dirac delta distribution limit, rather than an arbitrary phase.

The Euler-Lagrange equation for the free particle is $m\ddot{q}=0$. Its unique classical path satisfying the endpoint boundary conditions is $q_{\mathrm{cl}}(t)=q^{(i)}+\Delta qt/T$. Therefore its classical action isSubstituting into the operator result yields $\mathcal{I}=C(T)e^{i{S}_{\mathrm{cl}}/\hslash }$, with $C(T)=(m/(2\pi i\hslash T){)}^{1/2}$ independent of the endpoint positions.

The same independence is transparent directly in the path integral. Write $q=q_{\mathrm{cl}}+\eta$, with $\eta (0)=\eta (T)=0$. The cross term in the action is $m{\dot{q}}_{\mathrm{cl}}{\int }_0^T\dot{\eta }dt=0$, so $S[q]=S_{\mathrm{cl}}+(m/2){\int }_0^T{\dot{\eta }}^{2}dt$. The fluctuation integral depends only on $m,T,\hslash$. Because the action is quadratic, the factorization is exact; the semiclassical propagator already equals the full answer.

Use the unscaled Finite Fresnel integrals of this paper. Termwise integration of the absolutely convergent Taylor series for sine and cosine gives the full small-argument expansionsIn particular,These Taylor series converge at every finite argument, although cancellation makes them unattractive for large arguments.

For the large-argument asymptotic expansions, combine the two Finite Fresnel integrals into $C+iS$. The standard Fresnel integral givesFor the tail $J(\lambda )={\int }_{\lambda }^{\infty }{e}^{i{t}^2}dt$, integration by parts using $(e^{i{t}^2}{)}^{\prime }=2it{e}^{i{t}^2}$ givesFor example $a_1=1/4$, $a_2=-3i/8$, and $a_3=-15/16$. Repeated integration by parts bounds the remainder after any fixed number of terms by the order of the next term. Equivalently the Fresnel tail expansion isHere $(-1)!!=1$. The first three terms of the requested cosine Fresnel integral, counting its constant term, areFor comparison, retaining one further correction in both Finite Fresnel integrals givesUnlike the small-argument Taylor series, these large-argument asymptotic expansions must be truncated; adding indefinitely many terms at a fixed argument does not improve the approximation.

A calculator needs elementary arithmetic and its built-in sine and cosine, together with stored approximation coefficients. A convenient global construction uses compactified approximation from endpoint asymptotics. Define the exact real tail functions byThen $P(0)=Q(0)=A$, while $P\sim 1/(2\lambda )$ and $Q\sim 1/(4{\lambda }^3)$ at infinity. Consequentlyextend continuously to the compact interval $0\le u\le 1$, with $p(1)=1/2$ and $q(1)=1/4$. Choose stored polynomial approximations $p_N,q_N$ on that interval, constrained by the small-argument Taylor series and large-argument asymptotic expansions, and calibrated across the intervening interval. The resulting elementary formula isUniform errors ${\delta }_p,{\delta }_q$ in the stored polynomial approximations give $|C_{\mathrm{calc}}-C|\le {\delta }_p+{\delta }_q$ for every positive argument. Existence of arbitrarily accurate such approximations follows from the Weierstrass approximation theorem. A Padé approximant with a denominator having no positive zero is an alternative implementation. Matching the two end expansions alone is not an error guarantee: the middle interval must also be controlled. For very small arguments, switching to the short Taylor series avoids cancellation. Ordinary finite-precision limitations in evaluating ${\lambda }^2$ and reducing its trigonometric phase remain those of the calculator itself.

For small $x$, use $u=\cosh t$ to write the defining oscillatory integral asThe first integral is $-\mathrm{Ci}(x)$, where the cosine integral satisfies $\mathrm{Ci}(x)={\gamma }_E+\log x+O(x^2)$. Its logarithm follows by integrating the $1/x$ singularity of its derivative; its conventional constant is the Euler--Mascheroni constant. The primary normalization is given by dlmf.nist.gov/6.6.E6 .

The second integrand has an integrable inverse-square-root endpoint singularity and an $O(u^{-3})$ tail. The dominated convergence theorem therefore gives its limitSplitting at $u=1/x$ and using $|1-\cos (xu)|\le \min (x^2{u}^2/2,2)$ also bounds its error by $O(x^2|\log x|)$. ThusIn particular, the leading term is $(2/\pi )\log x$.

For large $x$, the phase $\cosh t$ has its only stationary point at the endpoint $t=0$. Its expansion is $1+t^2/2+O(t^4)$, so the contributing width is $t=O(x^{-1/2})$. The endpoint stationary phase method and the Fresnel integral yieldA smooth cutoff near the endpoint justifies this local replacement; the remaining nonstationary tail can be integrated by parts, with no stationary contribution at infinity. Taking the real part gives the Bessel function of the second kind asymptoticThe additive error formulation remains meaningful near zeros of the leading oscillation. Both limits were derived from the given integral, rather than identifying a special function and merely quoting its asymptotics.

On the top side put $z=x+i\pi$, with $x$ running from $R$ to $-R$. Since $e^{i(x+i\pi {)}^2/\pi }=-e^{i{x}^2/\pi }{e}^{-2x}$ and $e^{-2(x+i\pi )}=e^{-2x}$, the reversed orientation cancels the minus sign. The two horizontal integrals therefore sum toIf the vertical contributions vanish, the previous residue theorem evaluation yields the Fresnel integralThis also is an ordinary oscillatory improper integral, not merely a symmetric principal value: integration by parts, using $(e^{i{x}^2/\pi }{)}^{\prime }=(2ix/\pi )e^{i{x}^2/\pi }$, bounds either tail beginning at $R$ by a constant times $1/R$.

Use the prepoint prescription for the separable Hamiltonian. Its regulated exponent isEach momentum integration is an ordinary Gaussian integral:Therefore the measure written in the question gives exactlyThe resulting configuration integral is $\int \mathcal{D}{q}_{\mathrm{raw},N}{e}^{-S_{E,N}}$. This is Gaussian momentum integration in a phase-space path integral. Formally its exponent tends to the usual Euclidean kinetic-plus-potential action, but the slice-dependent factors in the measure must be retained.

For a normalized quantum-mechanical propagator, Fourier completeness uses $d{p}_i/(2\pi )$ rather than $d{p}_i$. With that normalization the configuration measure isThis differs from the raw measure by $(2\pi {)}^{-N}$. It makes the free single-step kernel integrate to one and tend to a delta distribution as the interval tends to zero. In these formulas $q_{N+1}$ is fixed and $q_1,\dots ,q_N$ are integrated for propagation from an initial wavefunction. If both endpoints are fixed, omit $d{q}_1$ while retaining its slice normalization factor. The continuum expression means the limit of these measures and exponents, not a flat product of $dq(t)$ with no time-step weights.

We now address the unheaded real-time continuation. Set $m=\hslash =1$. The normalized short-time Schrödinger kernel gives the final-slice recurrenceThe square-root branch is fixed by the usual damped Fresnel integral, or continuation from the Euclidean kernel. The hint's $d{q}^{\prime }/\sqrt{\Delta t}$ contains the essential time-step dependence; the constant $(2\pi i{)}^{-1/2}$ fixes the identity limit and is included at every step.